Monty Hall Problem... For Newbies

[ynot]

CORRECTION – Please ignore my last few posts as they are wrong. If you know Monty will reveal the middle door then you know that’s where a goat is and you have a 50/50 chance of correctly guessing the car from the other two doors. The 7/10 result I got was merely a “lucky streak” from a small sample (sorry about that).

But knowing where one of the goats is before you choose means the odds of the first choice are 50/50 not 1/3. So in this scenario the odds of the first choice don’t start being 1/3 and then change to being 50/50 as some seem to be claiming. In the real Monty Game the odds for the first guess are 1/3 and remain so throughout.

The only way the odds of the first choice could change would be if Monty revealed what door the car was behind.

 

[HumptyDumpty]

CORRECTION – Please ignore my last few posts as they are wrong. If you know Monty will choose the middle door then you know that’s where a goat is and you have a 50/50 chance of correctly guessing the car from the other two doors. The 7/10 result I got was merely a “lucky streak” (sorry about that).

But knowing where one of the goats is before you choose means the odds of the first choice are 50/50 not 1/3. So in this scenario the odds don’t start being 1/3 and change to being 50/50. In the real Monty Game the odds for the first guess are 1/3 and remain so throughout.

But you don't know where any of the goats are before you choose. Your 1st choice is 1/3, after Monty opens (his preferred) Door2 your 1st choice increases to 1/2

 

[ynot]

Your relevant knowledge can change, as explained above, but only if you are aware that Monty is not choosing randomly and know his preference.

If you know his preference before the first guess then you know where a goat is and the odds are 50/50 not 1/3. If you find out his preference after the first draw you still don’t know any better if the other goat is behind the door you have chosen or the remaining door and the odds of your first choice remain 1/3. If you find out where a goat is (Monty's preference) after the first guess but Monty reveals a goat behind another door for some strange reason then you would know 100% where the car is.

 

[lionking]

But you don't know where any of the goats are before you choose. Your 1st choice is 1/3, after Monty opens (his preferred) Door2 your 1st choice increases to 1/2

What's this got to do with switching?

 

[ynot]

But you don't know where any of the goats are before you choose. Your 1st choice is 1/3, after Monty opens (his preferred) Door2 your 1st choice increases to 1/2

Why and how? Please explain the mechanism by which you arrive at this conclusion.

By the rules of the game Monty has to open a door after the first guess that has a goat behind it. I don’t see how or why it matters whether the door he opens is a preferred choice or not. Please educate me with a practically detailed explanation of how a preference changes anything.

Obviously Monty knows where the goats and car are before the first guess is made. He knows if the first guess is a goat or not and if it is he doesn’t have any choice in which remaining door to open as only one hides a goat. If the first guess is the car I don’t see that it matters a toss which other door he opens or that he prefers to open one rather then the other.

ETA - Do you agree that if the remaining doors both hide goats, and Monty randomly chooses one to reveal a goat, that the odds of the first guess remain at 1/3?

 

[ynot]

What's this got to do with switching?

Because if the odds change from 1/3 to 1/2 as is being claimed, then switching looses it's advantage.

 

[lionking]

Because if the odds change from 1/3 to 1/2 as is being claimed, then switching looses it's advantage.

But the odds don't change to 1/2, as has been shown time and time again.

 

[GlennB]

What HumptyDumpty is saying is a close parallel to the way this puzzle affects the play in bridge. But it would require study of Monty's actions - suppose that when he has both goats he'll always open the rightmost door of two. Then, whenever he opens the leftmost door you can be certain it's because the car is behind the other one.

In bridge, when holding the bare QJ of trumps (say) a defender should follow at random. Many players automatically play the J on the first round of that suit, but this means whenever they play the Q first it's because it was singleton. Declarer can then finesse against the J remaining with the other defender (that's a somewhat cut-down explanation of the "principle of restricted choice" as it applies to bridge).

 

[jt512]

uppose that when [Monty] has both goats [ie, the player has chosen the door with the car] he'll always open the rightmost door of two. Then, whenever he opens the leftmost door you can be certain it's because the car is behind the other one.

Correct.

In bridge, when holding the bare QJ of trumps (say) a defender should follow at random. Many players automatically play the J on the first round of that suit, but this means whenever they play the Q first it's because it was singleton. Declarer can then finesse against the J remaining with the other defender (that's a somewhat cut-down explanation of the "principle of restricted choice" as it applies to bridge).

What do you suppose the probability is that a randomly selected reader will understand that paragraph; and given that probability, why would you possibly post it?

 

[GlennB]

What do you suppose the probability is that a randomly selected reader will understand that paragraph; and given that probability, why would you possibly post it?

Because it might be of interest to card players and has a bearing on MH.

 

[jt512]

Because it might be of interest to card players...

Which would make it a great post at a bridge forum.

...and has a bearing on MH.

You're using a point about MH to make a point about bridge, not the other way around.

 

[Brian-M]

(of course, your post is utterly wrong, and then you skip over the problem of explaining the problem in parentheses. Since you never know when he has one goat or two to choose from, we're back where we started.)

(It's odd that you criticize my post as being utterly wrong but completely fail to give any reason why. The stuff I added in parentheses was just to reinforce the fact that I was only talking about a situation that arises 2/3 of the the time, and is non-applicable 1/3 of the time.)

 

[GlennB]

Which would make it a great post at a bridge forum.

Sorry it didn't meet with your approval, but you'll find that illustrations and analogies like this occur quite a lot in these discussions.

You're using a point about MH to make a point about bridge, not the other way around.

No, I'm not. Both subjects involve 'restricted choice'

 

[jt512]

Sorry it didn't meet with your approval, but you'll find that illustrations and analogies like this occur quite a lot in these discussions.

Generally, the purpose of an illustration or analogy is to illucidate an obscure concept by reference to a more common concept. Instead, you are purporting to illucidate an obscure concept by reference to an even more obscure concept.

I know you'll fight this to the death, so just go ahead and have your last say.

 

[HumptyDumpty]

What's this got to do with switching?

It has to do with probabilities and the assumptions we (consciously or unconsciously) make in calculating those probabilities.

By the rules of the game Monty has to open a door after the first guess that has a goat behind it. I don’t see how or why it matters whether the door he opens is a preferred choice or not.

In the example given (you pick Door1, Monty reveals a goat behind Door2) many people assume the reason the probability the prize is behind Door1 is 1/3 is because that was the initial or 'a priori' probability, and Monty opening a goat door doesn't or cannot change that.
However that reasoning doesn't explain why, if Monty is ignorant of where the prize is and opens Door2 at random, the probability of Door1 hiding the car increases to 1/2, nor in the scenario where Monty has a preference for opening Door2 the probability of Door1 similarly increases from 1/3 to 1/2.

If you approach the problem from the viewpoint of "why did Monty open Door2?", then the answer to all variations becomes clearer.

In all varaitions the car is either behind Door1 or Door3, and the prior probabilities for both these doors was the same - 1/3.

(A) Classic Monty.
If the prize is behind Door3 there's a 100% chance Monty opens Door2.
(constraint of the game)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 and a 50% chance Monty opens Door3 (applying the Principle of Indifference).
Therefore it is twice as likely Monty opened Door2 because he had to than because he chose to , consequently the prize is twice as likely to behind Door3 as Door1.

(B) Ignorant Monty
If the prize is behind Door3 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
Both scenarios are equally likely, therefore it's 50/50

(C) Preference Door Monty
If the prize is behind Door3 there's a 100% chance Monty opens Door2 (constraint of the game)
If the prize is behind Door1 there's a 100% chance Monty opens Door2 (because it's his preferred door)
Both scenarios are equally likely, therefore it's 50/50

So, those saying that the initial probabilities don't change are assuming (whether they realise it or not) that given a choice of 2 goat doors to open, Monty picks one at random.

The mathematical solution also bears this analysis out. Bayes Theory simplifies to:
Prob(Prize behind Door1) = p/(1+p) and
Prob(Prize behind Door3) = 1(1+p)
where p=probability Monty opens Door2 give the prize is behind Door1.

 

[Didactylos]

Because it might be of interest to card players and has a bearing on MH.

Count me up for one of the interested, I did play a lot of some bridge about 30 years ago. I found this interesting article, I'm not sure I get it, and what is RHO and "drop"? I think I know "finesse". Maybe you'll elaborate for a rusty danish bridge player?

http://www.acbl-district13.org/artic003.htm

 

[ynot]

suppose that when he has both goats he'll always open the rightmost door of two. Then, whenever he opens the leftmost door you can be certain it's because the car is behind the other one.

Thanks – That makes the scenario much clearer.

If Monty always opens the rightmost door when it hides a goat then when he doesn’t open that door you know with 100% certainty that it hides the car. The odds of the first guess being correct would therefore change to 0/0 not 1/2 and the remaining door would have odds of 1/1.

Not sure it’s a credible scenario in practice however and I think the “spirit” of the Monty game is to analyse the odds without them being effected by a Monty bias.

 

[wollery]

The advantage of switching in the Monty Game can be quickly and easily demonstrated in practice using two black and one red playing cards. As Monty you place the cards randomly facedown and knowing their positions. You ask a “contestant” to guess which card is red by pointing to it. You then turn over one of the other cards you know to be black and ask the contestant if they want to stay with their first choice or switch to the other. Rinse and repeat and keep track of the contestant’s wins and losses. The advantage of always switching will become self-evident by the empirical results and convoluted intellectual arguments will become moot (Yep – I’m an optimist).

You mean like this?

http://www.internationalskeptics.com/forums/showpost.php?p=9816951&postcount=278

Funny how few people actually responded to that post.

 

[wollery]

It has to do with probabilities and the assumptions we (consciously or unconsciously) make in calculating those probabilities.



In the example given (you pick Door1, Monty reveals a goat behind Door2) many people assume the reason the probability the prize is behind Door1 is 1/3 is because that was the initial or 'a priori' probability, and Monty opening a goat door doesn't or cannot change that.
However that reasoning doesn't explain why, if Monty is ignorant of where the prize is and opens Door2 at random, the probability of Door1 hiding the car increases to 1/2, nor in the scenario where Monty has a preference for opening Door2 the probability of Door1 similarly increases from 1/3 to 1/2.

If you approach the problem from the viewpoint of "why did Monty open Door2?", then the answer to all variations becomes clearer.

In all varaitions the car is either behind Door1 or Door3, and the prior probabilities for both these doors was the same - 1/3.

(A) Classic Monty.
If the prize is behind Door3 there's a 100% chance Monty opens Door2.
(constraint of the game)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 and a 50% chance Monty opens Door3 (applying the Principle of Indifference).
Therefore it is twice as likely Monty opened Door2 because he had to than because he chose to , consequently the prize is twice as likely to behind Door3 as Door1.

(B) Ignorant Monty
If the prize is behind Door3 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
Both scenarios are equally likely, therefore it's 50/50

(C) Preference Door Monty
If the prize is behind Door3 there's a 100% chance Monty opens Door2 (constraint of the game)
If the prize is behind Door1 there's a 100% chance Monty opens Door2 (because it's his preferred door)
Both scenarios are equally likely, therefore it's 50/50

So, those saying that the initial probabilities don't change are assuming (whether they realise it or not) that given a choice of 2 goat doors to open, Monty picks one at random.

The mathematical solution also bears this analysis out. Bayes Theory simplifies to:
Prob(Prize behind Door1) = p/(1+p) and
Prob(Prize behind Door3) = 1(1+p)
where p=probability Monty opens Door2 give the prize is behind Door1.

Your scenario C is irrelevant, because in that case the positioning of the car is not random. The car can be behind any door, otherwise the original probability is not 1/3.

If you want to change the rules then that's fine, but don't pretend that the result you get applies to the original problem.

 

[This is The End]

No. Monty is omniscient and so will always eliminate 98 irrelevant doors. From the very outset, I know that no matter what, the car is either behind my first choice, or behind the door Monty will not eliminate. You start the game with a 50/50 chance of having chosen correctly among the only two doors that Monty will not open, and you finish with exactly the same 50/50 chance. That, of a whole population of contestants, 99/100 of them would win by switching just doesn't matter because you can only play once. You've still got a 50/50 chance of winning your one trial. I understand that this seems counter-intuitive.

You literally could not be more wrong.

It seems like you are perpetuating the marbles fallacy explained earlier.