Sciforums vs The Monty Hall problem

I'm not sure that this belongs in Philosophy... but in a way, I think it sort of should. Because it doesn't make sense to me at all--like it defies some kind of law of logic. And logic belongs here, I imagine... so here goes.

I was reading a Wikipedia article about Marilyn vos Savant and there's a section called the Monty Hall problem. Here it is.

Taken directly from the article:

Okay. I don't understand this at all. Not only that, but I don't think it's possible for me to understand it. I think if I understand it, the world might explode. How is this possible?

 

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Maybe this should be in a mathematics thread, since it deals with odds. I think it has to do with whether or not the odds change when one factor of choices in eliminated. I believe it does. With the first set of three doors, you have a 1 in 3 chance of getting the car, but after one door is revealed to not have the car, there is now a 1 in 2 chance.

 

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Okay, so originally, there's a 1 in 3 chance of picking the right door. Then the host removes one of the possible doors from the equation. Now there are only two doors. Only one of them has the car behind it. So now there are two doors. One door has the car behind it; the other door does not. 50/50. How is it not 50/50? What gives?

 

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Because when you made your decision of which door, you only had a 1 in 3 chance of getting the car. When the host removes a door that has no car behind it, the odds of choosing a door goes to 1/2. The odds the door you picked are still 1/3.

 

Yes, but now you are asked if you want to switch. That resets the odds, starts the thing all over again with different parameters. The odds now are not which of three doors, but which of two doors do you think has the car.

 

Exactly. I'm still stuck... why are the odds not 50/50? Maybe Absane could help with this one.

 

I'll say more or less exactly what I said in the last Monty Hall thread, because I think I explained it succinctly and accurately.

You have a higher chance of winning if you switch doors. There are 3 possible starting conditions:

1) you pick the car to begin with
2) you pick a goat
3) you pick the other goat

Imagine you start with case 1. It's a 1/3 chance of happening. When Monty opens the other door, switching would reveal a goat. That is a LOSS.

With case 2: Monty reveals the other goat. Switching would net you the car (WIN).

With case 3: Monty reveals the other goat. Switching would net you the car (WIN).

So TWO out of the three possible starting cases will get you the car if you switch doors. The only possible way you can lose on a switch is by picking the car to begin with (which is only 1/3). There's really no question about it. That is a 2/3 chance of winning the car when you switch doors. For you to win without switching would mean you had picked the car on the first guess (which is only a 1/3 chance).

 

Makes sense.

 

@RubiksMaster you make a home lotery for three. The first person looks at his/her ticket and shows it to the other two (he/she lost). If the other two switch their tickets do they encrease their chance of winning?

 

How does statistical analysis and probability affect our actions? Don't people tend to overlook these things when believing, for instance, in Astrology? ...or "weird" coincidences? How probable was any particular coincidence? Can one calculate the probability that your rain dance was the cause of a rainstorm? ...or that your disease was cured due to a prayer or the healing powers of holy person?

 

this explains it quite well.
i too was of the opinion it would be 50/50 but the above changed my mind.

 

Though it should be noted that the two choices have an equal chance of being right, only that your choice had a less equal chance.

 

If two people switch their tickets they can't BOTH increase their chance of winning.

Let's say you're one of those 3 people. one of the other two people shows everyone his lottery ticket. If he won.........then he won. If he lost, then switching tickets won't make any difference to your chance of winning...

BECAUSE you're picking a random person to reveal their ticket. The Monty Hall problem has 2 extra clauses:

1) The ticket revealed is always a LOSING ticket
2) The ticket revealed is always someone else's ticket (never yours)

When you add these two factors into the problem then the odds unbalance, and switching with your opponent gives you a higher chance of succeeding.

 

This explanation is wrong. You cannot simultaneously hold that you had a 2/3 chance of choosing a goat AND that you did not choose the goat which monty has already revealed. If you now know that you did not choose one of the goats, you must reevaluate the odds of your original choice. Attempted explanations like this show why the problem was debated so fiercly for so long - people try to explain why it is right using a false argument and then other begin poking holes in that argument only to find that it is right anyways later on but for a different reason.

Here is an example of an argument that shows why there is a greater probability of getting the car when switching:

This explanation also gives an alternate version of the monty hall problem which shows why the typical explanation of the problem is wrong. Here the chance that you chose each goat or the car is still 1/3 each at first but monty reveals one of the goats more often than the other when you choose the car and you know that he does this. Since it is logically impossible to have chosen a goat and then monty reveal the goat you have chosen, the odds are different depending on which goat monty reveals - even up to a 50/50 chance of getting the car by switching or not when monty always favors one of the goats over the other.

Note that everything said about the original problem still applies to this alternate version, but the conclusion is no longer true. That is because it is a logical fallacy to hold simultaneously that you had a 1/3 chance of choosing each object and that you did not choose the goat which monty has revealed. It is simply a coincidence that such reasoning brings you to the correct answer in the case of the original problem.

 

Considering more doors might make you feel more comfortable with the correct analysis.

Suppose there were 100 doors with a valuable prize behind one and a penny behind the other 99.

You pick a door and Monty Hall opens 98 other doors revealing 99 pennies. oats. You can stay with your origianl choice or swirch.

As with the three door version, there are only two unopened doors. Do you think it is an even bet between those two doors?

Without doing any serious analysis, it seems to me that the odds against your picking the correct door was about 100 to one against you. When you made your choice, the odds favored the prize being behind some door other than the one you picked. What door might that be jnow that you have seen 98 of the doors you did not pick.

 

Actually I think this question should be in the math section ....

If there were an almost infinite numbers of doors (and only one car) -
you choose one door - and the host (who knows where the car is ) opens all doors except 2 doors (one of which is the door you choose) - then you would be a fool not to change your door !!

HOWEVER, that is actually cheating - for believe it not , calculation of probability for large numbers are DIFFERENT from calculation of probability for small numbers ....... So Marilyn Vos Savant is actually cheating ....

If you do not believe me , just see these links :

http://en.wikipedia.org/wiki/Probability
http://en.wikipedia.org/wiki/Law_of_small_numbers
http://en.wikipedia.org/wiki/Law_of_large_numbers

If everything goes really difficult , you can always use quantum physics and get surprising results ...

So ..... by applying the probabilty theories for big numbers - you only have
1/3 chance for choosing the right door ....

By using the normal calculations for small numbers - you actually ends up with : Yes, you guessed it : 50 % ....

Scifo versus Monty Hall : 1:0

:m:

 

That depends on , if you think 3 doors are a large number or not ...

If you think it is a small number : then the chance is 50 %, so it doesn´t matter
first goat = A
second goat = B
car = C

for door 1:2:3 possibilities are :

A:B:C
B:A:C
A:C:B
B:C:A
C:A:B
C:B:A

By opening door 3 revealing a goat , the 2 first options are eliminated , leaving only the 4 lower possibilities : exactly 50 % chance - it really doesn´t matter which door you choose .... that is the normal math for a smallnumber problem ...

If you think it is a big number , then it does matter - and you should change the door !!

I actually think 3 is a small number .....

What I am trying to say is, that Marilyn is applying largenumber theories to a small number problem ...... I am not saying she is wrong - only explaining why so many people do not understand it .... meaning it is not a mystery anymore ...

 

Theres some interpretation trickery that goes into part of Marilyn's math. I'll explain the correct way to work this out and Marilyn's way.

When you select one of the 3 doors your chances of winning are 1/3. If a goat door is shown and you don't change your answer then your chances of winning are still 1/3. Not changing your answer is the same as selecting one of the 3 doors without knowledge of any goat doors.

By choosing a new door you up your chances of winning to 50% because now you're performing the equivelant operation of flipping a coin.

Marilyn's way differs in this last part. In her interpretation the system doesn't change once a goat door is shown therefore this knowledge increases your odds of winning to 2/3 if you choose to change doors. 1/3 (original odds) + 1/3 (knowledge of the goat door) = 2/3.

So Marilyn is correct in that it is better to choose again after knowledge of the goat door but she is incorrect in the objective odds that exist once the new door chosen.

Run it through a simulation... you'll see that re-choosing yields a 50% chance of winning (not 66.66666~% as Marilyn would have you believe). Not re-choosing yields a 33.33333~% chance of course (the same as not having the option to re-choose).

 

No - the moment the goat appears in door 3 - your door suddenly gets 50 % chance , by low number probability math.....

You are correct , the simulation with rechoosing the door yields 50 % chance,
of winning (not 66.666 % as Marilyn claims ) .....

If you run a correct simulation , without rechoosing ..... surprise, surprise
again 50 % chance (remember to deduct the possibilities with door 3 with a car - we KNOW it is not a car )