Are we able to find conclusive evidence that chance or randominity are actually absolute physical realities.( not just math constructs )

I would think that for the concept of chance to be true it would have to be absolute. Am I wrong in this assumption?

To say things occure by chance or are random events is fine but to prove that they are random and in fact chance is another.

Mathematics and as I have learned probability theory suggest absolutes to chance and probability ( based on pure randominity I think???)

But I question whether pure randominity can be in fact proven with evidence held in reality?

If I am being silly again I am sure you guys will let me know yes?:D

I believe there is no such thing are true randomness. As I can see, there is a system that causes many things to happen to appear what we like to call "random." Think about lava lamps. They look random. I have even read an article where some guy uses them to seed his random generator. However, thinking closely, you can see the lamp is not random.

Check out the hidden variable theory.

If randominity can not be proven as true then this would suggest that everything is fully determined or determinable leading on to philisophical questions of absolute determinism and concepts of free-will.

Originally posted by 4DHyperCubix
I believe there is no such thing are true randomness.Check out the hidden variable theory.

What about at the quantum level? What about radioactive decay for example?

If randominity can not be proven as true then this would suggest that everything is fully determined or determinable leading on to philisophical questions of absolute determinism and concepts of free-will.

No, this is incorrect. If some hypothesis cannot be proven to be true, it does not automatically prove that some comepting hypothesis is true. That would be a non sequitur.

Originally posted by Quantum Quack
If randominity can not be proven as true then this would suggest that everything is fully determined or determinable leading on to philisophical questions of absolute determinism and concepts of free-will.

I think from our perspective in time it is literally impossible to actually prove. Statistically, you can say that it is however likely, but you are never privy to the entire population of a distribution in question if it is time dependent, because future moments haven't occured. Know what I mean?

Originally posted by 4DHyperCubix
I believe there is no such thing are true randomness...Check out the hidden variable theory. And while you're at it, check out the Bell inequality.

Originally posted by Quantum Quack
Are we able to find conclusive evidence that chance or randominity are actually absolute physical realities.( not just math constructs )

I would think that for the concept of chance to be true it would have to be absolute. Am I wrong in this assumption? If you could find examples of systems that don't obey the laws of probability it would be a huge deal. So far, no one has been able to find one. It seems likely that mathematical probability does in fact reflect reality, since it always manages to predict it perfectly.

Just a question a little off topic that some one may be able to answer.

As i am no computer programmer.

How do you write a program to show randominity?

I assume that you have a heap of data and select from this data randomly but how is the randominity created and is it truely random?

First of all, I think the word is randomness.

I was under the impression that programming computers only affords one the ability to create pseudo-random number generators -- programmes that adhere closely, but not exactly, to the laws of probability.

randomness it is then.

The thing I find a little difficult to see is that if a computer is presented with say 100 numbers to choose from how would it choose randomly. In an absolute sense the choice would be impossible I would think.

I know as people we can think of the choice as random if we do it but we also know that the choice may be governed in some way in that maybe we subconsciously prefer even numbers to odd of we have some other bias that we are unaware of.

The first number chosen would be a difficult choice I would speculate if one wants to be truelly random.

You are correct, it's impossible to generate truly random numbers with a computer. When a program generates random numbers it is really very deterministic. You can only get random numbers if you use special hardware devices that plug into the computer.

Originally posted by Nasor
You are correct, it's impossible to generate truly random numbers with a computer. When a program generates random numbers it is really very deterministic. You can only get random numbers if you use special hardware devices that plug into the computer.

But even those are deterministic to some extent!

Does it matter if the set of the numbers to choose from is (1,2) or (1,infinity) Is the second set more random?

Originally posted by ProCop
Does it matter if the set of the numbers to choose from is (1,2) or (1,infinity) Is the second set more random?

Not more RANDOM but the PROBABILITY of choosing 1 or 2 is lower!

Originally posted by John Connellan
Not more RANDOM but the PROBABILITY of choosing 1 or 2 is lower!

Then I am inclined to think that "true" randomness would require an infinite set to choose from (only then you have no way of knowing what the choice will be: the probability to chose one identifiable number is 0 (in truly infinite set).

Then I am inclined to think that "true" randomness would require an infinite set to choose from (only then you have no way of knowing what the choice will be: the probability to chose one identifiable number is 0 (in truly infinite set).

Only when the set is continuous, if it is discrete, then an infinite series can have a finite limit, such as the sum of 1/2^n from 0 to infinity.

Not being all that sure of the math but from what you say about 1 to infinity wouldn't this make the concept of randomness intrinsically flawed. In that if you apply infinity then randomness is vitually zero as in unworkable.

Is this a correct way of looking?

If the choice is between 2 numbers say 1 amd 2 and infinity was applied to the choice then a choice could not be made. Infinity being infinite therefore undefinable except as an absolute.
At what point in infinity can you make a choice? (obvious answer)

Just musing:

A=10/3

set reals(1 to 10)

What is the chance of getting A on a random generator?:)

Procop I'm not sure u understand. Have a look at my post again. Randomness doesn't depend on how many u can choose from (other than one!). Probability does.

Originally posted by John Connellan
Procop I'm not sure u understand. Have a look at my post again. Randomness doesn't depend on how many u can choose from (other than one!). Probability does.

If we take all numbers then some numbers has 0 chance to be chosen eg A (where A=10/3) some (reals) have 1/infinity chance, (is an integer more likely to be chosen then a real?) . The problem is that to choose from set (1,2) you know before the (random) choice that it will have to be or 1 or 2. True randomness should be unpredictible in any way. As the thread question was about randomness in reality then ( eg from a point P in a 3d space you have an infinite number of directions which eg. a ray beemed from P could follow). I think numbers as a set do not reflect the randomness in reality. At the very beginning of the Big Bang the very first atom which moved (from nothing into something) had a true random character. The second one and all tah followed was not purely random any more.....

Just as an aside from your question about randomness, how do you propose to test randomness? I would have thought that randomness was unverifiable, except over an infinite period of time, which makes the difference between random and pseudorandom rather philosophical.

Originally posted by ProCop
Then I am inclined to think that "true" randomness would require an infinite set to choose from (only then you have no way of knowing what the choice will be: the probability to chose one identifiable number is 0 (in truly infinite set). Mathematically, a sequence of numbers is random if it's impossible to figure out when the next number in the sequence is by examining previous numbers in the sequence. A set of numbers made up only of 1 and 2 could be just as random as a set of numbers made up of 1-> infinity.

Randomness doesn't mean that you have no idea what the next number will be, it merely means that it's impossible to be certain. If a random sequence is made up entirely of ones and twos then yes, I know that the next number will have to be a one or a two, but it does not mean that I know which it will be – thus it is random.

As to the computer thing:
Sequences generated by a computer are not truly random because you can figure out what the next number in any sequence will be if you know the previous numbers in the sequence and the algorithm that the computer is using to generate the numbers. This is where hardware comes in. I could have my computer connected to, say, a digital thermometer. The computer could measure the temperature out to the billionth decimal place and use the billionth digit as the first number in a sequence, wait for a moment, measure the temperature again, etc. It doesn't even matter if your thermometer is that accurate or not. Since the billions place will fluctuate wildly, it will give you a series of random numbers that no one will ever be able to figure out. The numbers won't be based on any sort of deterministic algorithm and there will be no relationship between any previous numbers in the sequence and the next set of numbers in the sequence. There are all sorts of hardware devices that operate in a similar way to generate random numbers. I suppose you could argue that such numbers aren't really random, since if you know the exact conditions of the hardware device at the time the numbers were generated you could probably figure out what the sequence would be. Never the less, the numbers will be random enough for any application that requires random numbers, ie. most cryptography.

If you insist and having really random numbers, you can get a hardware device that generates random numbers based on non-deterministic quantum processes, like particle decay times. This is a bit over the top for most applications, so it's only usually used for special things like generating ultra-secure cryptographic keys or one-time pads.

Originally posted by Nasor
Mathematically, a sequence of numbers is random if it's impossible to figure out when the next number in the sequence is by examining previous numbers in the sequence. A set of numbers made up only of 1 and 2 could be just as random as a set of numbers made up of 1-> infinity.

Randomness doesn't mean that you have no idea what the next number will be, it merely means that it's impossible to be certain. If a random sequence is made up entirely of ones and twos then yes, I know that the next number will have to be a one or a two, but it does not mean that I know which it will be – thus it is random.




If the random generator per accident chooses 100x 1 without interruption than evidently the chance is higher (and higher) with every next round that the next number will be 2. Previous instances imply the result of the next step. That's way I proposed that known outcome (1 or 2) is not really random.

Originally posted by ProCop
If the random generator per accident chooses 100x 1 without interruption than evidently the chance is higher (and higher) with every next round that the next number will be 2.I don't understand what you're saying here. Are you saying that if a 1/2 random sequence generator gives you a series of several 1s that it increases the likelyhood that the next number will be a 2? That isn't true.

Originally posted by Nasor
I don't understand what you're saying here. Are you saying that if a 1/2 random sequence generator gives you a series of several 1s that it increases the likelyhood that the next number will be a 2? That isn't true. [/B]

If you had 100 x 1 then the chance is greater then 50% that in 101 round of the random generator it will produce 2. I think it could be tested too. It was tested by someone here at Sciforums that to get 20 x 6 in a row in the dice game woud take 3.500 years on P3 700MGH generator. Now you are proposing that after 19 x 6 in a row was generated the chance to get the 20th 6 in that row was 100/6 = 16 procent? (let me tell you that it took thousands of processor-hours to get it). Probability changes if it gets out of balance...the number of elements in the set and result history influence the future behaviour. Your set 1,2 is after the second run/through not purely random any more.

Originally posted by ProCop
If you had 100 x 1 then the chance is greater then 50% that in 101 round of the random generator it will produce 2.If it's really a random generator, there will still be a 50% chance of getting a 1 after generating a sequence of 99 1s. If the probability is not 50%, it means that your generator isn't truly random and you need to get a better generator.

Originally posted by Nasor
If it's really a random generator, there will still be a 50% chance of getting a 1 after generating a sequence of 99 1s. If the probability is not 50%, it means that your generator isn't truly random and you need to get a better generator. [/B]

You are probably right about the machine of the generator. Within the machine the chance is 50% (at each individual run). But <b> above</b> the machine sits the Zeus of Probability and he will overule the machine and put the odds differently. The test I described above sugests that.

I think one the interesting characteristics of randomness is that its unknown or possibly unknowable.

We say that a dice roll is random, what we really mean that it is unknown. We cannot predict how the combination of kinetic energy, air resistance etc... will produce the roll of a particular side of the dice. Thus, we call it random. Theoretically, if we could control all the components of a die roll it would become deterministic not probabilistic.

Some things may be literally unknowable, though. There is (or seems to be) no way to predict when a specific particle will undergo decay.

So is chance proveable? Not, IMO, unless you can eliminate all the potentially knowable unknowns. You may then be left with some unknowable quantities that can be proven to be random.

Cheers!

Originally posted by ProCop
If you had 100 x 1 then the chance is greater then 50% that in 101 round of the random generator it will produce 2. I think it could be tested too. It was tested by someone here at Sciforums that to get 20 x 6 in a row in the dice game woud take 3.500 years on P3 700MGH generator. Now you are proposing that after 19 x 6 in a row was generated the chance to get the 20th 6 in that row was 100/6 = 16 procent? (let me tell you that it took thousands of processor-hours to get it). Probability changes if it gets out of balance...the number of elements in the set and result history influence the future behaviour. Your set 1,2 is after the second run/through not purely random any more. Youp, that was me. However, you got something wrong.

Roll a dice (assuming an ideal dice), the chance of getting any of the six possible numbers is 1/6. This does not change depending on earlier rolls. SO, no matter how many sixes you just rolled, the probability for getting another six is 1/6.

I think this is what many people don't realize about probability: Probability figures are only relevant BEFORE the fact. If you decide to roll a dice 20 times, the probability of getting 20 sixes is (appr) 1/3.6562E15. But if you just happened to roll 19 sixes, the probability of rolling another is 1/6.

Take it another way: Any roll of 20 dice is equally likely (or unlikely). Roll a dice 20 times, noting the result. The probability of getting the same sequence has the same probability as rolling 20 sixes, or any other sequence.

Now for randomness:

In nature, random events exist. Of course, one can fabulate about hidden variables, but that is merely speculation. A quantum event is random. If you have an isotope with a half-life of 1 hour, there will, at any time, be 50% probability that it will have decayed within the next hour.

Since compurters are deterministic, barring faults, you cannot generate a true random number algorithmically.

Hans

Originally posted by MRC_Hans
I think this is what many people don't realize about probability: Probability figures are only relevant BEFORE the fact.

GREAT point. This is really a subtle thing though it seems blatant. It's so easy to apply the concepts of probability incorrectly when you don't keep this perfectly clear in your mind.

Side note: IMO, this directly nullifies the "monkeys at typewriters" (and all its relatives) approach to validating intelligent design.

Originally posted by ProCop
You are probably right about the machine of the generator. Within the machine the chance is 50% (at each individual run). But <b> above</b> the machine sits the Zeus of Probability and he will overule the machine and put the odds differently. The test I described above sugests that. Ah, if only that were true; how rich I could be! I would simply go to a casino and play roulette. I would wait until the ball landed on black several times, then bet on red. Since red would now be more likely to show up, I would win more often than I would lose!

Unfortunately (or maybe fortunately, if you're a casino) probability doesn't work like this. If I flip a coin four times then there is a .5^4 chance that I will get heads four times in a row. This does not mean, however, that if I have already flipped the coin three times and gotten three heads that the next flip is less likely to be heads. The probability of getting an HHHH sequence is only .5^4 before I do any flipping. If I've already gotten H three times, then the chance of getting HHHH for this particular sequence is now 50/50. Although getting any particular sequence is initially unlikely, the odds increase and I successfully work my way farther and farther into the sequence.

Originally posted by ProCop
The problem is that to choose from set (1,2) you know before the (random) choice that it will have to be or 1 or 2. True randomness should be unpredictible in any way.

That is very very wrong. As I said twice before, randomness is not probability. U know it is going to be 1 or 2 but u have no idea which! By your logic, since radioactive decay has two choices - it will or wont decay, then it is not random. Do u see where your problem lies?

Originally posted by Nasor
Ah, if only that were true; how rich I could be! I would simply go to a casino and play roulette. I would wait until the ball landed on black several times, then bet on red. Since red would now be more likely to show up, I would win more often than I would lose!



Actually Holand Casino (a chain of the Dutch entertainment) went to the court of law (a coulpe of years ago) to forbid a group of people the entry to the Dutch Casino's. They were doing what you described. The judge decided the casinos had to let them in. Arguing the decission the judge said that they didn't brake the rules of the casino and nowhere in the law was written that only casino owners could make money out of the game. There was also an interview with one of the group of the gamblers on the TV, who said that thay would accept a pay off from the casinos to go away, and he mentioned that casino tried to make their lives difficult in other ways. He desrcibed also how they played: They waited for a favorable moment (a row of couple of the sames eg 4 blacks and then they started betting doubling the bet if they had lost. They had to have some reserve money and it happend thet they lost big, but on averidge the group of about 20 made between 15 - 20 thousends a month per person. Possibly some Dutch posters can confirm that - it was taked over quite a lot at that time.

ProCop, I don't know what to tell you, except that you are simply wrong about this. The probability of getting any sequence of heads or tail when flipping a coin flip is exactly the same as the probability of getting any other sequence. The odds of my getting HHHH are exactly the same as the odds of my getting HHHT, or TTTT, or HTHT. Past flips have no effect on future flips; the coin has no way of 'knowing' whether the first flip was a heads or a tails when I flip it the second time.

Originally posted by ProCop
They waited for a favorable moment (a row of couple of the sames eg 4 blacks and then they started betting doubling the bet if they had lost. Congratulations, you've managed to combine two classic gambling fallacies into one.

Possibly Hans can test the following:

If you through a dice six milion times you should get on average 1 milion of sixes one of fives etc. If that is so than there is a probability law which causes this result.

If you get appr. 2 milions of 1's 200 of 2's 800.000 of 3's or something in that trend it would show that there is no probability law.

Hans?

Originally posted by ProCop
Possibly Hans can test the following:

If you through a dice six milion times you should get on average 1 milion of sixes one of fives etc. If that is so than there is a probability law which causes this result.

If you get appr. 2 milions of 1's 200 of 2's 800.000 of 3's or something in that trend it would show that there is no probability law.

Hans? What is this supposed to prove? Everyone agrees that there is a probability law, you just don't seem to understand how it works.

OK.

Lets get it stright.

Suppose we have 1.000 throughs of 1 or 2.

Optimaly we get 500 1's and 500 2's

now the situation

we made 500 throughs with result 490 1's and 10 2's.

The probability is then that in next 500 throughs we get more 2's.

Right?

They waited for a favorable moment (a row of couple of the sames eg 4 blacks and then they started betting doubling the bet if they had lost. They had to have some reserve money and it happend thet they lost big, but on averidge the group of about 20 made between 15 - 20 thousends a month per person. Possibly some Dutch posters can confirm that - it was taked over quite a lot at that time.

Strictly speaking this approach is not based on a fallacy, but it does not really have anything to do with the number of blacks in a row. If you double your bet after every loss assuming no limit on the maximum bet, eventually you will earn money ie lose 5, bet 10, lose 10, bet 20, lose 20, bet 40, win 40 gives net winnining of 5.

If you have enough patience and a big enough bankroll you can break the bank anywhere.

Not that I would recommend it as an investment strategy, though.

Cheers!

Originally posted by ProCop
OK.

Lets get it stright.

Suppose we have 1.000 throughs of 1 or 2.

Optimaly we get 500 1's and 500 2's

now the situation

we made 500 throughs with result 490 1's and 10 2's.

The probability is then that in next 500 throughs we get more 2's.

Right? This is incorrect. Probability only indicates that we will get 500 H and 500 T in my thousand coin flips before I start flipping. Once I actually start flipping, I must adjust the probability distribution for that particular sequence of flips.

If in my first 500 flips I get 500 Hs (which is very unlikely, but could happen) then probability now indicates that this sequence of 1000 flips will produce 750 H and 250 T. This is because there is a 50/50 chance of each flip being a H or T, and I have 500 flips left that will probably produce about 250 H and 250 T.

The whole point here is that past results of coin flips, roulette wheels, or whatever, do not have any influence on future results. In other words, if I flip a coin 999 times and it comes up heads each time, it does not mean that the next flip is any more likely to be tails. When I started flipping it was very unlikely that I would get 1000 heads in a row. Now, 999 flips later, most of the improbably stuff has already occurred; the odds of my getting 1000 heads in a row is now 50/50.

Originally posted by ProCop
OK.

Lets get it stright.

Suppose we have 1.000 throughs of 1 or 2.

Optimaly we get 500 1's and 500 2's

now the situation

we made 500 throughs with result 490 1's and 10 2's.

The probability is then that in next 500 throughs we get more 2's.

Right?

No, the past results don't affect the future probability. In any given 1000 coin tosses you would expect roughly 500 to be heads and 500 to be tails. After you toss a coin the odds of it going from being 0.50, to either 1.00 or 0.00. The factors causing the coin to come up in future rolls are exactly the same(shape of the coin etc...) The only way to affect the future probability of coin tosses as you suggest would be to alter the coin in some way.

Originally posted by contrarian
Strictly speaking this approach is not based on a fallacy, but it does not really have anything to do with the number of blacks in a row. If you double your bet after every loss assuming no limit on the maximum bet, eventually you will earn money ie lose 5, bet 10, lose 10, bet 20, lose 20, bet 40, win 40 gives net winnining of 5.

If you have enough patience and a big enough bankroll you can break the bank anywhere. This is always a losing strategy in the end. It is true that you will be more likely to win any individual sequence of bets, but the amount of money that will be lost should you lose the betting sequence is so large that it more than cancels out your winnings.

With this strategy I might only lose one out of every 100 betting sequences, but the money that I lose in my one lost sequence will be more than the money that I won in the 99 successful sequences.

The whole point here is that past results of coin flips, roulette wheels, or whatever, do not have any influence on future results. In other words, if I flip a coin 999 times and it comes up heads each time, it does not mean that the next flip is any more likely to be tails. When I started flipping it was very unlikely that I would get 1000 heads in a row. Now, 999 flips later, most of the improbably stuff has already occurred; the odds of my getting 1000 heads in a row is now 50/50.

I understand you now. If you flip a coin 999 times and it comes up heads then you can bet all your money on that 1000th flip will be head too. The coin is probably a fraud. :)

But OK I can see your point. But even though, there is eg. 10 milions of blacks and 10 milions of reds in a casino a year. If you wait four blacks in a row to bet on red you eliminate eg. 20% of blacks out of the game. In the long run your odds are in favor of red betting. The same inversed: four reds in a row you bet black.
(The complaint in the case of Holland Casino was that the group they tried to forbit to enter was that they were keeping the places at the roulette tables occupied all day long placing relatively small bets, and <b> winning</b>...)

Originally posted by ProCop
But even though, there is eg. 10 milions of blacks and 10 milions of reds in a casino a year. If you wait four blacks in a row to bet on red you eliminate eg. 20% of blacks out of the game. In the long run your odds are in favor of red betting. The same inversed: four reds in a row you bet black. I don't follow you here. Why would 20% of the blacks be eliminated? The ball can land on the same square twice in a row.

I think I see the cause of your misunderstanding, ProCop.

You have to realize that when we calculate the probability of a sequence, we are really only calculating the number of possible outcomes for the sequence. If I have a 3-digit sequence of 1s and 2s, then there are 8 possible outcomes:

111
112
121
211
212
221
122
222

We would say that the odds of getting the sequence 222 are 1 in 8, and we would reach this conclusion by doing the calculation 2^3. The purpose of this calculation is to figure out how many ways our sequence could turn out.

Now if we generate our first two numbers and get 22, this allows us to eliminate six of the possible outcomes. Since the probability of something happening is based on the number of possible outcomes, the odd of us getting our 222 is 50/50 because there are now only two possible outcomes, and only one of them us 222.

It's not as if there's some sort of mystical force that will drive the sequence toward one outcome or the other; it's all just a matter of counting possible outcomes. As the sequence progresses certain outcomes are eliminated, and so the probabilities change.

Note that this only applies to situations where all outcomes are equally likely. If we knew that our number machine was inherently more likely to produce 2s than 1s, we would have to account for that as well.

Lookedt at backwards you eg. can see that the colors alternation has a pattern. (there comes more sequences black, red, then black,black/(should be red),black,red) - betting on last red (in the long run) will use the pattern in your advantage (you skiped the losing on black in red pattern and so you can double the next bet). But I am not a gambler cannot compare this to the real wheel turns. Anyway you can turn the chance in your advantage: the casinos have a betting limits to counter that - otherwise why woud they have such a limit? Principally at the full table gamblers bet against one another and the casino gets the most of it....

Originally posted by Nasor
This is always a losing strategy in the end. It is true that you will be more likely to win any individual sequence of bets, but the amount of money that will be lost should you lose the betting sequence is so large that it more than cancels out your winnings.

With this strategy I might only lose one out of every 100 betting sequences, but the money that I lose in my one lost sequence will be more than the money that I won in the 99 successful sequences.

I don't follow you here. The odds of losing ten times in a row is 1/2^10 or 1/1024. Thus, the odds of winning at least once in ten times is thus 1023/1024. The total amount bet after losing for the tenth time is 1023(1+2+4+8+16+32+64+128+256+512), which is exactly equal to the amount you would've won.

The benefit to this approach is that you can always bet one more time to win money if you chose to. If there is no limit on the maximum bet, in theory any sequence will be a winner eventually.

cheers!

Originally posted by contrarian
I don't follow you here. The odds of losing ten times in a row is 1/2^10 or 1/1024. Thus, the odds of winning at least once in ten times is thus 1023/1024. The total amount bet after losing for the tenth time is 1023(1+2+4+8+16+32+64+128+256+512), which is exactly equal to the amount you would've won.

The benefit to this approach is that you can always bet one more time to win money if you chose to. If there is no limit on the maximum bet, in theory any sequence will be a winner eventually. This is only true if the odds of winning and losing are exactly the same. In all casino games the odds are more than 50% in the casino's favor.

Also, you have to keep in mind that you can't simply keep betting unless you have an infinite supply of money, which no one does. For any given real quantity of money, no matter how large, it is always most likely that you will lose everything before you will win as much money as you started with.

It's a question of risk verses reward. While it's true that with this strategy you will most likely win any given round of betting, the potential reward is not proportional to the huge amount of money that you must risk.

The question I ask about all this probability analysis is that does the Gambler influence the outcome in any way beyond that of pure chance?


Whilst probablity suggests not, I wonder?

If the casino played only games of two colours say red and black would it go bust, stay even or win.

The reality suggests that even though the odds appear even the casino will still win.

Possibly the amount of time you have and the number of bets you place and when you stop playing may be important.



say win red red red red black red black black black red black black

outcome neutral 6*red 6*black

But if the gambler withdraws early in this sequence the casino or the gambler could end up in front.

there was an article recently on TV that discussed the tricks of a card sharp.

The main point was that when people are playing for match sticks they play really well and win more often, but as soon as money is involved they invariably fail to win.

AS suggested in an earlire post the fear of loosing the money creates an environment of loss and loose they do

The question is why? Bad decisions? Pulling out early or too late?

The timing of quiting the game seems to be important. And maybe this is how the cassino wins?

I just did a quick and simple sample excel sheet on the issue of a gamblers timing
(I hope it uploads correctly)

Obviously a gamblers key to success is not so much probability but when he enters the game and when he leaves it.

The gamblers intuition not being governed by probability as such.

Well yes, obviously a person can win if they enter and leave the game at the right time. The problem is that since each round in the games is random, it's impossible to know when the 'right time' to bet is. Each spin of the roulette wheel (or whatever) is not effected by any of the previous spins, as has been stated here repeatedly. The games are carefully designed so that it will be the 'wrong time' more often than not.

So the key for a successful Casino is not so much the nature of chance but how they can effect the timing of the gambler. Could be as simple as how many times or at what time a waitress asks you whether you want a drink or not, or how comfortably the seats are or whether there is music playing etc etc.....

The gambler trying to hard to intuite where the luck is he makes himself vulnerable to "off-luck" timing suggestions.

( just exploring the idea a little here )

If one did a full probability analysis say for the roulette game in its full options to the gambler I would bet ( ha) that the probability of overall success is heavilly weighted towards the Casino any way. ( of course)

Except of course someone coming to the table and placing only one bet , say $10,000 on the number 13 , and walks away a winner with a couple of million.

So what is the probablity of a gambler always knowing he is going to win and actually winning?

First of all, ProCop: I'm sorry, but you are wrong. Probability does not change according to the past, period. And this can be proved. You can do it yourself. Roll a dice as many times as you wish, noting the result. You will observe two things:

1) The sums will converge on being equal.

2) Any random differences will stay in the figures, there will be no compensation.

Thus, if you happen to get, say, six sixes in a row, this surplus may well persist during the whole sequence. But as the numbers pile up such differences will be small percentage-wise.

About casinos: Obviously, casinos must operate with a maximum bet, that is simple risk management. The trick you mention does not mean that probability changes due to earlier events, but the row of events might influende players, so those taking advantage of that might get an unfair share. Anyhow, casino managements are not scientists, they might be as superstitous as the next person.

Quantum Quack:

Barring cheating, there is no way a gambler can know if they are going to win. But different betting strategies can affect the type of chance you have, that is you can bet on a good chance of a small win or a slim chance of a big win.

In the long run, the casino always wins, because stastistics are on their side.

Hans

Originally posted by MRC_Hans
First of all, ProCop: I'm sorry, but you are wrong. Probability does not change according to the past, period. And this can be proved. You can do it yourself. Roll a dice as many times as you wish, noting the result. You will observe two things:

1) The sums will converge on being equal.

2) Any random differences will stay in the figures, there will be no compensation.

Thus, if you happen to get, say, six sixes in a row, this surplus may well persist during the whole sequence. But as the numbers pile up such differences will be small percentage-wise.



It is general practice to use observations of past to make valid judgement about the future. Past deviations in the patterns will be "corrected" in the future behaviour of the dice...If you take a curriculum vitae of a dice in Monte Carlo it will look out as a pendullum. I agree with you cannot predict any particulair trhow but you can discover and consequently predict a pattern.

No, that is a belief you have. It is not based on facts. But yes, the rolling average score of an indefinite series of dice throws will swing around the predicted probability of a single throw, which exactly proves pure chance.

Hans

What is the the law of probability about, why we have it, what it explains, how there can be a law about completely random behaviour..please, please tell, I can hardly wait...

Originally posted by ProCop
What is the the law of probability about, why we have it, what it explains, how there can be a law about completely random behaviour..please, please tell, I can hardly wait...

what is wrong with you dude? hans is right, you're wrong. accept it, it's simple.

with cards, it's different because it is not random. You have a defined set of cards and as you remove them, you change the probability of the following cards. with dice, every time you throw them it's a random event. You have removed nothing from the possibilty of the next occurence, so the probability of any particular combination of values is the exact same every time you throw.

now you might get a lil particular and say "this dude throws dice well" and until i analyzed data from him I couldn't really argue because it's possible that he has something in his wrist so to speak. maybe somehow he can throw them the same every time or something.. I dunno.. but barring the skill of the thrower and any other kind of setup, it is as hans has said, the probability is the same every damn time. :p

Anything dependent on unknown(s) is chanced.

This is the definition, and the proof lies in the definition.

It also irrelevant if the variable space is known or not.

There is some rationality mixed in with a lot of nonsense here.

The table limit is used to prevent the results being skewed by a player betting huge amounts. If ten players bet the same amount, the casino is assured of a profit in every 100 plays. If one player is betting $10,000 and the others are betting $10, the casino must hope that the big bettor is not lucky.

BTW: The odds in 100 plays at a dice table are as follows: 40% of the players will win; 8% will break even; 52% will lose. The casino does not want a really big bettor to be one of the 40% who win. That is the reason for the table limit. An oil sheik who wanted to double up after every loss starting with a $1000 bet might be allowed to have no table limit. The casino only risks $1000 per series of bets, and is likely to win a million when the sheik runs out of money or nerve. They know he will hit a losing streak sooner or later.

If a coin, roulette wheel, whatever is not rigged, past behavior just does not influence what happens next. As pointed out by somebody, a casino does not want people sitting at a roulette table not betting while waiting for a run of Black or Red to occur. The casino probably would not have objected if they had made a minimum bet every play and then bet a huge amount on Black after a series of Reds. If their theory had any merit, this strategy would pay in the long run.

BTW: In Las Vegas (and probably in other casino towns), you can purchase roulette monitoring forms. I have seen them in use at Vegas casinos in off hours. The user stands close enough to the table to observe, but does not take a seat. One strategy is to record the results of 380 plays. Each number is expected to come up 10 times. The gambler then sits down and bets on the numbers which have occurred less than ten times, assuming they are now more likely. Oddly enough, some bet on the numbers which have come up more often, assuming that the wheel is unbalanced in favor of those numbers. Either strategy takes a lot of patience. 380 spins is about 30 minutes to an hour. The strategy is used at off hours because there are empty seats, allowing the player an unobstructed view, and also allowing him/her to take a seat when his bookkeeping tells him it is time to bet.

The expected result of a large number of plays is a bit counterintuitive. If you flip coins a huge number of times and count heads and tails, AbsoluteValue(Heads - Tails) is expected to grow without bound, while Heads/Tails is expected to approach ½. Consider the following. P(even in 10 flips) = .246 094 P(even in 100 flips) = .079 589 P(even in 1000 flips) = .025 225 In 10²⁴ flips if (Heads - Tails) = one million, Heads/Tails would equal ½ to 18 digits of precision. The more you flip a coin, the less likely an exact 50-50 match becomes. As you flip it many times, it becomes very likely that Absolute(Heads - Tails) will equal 100 or more.

Somebody made an erroneous post about the half life of a radioactive substance. The half life is the time in which half the atoms are expected to decay. If the half life is one minute, half are expected to decay in one minute; In two minutes 3/4 are expected to decay; In 3 minutes, 7/8; Et cetera. If there is only one atom of the substance, it will last one minute half the time; It will last two minutes 1/4 of the time; It will last 3 minutes 1/8 of the time; et cetera.

OK seing the vehemence of general disgreement I will try to reconsider.

Originally posted by ProCop
What is the the law of probability about, why we have it, what it explains, how there can be a law about completely random behaviour..please, please tell, I can hardly wait... As I already told you, the laws of probability (as being applied here) are about counting possible outcomes. It's not about a mystical force that causes future rolls of the dice to 'compensate' for statistical anomalies in past rolls. No such force exists. Past results do not influence future results. End of story.

Originally posted by Nasor
As I already told you, the laws of probability (as being applied here) are about counting possible outcomes. It's not about a mystical force that causes future rolls of the dice to 'compensate' for statistical anomalies in past rolls. No such force exists. Past results do not influence future results. End of story.

Well I agree (finally). I have thought it through and cannot come with a good argument supporting the God of Probability notion. I saw probability as a sort of pendullum which has some optimal position which when lost will come back in a (sort of) predictable way. It was a too mechanical concept. However I doubt now that 3.500 years on a random dice generator will get 20x 6 in a row. Hans run the program half an hour and calculated the rest. Though long unprobable runs are theorethically possible (every through has equal chance) they probably do/will not happen <b>because</b> of the probability law.

if the probability of combination of two dice is supposed to be equal then why in my 100 throws experiment does there seem a bias to the combination of 6 and 7?

Why not 2 or 12?

Originally posted by Quantum Quack
if the probability of combination of two dice is supposed to be equal then why in my 100 throws experiment does there seem a bias to the combination of 6 and 7?

Why not 2 or 12? Because more of the possible combinations of dice faces add up to 7 than any other number…

ahhh..IC thanks

Originally posted by ProCop
Well I agree (finally). I have thought it through and cannot come with a good argument supporting the God of Probability notion. I saw probability as a sort of pendullum which has some optimal position which when lost will come back in a (sort of) predictable way. It was a too mechanical concept. However I doubt now that 3.500 years on a random dice generator will get 20x 6 in a row. Hans run the program half an hour and calculated the rest. Though long unprobable runs are theorethically possible (every through has equal chance) they probably do/will not happen <b>because</b> of the probability law. OK, so far so good :).

I have to review the dice program, I don't remember the details. However, in my tests, I found various instances of rows of sixes, the program was made to record that, and they all appeared in the way predicted by probability. So, while this has not been tested empirically, can you present any reason why this prediction should not be extendable to any number of sixes in a row?

Basically, on each "throw", the program does not know what the outcome of the earlier throws were, so conditions for the n-th throw must be EXACTLY the same as for the N-1 th.

Hans

correct me if i am wrong but
say you throw a dice and up comes a six

The probability of throwing another six is compounded yes?

And yet because the two throws have and equal chance of being a six how can the compounded probability be true?

( I know very little about probability law)
May be a silly question with a simple answer

I have to review the dice program, I don't remember the details. However, in my tests, I found various instances of rows of sixes, the program was made to record that, and they all appeared in the way predicted by probability. So, while this has not been tested empirically, can you present any reason why this prediction should not be extendable to any number of sixes in a row?


You have a pattern e.g: it takes more and more time to get the next six (to prolong the already achieved row eg 5x 6 to 6x 6 on a random generator it takes twice as much time as it takes to get from 1x 6 to 5x 6). Therefore I think the intervals between the sixes will get longer and longer (in a pattern) so eg. to get the 18th six it would take infinitely long (in other words you would never get it)

(The thread where we discussed this before was based on a notion that the universe (with people) could not have arised by accident because it would take 20 billion years to get 20x 6 on a dice game and in the arisal of the universe more coincidence - immprobabilities were involved than in the dice game. In other words how can man of now exist (developed " accidentally" in 20 bilion years) if 20 billions years is just enough to get 20x 6)

editted spelling

Just because we may suggest that the universe was generated with out pure chance does not rule out coincedence or successful integrations that lead on to more successful integrations.

Possible at some stage absolute chance existed for only a zillionth of a second and from there on the probability aspect just kept reducing. In the next 2000 zillionsths of a second the chance aspect just exponentially reduces and so on

So we get a situation that suggests that life emerged with purpose and not just by accident. Looking at DNA for example, so clever one wonders how seemingly random events could generate such cleverness but may be it can all be put down to more a coincidence of success ( Darwinism sort of) than chance. The more successfull the more successful the future is etc.

Reducing the chance aspect also doesn't validate a religious arguement either.

I'm not sure my arguement is all that clear but it was worth a shot.

ProCop:

Yes, however, the probability for an increasing row of sixes is decreasing linearly with the number of sixes. Thus getting seven in a row takes, an average, six times as long as getting six, etc. So we can safely extrapolate the probability of getting 20, and it is nowhere near infinity.

Using this for discussing the probability of man is totally moot, for two reasons:

1) Probability is irrelevant after the fact. Man DID appear.

2) Since we do not know the alternatives, we cannot know what the probability was before Man appeared. As I noted earlier, ALL sequences of 20 dice throws are equally probable. To Nature, there is nothing special about 6-6-6-6-6-6-6-6-6-6-6-6-6-6-6-6-6-6-6-6, 2-5-3-1-2-6-3-4-1-2-6-4-3-5-1-4-3-1-6-3 is just as unique.

Hans

So a guy starts work as a salesman, he learns over time that certain behaviours and plans work out and others don't. The success he has compounds and he continues to learn how to be a better salesman. This is obviously not an issue of chance or probability. Success is not able to be achieved with out success.
The salesman wasn't just a victum of chance in that he worked for his success. and therefore made himself succeed.

Whether someting is a succeess or not can only be determined over time. The longer it exists the more successful it is.

Unitl now we have a universe that has existed for a long time. Is the universe a success?

A single celled animal finally successfully meets another and they join and become a two celled animal, the relationship is a success.
So they go on based on success. The universe is intolerant of failure.

If a coin has equal chance to go head or tail, actually, both should/(must?) happen. Everett's multiverse theory accounts for that. In this theory the coin goes head in one universe and tail in another one. Chance is then an atribute of a partial reality (all that is possible to happen happens in complete reality). If so then we are conscious only of a small part of the reality. Some of us are more succesful then others (in this universe).

I do not care what credentials Everett & Wheeler have, the multiverse theory has to be crazy or deliberate nonsense created due to the Publish or Perish syndrome.

Flip a coin and all at once there are two universes: One with the coin landing heads and one tails??? Worse yet, every quantum event that has two alternatives results in two complete universes?? Where is all that matter and evergy coming from? the univrse we have seems to have existed for billions of years. Flipping a coin creates another one in a few seconds?

The Big Bang with one universe jumping out of a quantum vacuum or some other state is tuff enough to believe.

Quantum Quack:

correct me if i am wrong but
say you throw a dice and up comes a six

The probability of throwing another six is compounded yes?

And yet because the two throws have and equal chance of being a six how can the compounded probability be true?


Before you throw the dice, the probability of getting two sixes is 1/36=1/6*1/6
If you have thrown one dice and got a six, the probability of throwing two sixes is 1/6, because you already know the outcome of the first throw.
This is written P(throw1+throw2=12)=1/36
alternatively P(throw1+throw2=12 | throw1=6)=1/6
Basically, you cannot say the two are the same, because the conditions are different.

I think I understand what you are saying however to me It seems that we have two levels of probability at work.

If the second throw is in no way conditioned by the first throw how can probability of 1/36 be valid?

It strikes me that this 1/36 is just an abstract and not relevant to the reality of the dice. Being that each throw is equally 1/6
How does probability law deal with this paradox?:)

Originally posted by ProCop
If a coin has equal chance to go head or tail, actually, both should/(must?) happen.

Why must it happen? That is nonsense. Hald of the tomes one happen, half of the times, the other. In each single coin-throw, only one outcome is possible (disregarding the slight possibility of the coin ending on its edge).

Everett's multiverse theory accounts for that. In this theory the coin goes head in one universe and tail in another one. Chance is then an atribute of a partial reality (all that is possible to happen happens in complete reality). If so then we are conscious only of a small part of the reality. Some of us are more succesful then others (in this universe).

The idea that every chance vent should spawn an alternative reality is really very very silly. Just consider your coin throw: Heads or tails are not the only two possiblities (just the ones we have chosen to consider), the coin can land in an almost infinite number of positions and orientations, and exactly when does it land? If all chance events should create alternative realities to exhaust the potential possiiblities, a near infinite number of parallel ralities should be created in every instance. That is simply absurd.



Hans

Just a fantasy about alternate realities.

Say Pres. Bush Is trying to make a decision whether to go into Iraq or not. All evidence is inconclusive but the public want action.

In the end he reaches into his pocket and produces a coin. Heads we go in tails we don't.

Heads won.....new reality.

Originally posted by Quantum Quack
I think I understand what you are saying however to me It seems that we have two levels of probability at work.

But we have just that: One level when we try to predict the outcome of two consequtive throws, another when we try to predict one throw.

If the second throw is in no way conditioned by the first throw how can probability of 1/36 be valid?

Because 36 is the number of possible outcomes of two (independent) dice throws. Two dice throws can can yield exactly 36 different outcomes (1-1, 1-2, 1-3, etc. through 6-6, - count-em ;)), so before we start, there is a 1/36 probability of getting any particular sequence.

It strikes me that this 1/36 is just an abstract and not relevant to the reality of the dice. Being that each throw is equally 1/6
How does probability law deal with this paradox?:)

What paradox? You can count the possible outcomes, there are 36. And (1/6)/(1/6)=1/36. The math fits and all. :)

Hans

Hans, is there a difference inprobability if you throw the two dice together as a pair of dice and when you throw them separately as single dice?

Originally posted by Quantum Quack
Hans, is there a difference inprobability if you throw the two dice together as a pair of dice and when you throw them separately as single dice?

You will have to explain. Probability of what? If u mean probability of getting two numbers that u have predefined then no (unless the dice interact in such a way as to bias the throw). The probability for each die is independent on the other die.

Originally posted by MRC_Hans
The idea that every chance vent should spawn an alternative reality is really very very silly. Just consider your coin throw: Heads or tails are not the only two possiblities (just the ones we have chosen to consider), the coin can land in an almost infinite number of positions and orientations, and exactly when does it land? If all chance events should create alternative realities to exhaust the potential possiiblities, a near infinite number of parallel ralities should be created in every instance. That is simply absurd

Possibly all these (parts) of universe(s) take place at the same time and place "entangled" in a full universe not observable in its fullness" with the same energy pool, a sort of <a href=http://www.mala.bc.ca/~lanes/english/hemngway/picasso/guernica.htm>Guernica</a> in physics.

Quantum Quack: If you are really want to understand probability, you should take a course or read a good book on the subject.

Asking questions here when you have hardly any background knowledge is unlikely to result in your understanding the subject. You need a more organized approach.

Originally posted by Quantum Quack
Hans, is there a difference inprobability if you throw the two dice together as a pair of dice and when you throw them separately as single dice?

When asking about probability you have to add "of what".

In terms of probability, there is no change for the outcome of each die if you add dice to the mix.

Originally posted by Quantum Quack
Hans, is there a difference inprobability if you throw the two dice together as a pair of dice and when you throw them separately as single dice? No. Of course there might be some practical issues, like the dice hitting each other and such, but otherwise, throwing one dice n times or throwing n dice once will give the same results. Obviously, if you throw several dice, you have to somehow mark them to decide the sequence. Otherwise, the throw of two dice does not yield 1/36 probability but 1/21 (6+5+4+3+2+1) because you cannot distinguish between, e.g 1-2 and 2-1.

Confused? You won't be after this week's episode of SOAP!

Hans ;)

MRC_Hans: The following confuses me a bit.Obviously, if you throw several dice, you have to somehow mark them to decide the sequence. Otherwise, the throw of two dice does not yield 1/36 probability but 1/21 (6+5+4+3+2+1) because you cannot distinguish between, e.g 1-2 and 2-1.If I roll two dice simultaneously, I expect the following probabilities for various totals. P(2) = 1 / 36 P(3) = 2 / 36 or 1 /18 P(4) = 3 / 36 or 1 / 12 P(5) = 4 / 36 or 1 / 9 P(6) = 5 / 36 P(7) = 6 / 36 or 1 / 6 Probabilities for other totals are symmetric around P(7): P(8) = P(6), P(9) = P(5), et cetera.From whence comes 1 / 21 ?

Any book including an analysis of dice throws will verify the above probabilities, which are easy to understand if you make up a 6X6 array of equally likely possibilities for two dice.

If we take many dice and compare it to the constalation of the universe supposing the dice were in an ordered state and we were disturbing this ordered state by throwing them - then a proces similair to entrophy should take place. (The more we move the dice the more disorder in dice..) and still there would be some chance that they would reenter the ordered state by an accident. This is againt the second law of thermodynamics: the disorder gets bigger and bigger. How come?

A question.. if I threw two die 36 times what odds would you give (if I wanted to place a bet) on 3 coming up exactly twice, as predicted by the probabilities?

Say I wanted to bet £1000 that it would not be twice. I can't see how to work this one out.

The problem may simply be you are betting in Pounds and not USAD:D

I think that binomial probabilities are applicable to the question about a total of 3 occurring exactly twice in 36 rolls of a pair of ordinary dice. The odds are as follows. P(> 2) = .323 250 P(= 2) = .278 480 P(< 2) = .398 270P(=2) = C³⁶₂p²q³⁴, where p = 1/18, q = 17/18, and C³⁶₂ = the number of combinations of n items taken 2 at a time.

In general, evaluate the terms of (p + q)ⁿ to determine the probability of various multiple occurrences in n trials. p is the probability of a particular event happening in one trial, while q is the probability that it will not happen. (p + q) = 1

P( k occurrences in n trials ) = Cⁿ_kp^kq^n-k

Note that Cⁿ_k = n! / k!(n - k)!

Sorry: You asked about odds, not probabilities.

Probability = .278 480 is equivalent to odds of 72,152 to 27,848 against.

Approximately 18 to 7 against.

ProCop:
If we take many dice and compare it to the constalation of the universe supposing the dice were in an ordered state and we were disturbing this ordered state by throwing them - then a proces similair to entrophy should take place. (The more we move the dice the more disorder in dice..) and still there would be some chance that they would reenter the ordered state by an accident. This is againt the second law of thermodynamics: the disorder gets bigger and bigger. How come?
Firstly, there are some important differences between the dice model and the Universe. If we take an 'ordered' dice to have a value of six, there is no tendency for the dice to enter a lower energy state - 1 is as likely as 6.
However, if you're talking about taking ordered dice, randomizing them and then regaining the order at random as a real life example of 'entropy reverse', remember that to regain the order, you have to put a lot of energy into throwing the dice untilthey come up ordered, so although the entropy of the dice is reversed, the entropy of the system has increased.

Originally posted by geodesic


......, so although the entropy of the dice is reversed, the entropy of the system has increased.

Haven considered that.

(Brought to this "dice entrophy" idea by the Maxwell’s Demon metaphor
<i>
Pynchon is more “scientific” than many literary novelists, treating the impact of computers as early as 1965. The Maxwell’s Demon metaphor, whereby it is posited that one could theoretically achieve a perpetual motion machine by utilizing an effortless sorting of the molecules into hot and cold, is an apt metaphor for the book. Computers realize this efficient sorting, and so achieve the perpetual motion machine of the information age. Oedipa herself lives inside a sort of Maxwell’s Demon—all the clues she must sort out. </i>

<a href=http://www.thesatirist.com/books/TheCryingOfLot49.html>The Crying Of Lot 49</a>)

Dino: Your probabilities for the sums are correct, but I was talking about combinations. If the order of the dice is deemed insignificant, there is 21 possible combinations (black numbers):

1+1, 1+2, 1+3, 1+4, 1+5, 1+6,
2+1, 2+2, 2+3, 2+4, 2+5, 2+6,
3+1, 3+2, 3+3, 3+4, 3+5, 3+6,
4+1, 4+2, 4+3, 4+4, 4+5, 4+6,
5+1, 5+2, 5+3, 5+4, 5+5, 5+6,
6+1, 6+2, 6+3, 6+4, 6+5, 6+6.

If the sequence is deemed significant, then the reverse combinations also count (orange numbers) giving 36 combinations (orange+black numbers). The probability of the sums, of course, stay the same.

Yeah, just trying to make it complicated ;).

Hans :p

Dinosaur

Thanks. A lot of that was beyond me but I get the idea. Can you explain the principle behind the maths?

Canute: The following is not at all rigorous, but it conveys the idea.

Assume (p + q) = 1, where p is the probability of an event happening (a Hit)and q is the probability of it not happening (a Miss). This is necessary for an event with only two outcomes. Using percentages, the sum must be 100%. Using fractional probabilities, the sum must be one.

It is claimed that terms in the expansion of (p + q)ⁿ provide the probabilities for various outcomes. Consider the following. For any given sequence, the probability is the product of individual probabilities.
For example: P(HMMHM) = pqqpq, or p²q³

(p + q)¹ = p + q. this one seems obvious.

(p + q)² = p² + 2pq + q²
Now consider the possible sequences for two trials: HH, HM, MH, MM with corresponding probabilities pp, pq, qp, qq. Since pq = qp, we see that the binomial expansion gives the probabilities for the possible outcomes (two Hits, one Hit & one Miss, or two misses).

(p + q)³ = p³ + 3p²q + 3pq² + q³
The possibilities for three trials are as follows: HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM with corresponding probabilities: ppp, ppq, pqp, pqq, qpp, qpq, qqp, qqq. If you check it out, the binomial expansion provides the probabilities. Note that ppq = pqp = qpp (thus the term 3p²q).

Since (p + q) = 1, (p + q)ⁿ = 1, indicating that all the possibilities are included. Note that a check on the validity of probability calculations is to make sure that the sum of all the probabilities is one.If you check out some other expansions, you will find that they indicate the validity of the method.

BTW: For an event with three outcomes, expand (p + q + r)ⁿ where (p + q + r) = 1

I hope the above helps.

I think I get that. Still, if you were a bookmaker, would it really allow you to calculate the right odds for the bet?

Isn't there just yet another level of regression where we have to calculate the probabilities of the outcomes behaving according to the mathematics of the binomial expansion?

Canute: Binomial & multinomial expansions are only one tool of the trade, and they are only applicable to some of the common situations.

Oddly enough, bookmakers do not worry a lot about odds. Also, casino games have rules determined by historical experience not by mathematical calculations. Some examples follow.

When betting on a football game the bookmakers tell you (for example) that Ohio State is favored over Michigan by 7 points. The fairer bookmakers will say 6.5 or 7.5, eliminating the possibility of a tie (the bookie wins tied bets and this is a big edge). You bet $6 to win $5, but have your choice of team. If 50 people bet $6 on each team, the bookie collects $600 (100*12) and pays back $550 (50*11). The bookie does not care about true probabilities relating to the game. He wants his bettors divided: Half betting on Ohio State and half on Michigan.

A bookie I once knew said the following.A bettor cares who wins a particular game. The bookie makes sure that he does not care who wins. That is how you tell a bookie from a bettor.He said this to a small-time bookie who often took big bets and did not lay them off with a bigger bookie. He claimed that bookie had a bettor’s mentality. Several years latter, that bookie worked as a runner for my friend.

Some games have erroneous spreads. This happens due to regional prejudices or mismatched teams. The point spreads are calculated by a team of experts in Las Vegas (50-60 years ago, it was a Chicago operation), and sent to local bookies.In a mismatched game, the spread might be Ohio State by 49 over Northwestern and Notre Dame by 43 over Purdue. The local bookie will omit the Ohio State game, but Notre Dame versus Purdue is a traditional rivalry and his customers will want to bet on that game. He knows that everybody will bet on a 43 point underdog, so he changes the line to Notre Dame by 28, which he knows (from experience) will divide his bettors in half. I used to make modest amounts of money by betting on favored teams when I recognized a mismatched game. If you want to try your hand at this, do not bet too much, the bookie will quickly realize what you are doing.

Local prejudices are easy to understand. Many years ago, U of Pennsylvania tended to be under rated by local fans in South Eastern Pennsylvania. Navy tended to be over rated by fans in the Baltimore/Washington area. Philadelphia bookies regularly changed the spread on U of P, and similarly bookies in Baltimore/Washington changed the spread on Navy. When I worked in Washington DC and still had friends in Philadelphia, I became aware of the spreads being different on these two teams. Philadelphia bookies would have Princeton over Penn by 7, while Washington DC bookies called the game almost even. In Washington, it would be Navy over some opponent by 14 when Philadelphia bookies only gave the underdog 7 points.BTW: If his book is not balanced properly for some games, a bookie will lay off some of his bets by placing bets with a bigger bookie or a syndicate of bookies. He makes sure he will have a profit no matter who wins.

Casino operators want to maximize total profits, not the house edge. Craps is a very fast game, with a house edge of about 1.4% If the edge was less, the house would attract more players, but would have less total profit. If the edge were as much as 5% (typical at American roulette), an average bettor would be wiped out fast and hardly ever walk away a winner if he played for more than 30 minutes. With a bigger edge, the amount taken from each player would be more, but there would be very few players, and the total house profit would be less that it is currently. At a house edge of 1.4%, about 40% of the bettors will be winners after 100 plays (10% will break even, and about 50% will lose).

Roulette is a much slower game than craps, and a 2.5% (European) or 5% edge (American) will allow a bettor to last for at least an hour or so before being wiped out. While playing, he will be encouraged by frequent wins on the even bets like Black/Red & Odd/Even, and occasional big wins on an individual number (35 to one). At an American roulette table, about 27% will be winners after 100 plays (7% will break even, and about 66% will lose). 100 plays take longer than 100 plays at craps, so roulette players do not seem to mind (or notice) the poorer results.

At some casinos, Blackjack is almost an even game if you know how to play correctly (I am not referring to card counters who can get an edge on the house). A bigger house edge would be disastrous for the average player because most players make terrible decisions at the Blackjack table. When gambling first started in Atlantic City NJ, the game actually favored a good player (by about 0.2%) due to an option known as Early Surrender. It was even more favorable for a while if you were astute at reading the dealer. Most of the dealers were inexperienced for the first year or so at AC. When they had a ten and checked for Blackjack, you could usually tell if they had a bad hole card and make decisions accordingly. It did not help much when they had an Ace showing because players decisions are not so critical when the dealer has an Ace, and bad situations are not obvious to an inexperienced dealer when he holds an Ace.

Casino rules and payoffs for craps, roulette, blackjack, et cetera have been tinkered with for over 200 years to arrive at optimal house edges. Mathematicians are used for other purposes at a large casino, but no mathematician ever made up the rules for the games.

BTW: The casino desire to maximize total profit is not unlike the corporate desire to do likewise. I have heard many people claim that a monopoly would raise prices dramatically, a strategy which maximizes profit on an individual sale. Actually, even a monopoly would try to maximize total profits, which is a different strategy. In some cases, charging a very high price might reduce sales to the point where you would not even maximize unit profit due to higher units costs. To maximize total profits you must maximize the product of UnitProfit and TotalUnitsSold.

*kowtows to Dino*

YOU sir, are a freakin wealth of information.

We are not worthy, yet you grace us.

Far too kind sir, many thanks.

Seriously, I freakin love your posts.

Dinosaur

Brilliant and very interesting. I've lost enough in casinos to appreciate what you're talking about. (I first joined a casino because at around 2.00 am after gigs I could get free sandwhiches and coffee. What an idiot. They became the most expensive snacks in recorded human history.)

But it sort of sidestepped my question. I'm trying to get at the regression of calculations that occurs when trying to calculate what the chances are of events turning out as per their calculated probabilities. Is it just an endless regression of binomials?

Originally posted by Canute
But it sort of sidestepped my question. I'm trying to get at the regression of calculations that occurs when trying to calculate what the chances are of events turning out as per their calculated probabilities. Is it just an endless regression of binomials?

You're basically asking "how do you know that the binomial distribution is applicable"? I'm not sure if there is a "regression of calculations" per se, unless you are questioning the pertinence of mathematics itself, or how math is applied.

If I'm not mistaken, this is more matter of determining if your scenario fits a known distribution. For instance,
The binomial distribution gives the discrete probability distribution of obtaining exactly n successes out of N Bernoulli trials (where the result of each Bernoulli trial is true with probability p and false with probability ). (http://mathworld.wolfram.com/BinomialDistribution.html)

This is just basic statistics stuff. My guess is that Dino has a much better take on it. :)

WesMorris: Thanx for the compliment. If you live to be as old as I am, you pick up a lot of information along the way. If you do not go senile, you manage to hang on to most of it.

Canute: I do not understand your question.But it sort of sidestepped my question. I'm trying to get at the regression of calculations that occurs when trying to calculate what the chances are of events turning out as per their calculated probabilities. Is it just an endless regression of binomials?The mathematics can compute the probability of each possible total for pair of casino dice. When you actually roll a particular total, the calculated probabilities become meaningless for that throw.

All the calculations tell you is the expected percentage for each total if you throw the dice a large number of times.

Laymen talk about the Law of Averages. Experts refer to it as The Law of Large Numbers. The calculations do not mean much for a few dice tosses, only for hundreds/thousands or preferably millions.

If 100 people walk up to the dice table, make one bet and walk away, 49 are expected to win and 51 are expected to lose. The probability of exactly 49 winning is only about .08 (8%) even though this is the so called expected result. Oddly enough the so called expected result is not very likely.

The probability for the range of winners being 45-55 inclusive is about .724 (or 72.4%), meaning that there is better than one chance in 4 that a given trial will be outside that range, which is definitely not expected.

I am reminded of the following.The race is not always won by the swiftest runner and the fight is not always won by the fiercest warrior, but that is the way the smart money always bets.

Originally posted by Dinosaur
Oddly enough the so called expected result is not very likely.
This is what I was getting at. In effect my question was - just how unlikely is the expected result?

It's a back to front question, but I was thinking of it in wider terms than betting.

It seems to me that every time you calculate it you get another set of probabilities. (This is what I meant by a regression of binomials).

I'm not yet completely sure that I'm asking a sensible question. It's something to do with the relationship between probability and chance.

Canute

If it's a random event, you can't predict it. You can only estimate its likelihood with consideration to a representative sampling of the population. The "population" in terms of throwing two dice would be the results of every time they were ever thrown. That could be hypothetically extended to every time any dice were thrown.

I would imagine that if you took any two dice you would find that the population of the results of throwing them would vary slightly (statistically) than any other pair. In other words, i'd guess them to slightly favor a result over millions and millions of throws, but that's just a guess. It's probably almost impossible to discern the difference between most pairs of dice simply because the action of throwing them is so random to begin with. To correlate a favoring of a particular result (unless gross) with a slight physical variation in their manufacturing process would probably be tough.

Canute: It is still not clear to me what you are asking.

In some sense, the expected result for many situations is not likely to happen. If you flip a true coin one million times, the expected result is 500,000 heads & 500,000 tails. This result is more likely than any other specific number of heads & tails. The probability of it actually occurring is quite small. I cannot calculate it without a lot of trouble. The following are probabilities of exactly half heads and half tails for various numbers of coin flips. P(10 flips) = .246 094 (about one chance in 4) P(100 flips) = .079 589 (about one chance in 12) P(1000 flips) = .025 225 (about one chance in 40)As the number of trials goes up, an exact 50-50 division becomes less likely, but the percentage of heads gets closer and closer to 50%. Note that for 10 flips, you will get 60% or more heads about 37% of the time. For 1000 flips, there is hardly any chance of getting 60% or more heads. For 1000 flips the percentage of heads will round to 50% over 99% of the time.

Ok, thanks. I'll give it some more thought.

Is there anyone here that would suggest that "Lotto" or "Bingo" are not truely random?

Also this reminds me of an old parlor trick. There are 366 dates per year possible (including February 29th).

Yet if you are at a gathering of 21 or more persons you can safely bet that at least two persons will have the same birth date (year excluded).

IT seems counter-intuitive but it works. Ive done it many times. YOur odds of winning the bet go up rapidly as the number of persons in the group increase above the 50/50+ mark of 21.

You can prove this in an indirect way by having two people independantly write down 21 dates (days+month) and then compare the list.

You could do it here by having persons post their birth dates and taking the first 21 posts or persons making a date list and then comparing lists one against the other. that is take one list and then compare it to all other lists and see how many lists have matching dates.

Collectively doing that with each list and totalling the number of hits in the end.

Should you be game for a test of this claim here is mine:

1 February

NOTE: Having two persons write down 21 random numbers from 1 - 366 is not the same odds. Breaking 366 down into two required groups limits the choices. i.e. 1 - 31 days and 1 - 12 months.

MacM: As far as I know Lotto, Bingo, and state-run lotteries are random selection games, or at worst a very close approximation to random selection games. At least I have no reason to think otherwise.

You seem to have misremembered the birthday problem mentioned in your previous post. Your figures are a little bit off. When I first encountered it, I was surprised at some of the probabilities. I do not know how to do the calculations if Leap Day birthdays are considered. This is a peculiar problem because some of those born on 29 February celebrate on 1 March (one day after 28 February) in non Leap Years, while others celebrate on the last day of February every year.

Perhaps the correct way to consider Leap Days is to calculate two probabilities: One using 365 days per year; The other using 366 days per year. Then interpolate to the value for a 365.242 day year.

The following calculations are based on a 365 day year. Adjusting for Leap Years would not change the values significantly. Probability(No Hit) = Permutations(365, n) / 365ⁿ This is the probability that no two out of a group of n people will celebrate their birthday on the same day of the year.

For a group of 3, this is 365*354*363 / 365³ = .991 796

Define P(n) = 1 - P(No hit) This is the probability that two or more out of a group of n people will celebrate their birthday on the same day of the year.

P(21) = .443 688
P(22) = .478 695 (Not quite an even money bet)
P(23) = .507 297 (A bit better than an even money bet)
P(30) = .706 316 (About 7 to 3 in favor of at least one hit)
P(40) = .891 232 (Almost 9 to one in favor of at least one hit)I still find the above probabilities surprising.

I do not believe that birth dates are randomly distributed over a calendar year, although I would expect them to approximate a random distribution in the Western technological cultures. Perhaps the above would not quite be matched by probabilities derived by sampling actual groups of people, but I would expect them to be close.

I hope I have made no typo’s or errors in calculations.

Dinasuar,

I find your posts most amazing. I have little understanding of the mathematics you demonstrate. You are obviously quite masterful of what you do. You articulte your results extremely well.

Can I ask, do you work professionally in the fields of probability and chance? If not why not?

Quantum Quack: I majored in mathematics at a liberal arts college which required that you take at least one semester of work in almost every department. Mathematics has always been a kind of hobby for me.

When I was 7- 22 years old, the local bookie was a childless man who took a liking to me. He talked about his business a lot, and taught me much about the details of the gambling world. This made me especially interested in probability theory when I went to college.

In the early 1950's I became a programmer and loved to fool around in addition to doing my job. Several of us recognized that Blackjack was different from other casino games and used computer time at night to analyze the game, discovering that there were methods (card counting) that could overcome the house edge. We became fanatically interested in probability theory.

A few of us (not me) had delusions of grandeur and set out to make a fortune beating the casinos. That is not really possbile, because card counting only provides a slight edge and for the same effort you can make more money moonlighting at a high tech job. I enjoyed making very modest amounts at the Blackjack table and sometimes losing. Having fun at casinos got me interested in analyzing other gambling games.

BTW: Card counting at Blackjack can get you small edge on the house. There are system which do very well in theory, but are almost impossible to apply in real life. When Thorp published the first book on card counting, the casinos became paranoid. They should not have become excited. Most card counters make enough mistakes to give away their edge, and the ones who are letter perfect cannot make enough to hurt the house.

A practical card counting system can result in your winning 51 out of 101 bets, while losing 50. Since the house pays 3 to 2 for Blackjack and you sometimes are allowed to double your bet in favorable circumstances, the effect is a bit better than one unit bet profit per 101 bets. Since it takes an hour or more to play 101 hands, you can win about one unit bet per hour, maybe less. If you bet $20 per hand, this is no more than $20 per hour, and you need a bankroll to allow you to continue playing when you have a run of bad luck.

There are reasons why you should have a bankroll equal to at least 50 times your average bet. If you bet $20 per hand, you need at least $1000 as a bankroll. Actually a bankroll 100 times your unit bet is preferred, meaning $2000 if your average bet is $20. If you want to make more per hour, you need an even bigger bankroll to start. If you have a big bankroll, why not continue doing whatever it was that got you that bankroll, and not risk your money at a casino?

Card counting is a lot more work than you might imagine. You must concentate in a distracting environment, which is mentally exhausting. One or two errors and you have given away your edge on the house for the current 1-2 hours. It can be fun if you are not making a serious effort to make big money, but is a tuff job if you want to do it for a living.

I used to play at tables with a $2 to $5 minimum and considered it entertainment. I often played without bothering to count and lost modest amounts, but had fun. At a casino with decent rules, you figure to lose only $5 per $1000 in total bets if you know basic strategy and do not bother counting cards. If you do not do anything crazy at the craps table, you only figure to lose about $15 per $1000 in total bets. I hardly ever play anything but Blackjack. When I go to the casino with friends, I will sometimes make a few plays at games other than Blackjack.

At the above prices, casino gambling can be a cheap amusement. Besides, you can meet some very interesting members of the opposite sex at a casino (take care to avoid the professionals and the con artists). I consider a good piano bar and apres ski the only places better for prowling than a casino or a cruise ship.

You can make shitloads of money counting cards. Shitloads.

http://www.wired.com/wired/archive/10.09/vegas.html

- Warren

Dinasour,

I don't doubt your numbers and I could indeed be a couple of people off but I have been playing the wrong odds for years.:D

I do recall the explanation I saw had something a bit different.

It involved i.e. 365 - 1, then 365 - 2, 365- n for the number of persons in the group. In other words the odds kept changing for each calculation and then they averaged the sequence.

Sorry I don't remember the actual formula but for 21 people and a 365 day year you are left with only 344 maximum unused days.

Does that make sense.?

Glad you were surprised in any case. It always blew my mind as well.

Also, I always thought that the odds had something to do with the fact that the dates are blocked into two groups. That is 1 - 31 days and 1 - 12 months. I know that still works out to 365 combinations. But:

For example with 21 people your must have a minimum of 1.75 people with the same months. Now what are the odds that with 1.75 assured hits and most likely 4 - 5 (guessing here) having the same month then also having a 1 - 31 hit.

Can you run your numbers that way and see if it makes a difference?

Hey Dinasaur,

Programing in the1950's hey man that makes you a "Dinasaur".

Makes you over what, 70 years old.....my goodness..you have a nimble mind. Excuse the compliment.....

MacM: There is probably some clustering of birth dates which results in the real life experience being slightly different from calculated probabilities. I do not think this effect is significant in modern technological cultures.

The formulae I posted are the correct way to calculate the probabilities assuming birth dates randomly distributed over a 365-day year. I did calculations for a 366-day year, discovering that the third decimal digit was usually off by one when compared with calculations for the 365-day year. Interpolating for a 365.242-day year would result in probabilities very close to the 365-day probabilities, so I consider the 365-day year computations reasonably accurate.

BTW: There is a typo in my previous post. P(22) = .475 695, not .478 695

The logic is as follows. First imagine asking each person entering a room to tell his birth date, which you record. When the second person enters, Probability(No match) = 364 / 365 (there are 364 days out of 365 not matching the first birth date). When the third person enters, P(No Match) = 363 / 365 (there are 363 days out of 365 not matching the first two dates). When the fourth person enters, P(No Match) = 362 / 365 The trend here is obvious. The above probabilities must be multiplied to obtain the probability of no matching birth dates. This results in Probability(No Match) = Permutations(365, n) / 365ⁿ for a group of n people. A dummy multiplier of 365/365 is included to make the formula work slightly easier to express.From the above, P(n) = 1 - P(No match), where P(n) is the probability of no two people with matching birth dates in a group of n people.

Note that calculating the probability directly is a formidable task. The probability of exactly two people with matching birth dates would be required. Then the probability of two pairs with matching birth dates. Then the probability of three pairs. Et cetera. Then you would have to compute the probability of three people with the same birth date. Then the probability of two triplets with the same birth date. Then the probability of a pair and a triplet, et cetera.

It is relatively easy to calculate the probability of no matches and subtract from one. There are only two possibilities when viewed this way: Either there are no people with matched birth dates or there are various groups (perhaps only a group of two) with matching dates. The sum of the probabilities for all possible mutually exclusive events is one (certainty). Hence, subtract P(No Match) from one to get P(Some Match).

There is no other way to calculate the probabilities if you assume birth dates randomly distributed over a 365-day year.

Please correct me if I am wrong,

The use of averaging is primarilly to predict future events based on the Averaging of historical data.

so we have events say 2, 9, 4, 7, 2, 3 the average equals 4.5.

The figure we gain is only valid if future circumstances remain unchanged and the "now" is consistant with the past.

So we can predict that if all things remain equal the average of 4.5 holds true.

The average of 4.5 in no way controls the future outcomes in that as time passes the average may climb or diminish according to actual circumstances and new data.

So the question of whether chance is provable comes to mind.

Is probability or chance only provable with hindsight, then used to predict but in no way determine the future? ( a philisophical perspective)

Possibly in a closed environment of relatively contained variables prediction is more valid but still does it reflect the reality or simply analyse the reality?

I know this may be an exxtreme example but a casino for example is a fairly closed envioronment. The house relying on such to give itself it's confidence as a profit making entitiy.

"However in the middle of a profitable phase there is a power surge and all the poker machines shut down for an hour or two."
or
"one of the staff have a secret arrangement with a customer who consistantly walks away with a big win"
Etc etc

The house although a closed environment is now seen to be a lot less closed that first thought.

Btw is averaging just another probability tool or is it considered as somethng else?

I guess in the end it's a question of whether chance randomness is actually provable. Does chance exist or is it just a perception of our imagintions, ( a construct )?

Dinosaur,

I am in agreement with your process. The 365*364*363, etc is 365, 365-1, 365-2 etc. that I mentioned. However, my experience is that the odds are in fact slightly beeter than the direct mathmetical assumptions.

I used to think it was due to the bifurcation of the 365 into the 1 - 12 and 1-31 choice ranges but I agree that apears not to alter the end numbers.

However, I think you may have hit the cause for the odds being higher and that would be due to clustering of birth days. That would be something enfluenced perhaps more by weather cycles than anythingelse.

Cold winter nights makes for warm snuggles, etc. June weddings make for March babies, etc.

I claim the odds are better than calculated since over the years I have had many cases of two pairs of matches, and a couple of triple pair matches. Occasionally you will get 3 having the same birth date, etc.

Quantum Quack: There is a reason I call myself the Dinosaur. I actually did mathematics in the prehistoric era prior to the existence of computers.

Chroot: The URL you posted makes very interesting reading. I consider the story to be possible, but not likely to be true. Over the years, I have read many stories about people beating the casinos for huge amounts of money. The stories were usually about bright college students and/or professors who discovered some method to beat the house edge.

Some of the stories related to craps, roulette, and other independent trials games. There just is no way to beat such games. I always suspected that these stories were planted by casino marketing personnel when they had reason to believe that various types of system players had become discouraged and inactive.

Games using paramutual odds (race tracks, dog tracks, Jai Alai, et cetera) can theoretically be beaten, but in practice it is not possible. The betting public determines the payoffs. To beat such a game, you must recognize situations for which the betting public has caused a favorable bet to exist. The house edge at such games is typically 17 to 25%, making it almost impossible for favorable bets to exist. There is some possibility of beating the race track due to having inside information about fixed or otherwise peculiar races. My bookie friend was willing to take wagers from jockeys, owners, trainers, et cetera (except at one or two tracks). Due to his willingness to take such bets, I do not think there is much inside information available.

Blackjack is a dependent trials game, meaning that what has already happened changes the probabilities relating to what is going to happen. Such games can theoretically be beaten. It is known that card counting can result in winning at casino Blackjack. If allowed to use a modern high speed computer with a properly designed 11-12 key interface, I think you might be able to get a 2 to 5% edge on the house (this is a SWAG). I doubt that you can get much more than a 1% edge via any card counting method (this is another SWAG).

BTW: A SWAG is a Sophisticated Wild Ass Guess which is more reliable than a WAG.

The very interesting story cited does not seem believable. It looks like exaggerations of the experiences of Ken Uston and Stanford Wong, both of whom I believe won money from the casinos. I suspect that the stories about Ken Uston are themselves exaggerations of his actual experiences, told to provide PR for his books. I am very suspicious of the claims of other authors. One author of a popular book on card counting behaved obnoxiously at half a dozen casinos, resulting in his picture being posted as a person banned from almost all the Vegas casinos. His book brags about having been banned, implying that it was due to his success at the tables. His book is a rehash of various other books on the subject.

For those interested, the book by Wong is probably the best book on card counting, while a book by Julian H. Braun is an excellent description of Blackjack probabilities. The books by Thorpe (1962 & 1966) are only of historical interest (He was the first to publish, although not the first to use card counting methods). A book by Peter A. Griffin is the best book on the mathematics of card counting.

Griffin is the only author who claims to have lost money counting cards, and has an interesting humorous style. He has an interesting chapter on applying card counting methods to a much simpler game than Blackjack. This chapter provides some insight into the effectiveness of card counting. His book puts all of the heavy mathematics on a few pages at the end of each chapter, making the book easy to read even if your mathematics is weak. You can skip the heavy mathematics without losing much.

The story cited mentions a Hi Lo count and attributes it to Thorpe. His original book mentioned a system based on counting tens and others (A fresh deck has a 16 to 36 or 4 to 9 ratio). He also advocated keeping track of Aces. His later book described the more modern Hi-Lo count, but admits that it was originated by another researcher. The Ten-count method is a bit harder to use and probably not as effective as the Hi-Lo count, although also keeping track of Ace-Richness might make it more effective that the Hi-Lo count (at the expense of more complication)

Wong describes the modern Hi-Lo count, which is very simple to use and almost as effective as far more complex methods. He was the first to advocate playing exclusively versus favorable decks, a very effective strategy. His book seems more practical than the others I have read, suggesting that he might be the author who has spent more time using his methods than anybody else.

Julian A. Braun (from IBM) is responsible for the basic data in almost every modern book on Blackjack. He simulated millions of deals to verify the effectiveness of card counting methods, as well as computing some basic probabilities relating to the first hand dealt from a deck. His simulations are critical for verifying card counting methods. It is impossible to analytically determine the true probabilities for Blackjack. The best you can do is calculate the probabilities relating to the first few hands dealt from a deck.

For those interested in studying the game, I might mention the following. Most (perhaps all) books on card counting indicate that the starting count is zero. When the count is negative, the deck is unfavorable, while a positive count indicates a favorable deck. It seems easier to me to consider 100 as par and start the count at 100 when the deck is fresh. In the distracting atmosphere of a casino (even at off hours), it is much easier to keep track of an always positive count. This is a practical tip I have never seen in a book on the subject. The Hi-Lo count ignores sevens, eights, and nines. In theory, a seven should count plus 1 / 2, while a nine should count minus 1 / 2, but this only slightly improves the effectiveness of the Hi-Lo count. It adds complexity due to fractional counts, and is likely to result in errors and/or losing the count. Note that the absolute count is not the critical value it is the count adjusted for how many cards remain in the deck. A count of plus 5 when there is only about 2 decks left is more significant than the same count after only one hand has been dealt from an 8-deck shoe. However, counting red sevens as plus one and red nines as minus one does not require the use of fractions and is no more difficult than the recommended Hi-Lo count. This is a better approximation than ignoring all sevens and all nines. I have never seen this mentioned in a book. A single deck is very sensitive to the removal of a few cards. For example, it is always recommended that you draw a card to hard 14 when the dealer shows a ten. If the 14 is made up of two sevens, you should not hit in a single deck game. Also, a sixteen made up of combinations like (ten, 4, 2), (4, 4, 5, 3) should not be hit when the dealer shows a ten, but hands like (ten, 6) and (9, 7) should be hit versus a dealer ten. This and various other special cases are mentioned in some (not all) of the better books, and should be memorized for single deck play. Do not play at any casino that does not allow double down after splitting. If you find a casino (unlikely) with an Early Surrender option and which allows double down after splitting, the game is very slightly favorable to a player who knows the correct basic strategy (even without card counting). Late surrender is good, but not a great option. Do not play at a casino that only allows double down on a total of hard ten or eleven. The option of doubling down on various soft totals is important. There are a few (at least one that I know of) small casinos in Las Vegas with very favorable rules. One I encountered allowed double down on any hand, even hands like (5, 4, 2). Most casinos only allow double down on the initial 2-card hand dealt.Good luck to any who decide to try their luck and/or skill at the Blackjack table. It can be fun and not expensive if you have some discipline and are willing to learn basic strategy.

BTW: If you are a skier, go to Tahoe. It is close to Reno and has some casinos of its own on the Nevada side of the lake.

Quantum Quack:I am not sure what you mean by provable in the following context.So the question of whether chance is provable comes to mind.

Is probability or chance only provable with hindsight, then used to predict but in no way determine the future? ( a philisophical perspective)Probability Theory is a well established branch of mathematics. Its basis is as sound as any of the well known disciplines like calculus, analytical geometry, theory of equations, et cetera. It is used for different purposes than other disciplines.

Suppose somebody asks you to roll a seven and says he will give you four to one odds. Should you take this bet? Probability theory says no, but this does not mean that you will not roll a seven. It only means that it is a bad bet. Make a lot of such bets and you will win a few, but lose money. What is called fair odds for rolling a seven is five to one. If somebody offers you six to one take the bet, but do not bet more than you can afford to lose on one roll (you only figure to win one such bet every six rolls).

The probability of rolling a seven is 1 / 6, which translates to 16.6667%, once in 6 times, five to one odds against. There is a probability concept called expectation. It computes the theoretical win or loss for one bet. If you bet $10 at various odds on rolling a seven, the following is your expectation. 3 to 1 odds: Expectation = - $3.3333, meaning that you can expect to lose over $3 per bet. Obviously, you cannot lose $3.33 on a single bet. You can lose $10 or gain $30 if given 3 to 1 odds. Probability theory says that 3 to 1 is a terrible bet. You figure to lose over $300 if you make 100 such bets. 4 to 1 odds: Expectation = - $1.666667, a better bet. Still a bad bet. 5 to 1 odds: Expectation = zero, indicating that you will neither win nor lose money if you make a lot of such bets. 6 to 1 odds: Expectation = +$1.66667 Now this is a good bet that nobody is likely to offer you. If allowed to bet $10 per roll as often as I wanted at these odds, I would expect to live an affluent life.The above expectations are calculated as follows. The probability of rolling a seven is 1 /6 and the probability of not rolling a seven is 5 / 6 Note that these add up to one, indicating that there is no other possibility. Mathematicians say that you are not allowed to throw the dice into the ocean and not see the result. There are rules assumed he