If you don't understand that, there is nothing I can say to change your mind.
The Monty Hall Problem Revisited
- Thread starter raydpratt
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DaveC;
Staff positions car in 1 of 3 doors.
Player always guesses door 1.
Rule 1 prohibits host opening door from players 1st guess.
Rule 2 prohibits host opening door containing the car.
Host opens door 2 or door3 for events 1 to 4.
When car is not behind door 1, rule 2 applies.
Host opens door 2 and door 3 for events 5 to 8.
When car is behind door 1,rule 2 is not needed, but
host needs 2 games to choose both doors.
Savant visualized an oversimplified game, using above rules.
Host opens (red), result 'r' is + for win - for lose.
e 1 2 3 r
1 0 0 1 +
2 0 1 0 +
3 1 0 0 -
She also used generic goats noted as '0', and half of the results of e3.
If she had an [e4 1 0 0, -], there would be no difference for switch vs stay.
No startling revelation that switching wins 2/3 vs staying 1/3
The general list shows 8 possible events.
There are 4 possible wins for door 1 and door r.
No advantage.
If the list is reduced to 6 games by excluding e6 and e8,
there are 4 wins for door r vs 2 wins for door 1, an advantage.
Marilyn Savant differs with you.
"So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!)."
The host can open both doors but not in the same game. The purpose of the host is to prolong the game offering the player a 2nd guess, not to end it.
Logic/reason proposes an additional game for the additional opening.
That produces a consistency in the definition of a game.
You haven't comprehended the fact that if you perform the game in the same manner she proposes, game list, rules, etc., you will get the same results.
That does not prove she is correct, only that you can follow instructions.
Staff positions car in 1 of 3 doors.
Player always guesses door 1.
Rule 1 prohibits host opening door from players 1st guess.
Rule 2 prohibits host opening door containing the car.
Host opens door 2 or door3 for events 1 to 4.
When car is not behind door 1, rule 2 applies.
Host opens door 2 and door 3 for events 5 to 8.
When car is behind door 1,rule 2 is not needed, but
host needs 2 games to choose both doors.
Savant visualized an oversimplified game, using above rules.
Host opens (red), result 'r' is + for win - for lose.
e 1 2 3 r
1 0 0 1 +
2 0 1 0 +
3 1 0 0 -
She also used generic goats noted as '0', and half of the results of e3.
If she had an [e4 1 0 0, -], there would be no difference for switch vs stay.
No startling revelation that switching wins 2/3 vs staying 1/3
The general list shows 8 possible events.
There are 4 possible wins for door 1 and door r.
No advantage.
If the list is reduced to 6 games by excluding e6 and e8,
there are 4 wins for door r vs 2 wins for door 1, an advantage.
Marilyn Savant differs with you.
"So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!)."
The host can open both doors but not in the same game. The purpose of the host is to prolong the game offering the player a 2nd guess, not to end it.
Logic/reason proposes an additional game for the additional opening.
That produces a consistency in the definition of a game.
You haven't comprehended the fact that if you perform the game in the same manner she proposes, game list, rules, etc., you will get the same results.
That does not prove she is correct, only that you can follow instructions.
DaveC426913
Valued Senior Member
That 'not understanding it' door swings both ways, and it's kind of hit you on the ass.
Your decision to simply ignore counter-evidence and counter-challenges is damning. I think it is apparent that you know you're fighting a lost battle but have resigned yourself to die on this hill.
DaveC426913
Valued Senior Member
You are continuing to beat a straw man. We have moved on from Savant. What they think is not relevant. Address the arguments made here, in this thread.
Show in the simulated game a single row that does not follow the rules as you understand them.
Prove to us concretely that my results are flawed - not any logic you are ascribing to me or the irrelevant Savant, show us - in the data. it is an accurate simulation of the game as defined.
Here it is again:
JamesR;
post 65
In terms of reality/possibility the 5 doors can be represented as
1 2 3 4 5
1 0 0 0 0
with 1 meaning contains a car and 0 meaning contains no car.
The car is behind door 1. Player guesses door 1.
As the host reveals each door from right to left, n changes.
1 0 0 0 0_s/n=1/5
1 0 0 0__ s/n=1/4
1 0 0____ s/n=1/3
1 0______ s/n=1/2
The car is behind door 2. Player guesses door 1.
As the host reveals each door from right to left, n changes.
0 1 0 0 0 s/n=1/5
0 1 0 0__ s/n=1/4
0 1 0____ s/n=1/3
0 1______ s/n=1/2
The host can always reveal n-2 doors.
Using the general game list for 3 doors,
switch wins if the car is not behind the players 1st guess, which is 4/8, and
switch loses if the car is behind the players 1st guess, which is 4/8.
For each reveal, the probability s/n has to be recalculated for current conditions.
Removing a door from play does not alter the location of the car.
That looks like some form of mathmagical manipulation.
From Wikipedia:
"The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the belief that, if an event (whose occurrences are independent and identically distributed) has occurred less frequently than expected, it is more likely to happen again in the future (or vice versa)."
"The fallacy leads to the incorrect notion that previous failures will create an increased probability of success on subsequent attempts."
The 2nd case fits Savant's interpretation for the million door example, where the probability of M-1 doors accumulates onto door 777,777.
She states
"Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?"
Let's apply the Savant interpretation to the 5 door example.
She assigns .20 value for each door.
The host opens empty doors, 1 per game, right to left.
1 0 0 0 0
1 0 0 0__
1 0 0____
1 0______
There are 6 possible sequences of any empty door surviving to row 4, probability = .17, small compared to 1.00.
She concludes door 1 has retained its value of .20, but the surviving empty door has (magically) inherited a value of 4*.20=.80.
Why wouldn't the player switch?
Because her conclusion is false, per the 'fallacy' concept, and the player can only guess since they don't know which door contains the car.
The patterns/arrangements of prizes for each game is random. The car could appear behind door 2 7 times in a row. With ‘random’ meaning unpredictable, there is no method of equal placement. That’s why it’s referred to as a game of chance.
post 65
In terms of reality/possibility the 5 doors can be represented as
1 2 3 4 5
1 0 0 0 0
with 1 meaning contains a car and 0 meaning contains no car.
The car is behind door 1. Player guesses door 1.
As the host reveals each door from right to left, n changes.
1 0 0 0 0_s/n=1/5
1 0 0 0__ s/n=1/4
1 0 0____ s/n=1/3
1 0______ s/n=1/2
The car is behind door 2. Player guesses door 1.
As the host reveals each door from right to left, n changes.
0 1 0 0 0 s/n=1/5
0 1 0 0__ s/n=1/4
0 1 0____ s/n=1/3
0 1______ s/n=1/2
The host can always reveal n-2 doors.
Using the general game list for 3 doors,
switch wins if the car is not behind the players 1st guess, which is 4/8, and
switch loses if the car is behind the players 1st guess, which is 4/8.
For each reveal, the probability s/n has to be recalculated for current conditions.
Removing a door from play does not alter the location of the car.
That looks like some form of mathmagical manipulation.
From Wikipedia:
"The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is the belief that, if an event (whose occurrences are independent and identically distributed) has occurred less frequently than expected, it is more likely to happen again in the future (or vice versa)."
"The fallacy leads to the incorrect notion that previous failures will create an increased probability of success on subsequent attempts."
The 2nd case fits Savant's interpretation for the million door example, where the probability of M-1 doors accumulates onto door 777,777.
She states
"Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?"
Let's apply the Savant interpretation to the 5 door example.
She assigns .20 value for each door.
The host opens empty doors, 1 per game, right to left.
1 0 0 0 0
1 0 0 0__
1 0 0____
1 0______
There are 6 possible sequences of any empty door surviving to row 4, probability = .17, small compared to 1.00.
She concludes door 1 has retained its value of .20, but the surviving empty door has (magically) inherited a value of 4*.20=.80.
Why wouldn't the player switch?
Because her conclusion is false, per the 'fallacy' concept, and the player can only guess since they don't know which door contains the car.
The patterns/arrangements of prizes for each game is random. The car could appear behind door 2 7 times in a row. With ‘random’ meaning unpredictable, there is no method of equal placement. That’s why it’s referred to as a game of chance.
Pinball1970
Valued Senior Member
Everyone else has tried so I will have a go.
Once you have picked a door and stick with it you have 1/3 chance of winning.
You accept that? That is simple.
Once you have picked a door then switch you chances go up from 1/3 to 2/3rds.
Why? Because 1/3 goat has just been removed and all your options of switch now regardless of starting door are now 2/3.
Car - switch lose
Goat - switch win (because the other goat is gone)
Goat - switch win. (Because the other goat is gone)
Those are the only options.
This is just weird. It's like "debating" whether or not 2 plus 2 equals 4. This is not a contested matter. The problem was presented some 50-odd years ago (Savant's rendering was drawn from a previous text) and resolved some 50-odd years ago.
It's one thing to walk through the problem, present different ways of conceptualizing it, and to discuss the various ways in which people are apt to go about it the wrong way (and the reasons for such confusion). But there's only so many ways the same thing can be said over and over again.
You are simply incorrect, Phyti. You are not only focussing too much upon the (intentional) red herrings within the problem, you now seem to be manufacturing your own red herrings.
It's one thing to walk through the problem, present different ways of conceptualizing it, and to discuss the various ways in which people are apt to go about it the wrong way (and the reasons for such confusion). But there's only so many ways the same thing can be said over and over again.
You are simply incorrect, Phyti. You are not only focussing too much upon the (intentional) red herrings within the problem, you now seem to be manufacturing your own red herrings.
DaveC426913
Valued Senior Member
You have not addressed the actual data.
Pinball1970
Valued Senior Member
Did you read #186?
DaveC;
The 1/3 win applies to the basic game with only the player.
The modified game includes host participation, which reduces the players choices to 1 of 2.
Marilyn Savant is the person who claimed she could beat the odds of the game, as a result of her misconceptions of probability.
She insulted the intelligence of many professionals who use statistics on a regular basis in their work.
Einstein with his great analytical ability still made mistakes, but acknowledged them with published revisions.
The 1/3 win applies to the basic game with only the player.
The modified game includes host participation, which reduces the players choices to 1 of 2.
Marilyn Savant is the person who claimed she could beat the odds of the game, as a result of her misconceptions of probability.
She insulted the intelligence of many professionals who use statistics on a regular basis in their work.
Einstein with his great analytical ability still made mistakes, but acknowledged them with published revisions.
The game "in the same manner she proposes" is how the TV show worked.
If you program in any language, I recommend formalizing your logic in code. Or play the actual game with someone, you playing the role of Monty. It will become obvious to you.
I will study it before a reply.
DaveC426913
Valued Senior Member
There is only one Monty Hall game. There is no such thing as a modified game. This may be part of your problem.
Don't care. Savant is a straw man. This may be another part of your problem.
ADDRESS THE DATA. The data shows empirically that switching increases the odds. No giant tables posted over and over can refute that.
NONE OF ANY OF THIS ADDRESSES THE DATA.
Reported.
Any such modified game is irrelevant--under discussion here is the Monty Hall Problem as presented in the Savant column. In this game there is also host participation: the host always picks a door with a goat from the remaining 2 doors, whether both doors conceal a goat or just one door conceals a goat and the other a car. And that is the crux of the game. Given that you know that Monty will choose a goat door, it is statistically in your best interest to switch.
Pinball1970
Valued Senior Member
Just bear in mind that the first part is a given regardless of what Monty does. If you have three doors, pick one and stick with it then you have a 1/3 chance of winning.
Your intuition should tell you that your odds increase when a goat is revealed, what is not intuitive is by how much.
You only see that explicitly when you go through the options for switching.
I believe phyti knows that it goes up from the original 1/3.
Phyti has, at least as far as he/she has been able, gone through the "options". Unfortunately he/she analyses them incorrectly, ignoring that not all options occur with same frequency.
Until he/she adjusts their analysis for that, don't expect much.
DaveC;
Will do this again hopefully with sufficient clarity.
One of the first things you look for in analysis are patterns.
Your data was sorted into player chooses door1, while the car is located in each of 3 doors.
Case 1 differs from case2 and case3.
The host is switching between opening door2 and door3. How and why?
The host could toss a coin but that is prohibited.
Marilyn Savant correctly identifies the problem:.
"So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!).
She could have corrected it by allowing a 4th game.
Case1 is separated into case1a and case1b. Now all 4 games produce output in the same manner with the host opening the same door per game. The most significant change is the ratio of wins/games played.
Playing games 1, 2, and 3 approx. the same number of times yields an excess of goats.
case 1. 22 c, 00 g
case 2. 00 c, 20 g
case 3. 00 c, 18 g
total__ 22c, 38 g, twice as many g's as c's.
An apparent advantage with biased data.
Playing games 1a, 1b, 2, and 3 approx. the same number of times yields an equal number of g's and c's, with no advantage.
case 1. 44 c, 00 g
case 2. 00 c, 20 g
case 3. 00 c, 18 g
total__ 44c, 38 g
No bias.
The player's 1st guess is 1 of 3 doors.
The host removes 1 door containing a goat.
The player's 2nd guess is 1 of 2 doors, a car or a goat.
Since all games end with the player's 2nd guess, the probability can only be 1/2, same as the coin toss which has zero bias.
Will do this again hopefully with sufficient clarity.
One of the first things you look for in analysis are patterns.
Your data was sorted into player chooses door1, while the car is located in each of 3 doors.
Case 1 differs from case2 and case3.
The host is switching between opening door2 and door3. How and why?
The host could toss a coin but that is prohibited.
Marilyn Savant correctly identifies the problem:.
"So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!).
She could have corrected it by allowing a 4th game.
Case1 is separated into case1a and case1b. Now all 4 games produce output in the same manner with the host opening the same door per game. The most significant change is the ratio of wins/games played.
Playing games 1, 2, and 3 approx. the same number of times yields an excess of goats.
case 1. 22 c, 00 g
case 2. 00 c, 20 g
case 3. 00 c, 18 g
total__ 22c, 38 g, twice as many g's as c's.
An apparent advantage with biased data.
Playing games 1a, 1b, 2, and 3 approx. the same number of times yields an equal number of g's and c's, with no advantage.
case 1. 44 c, 00 g
case 2. 00 c, 20 g
case 3. 00 c, 18 g
total__ 44c, 38 g
No bias.
The player's 1st guess is 1 of 3 doors.
The host removes 1 door containing a goat.
The player's 2nd guess is 1 of 2 doors, a car or a goat.
Since all games end with the player's 2nd guess, the probability can only be 1/2, same as the coin toss which has zero bias.
DaveC426913
Valued Senior Member
And since that is not, in fact, what actually happens in real life, it proves that all your logic is wrong.
It doesn't matter how many times you rationalize it, it doesn't matter how many tables you produce. If it doesn't match reality, it's wrong.
You still have not actually addressed the data. Find any flaw in it. Find any entry in any cell that violates either probability or game rules.
But you won't. Because you know you can't.
Reported as trolling, on the following site criteria:
Trolls tend to follow certain patterns of behaviour that may include:
- Posting of similar responses and topics repeatedly.
- Avoiding giving answers to direct questions put to them.
Last edited:
Pinball1970
Valued Senior Member
Did you go through it? It's much simpler than lots of simulations.
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