The Monty Hall paradox...
Nov. 8th, 2007 10:47 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
alexpgpWhile cleaning out stuff, I ran across the source code of a BASIC program I wrote in... 1991 (!) that tests the "Monty Hall paradox."
Who is Monty Hall, you ask? And what's the paradox?
Monty Hall is (was?) a well-known game show host, and though I don't recall his ever doing anything along the lines of the setup to the problem, the name has stuck. The setup is this:
The other view is this: There was a 1/3 chance of winning before the worthless prize was revealed. This means there was a 2/3 chance the prize was behind one of the other two curtains. Now that one of the curtains has been shown to be worthless, the chance of winning is now "concentrated" in the other curtain, so the contestant should switch.
Although it turns out that the "should switch" argument is correct, there are nevertheless die-hard partisans, including (supposedly) mathematicians, who insist it doesn't matter what the contestant does.
The program I wrote simulated the scenario a thousand times and printed the result (which shows that switching is the preferred strategy). However, simulations aren't universally compelling. Since then, I've formulated a line of reasoning that I think does the trick. It runs as follows:
Consider an immense game show stage where there are a million curtains, and only one prize. After you pick a curtain at random, the host causes 999,998 of the remaining curtains to be opened to reveal... nothing. The prize is now either behind the curtain you picked, or the one other curtain left.
Should you stick or should you switch? Has your initial one-in-a-million shot turned into an even-money proposition because there are now only two curtains to choose from, or is it virtually a certainty that the prize is behind the one remaining curtain of the ones you didn't pick?
Me, I'd switch.
Cheers...
Who is Monty Hall, you ask? And what's the paradox?
Monty Hall is (was?) a well-known game show host, and though I don't recall his ever doing anything along the lines of the setup to the problem, the name has stuck. The setup is this:
A game show host offers a contestant the choice of whatever is behind one of three curtains. One of the curtains hides an expensive prize; the other two curtains hide nothing. Once the contestant picks a curtain, the host orders one of the other two curtains drawn aside, revealing nothing. The host then offers the contestant an opportunity to change his or her mind.The key question now is:
Should the contestant switch or stay with the original choice?One way to look at the situation is this: The contestant had a 1/3 chance of picking the prize before, and now, it's 1/2, since there are now only two choices and one prize, so it doesn't matter if the contestant switches or stays.
The other view is this: There was a 1/3 chance of winning before the worthless prize was revealed. This means there was a 2/3 chance the prize was behind one of the other two curtains. Now that one of the curtains has been shown to be worthless, the chance of winning is now "concentrated" in the other curtain, so the contestant should switch.
Although it turns out that the "should switch" argument is correct, there are nevertheless die-hard partisans, including (supposedly) mathematicians, who insist it doesn't matter what the contestant does.
The program I wrote simulated the scenario a thousand times and printed the result (which shows that switching is the preferred strategy). However, simulations aren't universally compelling. Since then, I've formulated a line of reasoning that I think does the trick. It runs as follows:
Consider an immense game show stage where there are a million curtains, and only one prize. After you pick a curtain at random, the host causes 999,998 of the remaining curtains to be opened to reveal... nothing. The prize is now either behind the curtain you picked, or the one other curtain left.
Should you stick or should you switch? Has your initial one-in-a-million shot turned into an even-money proposition because there are now only two curtains to choose from, or is it virtually a certainty that the prize is behind the one remaining curtain of the ones you didn't pick?
Me, I'd switch.
Cheers...
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
no subject
Date: 2007-11-08 06:50 pm (UTC)I didn't mean to...and he laughed about it when I called back.
no subject
Date: 2007-11-08 06:59 pm (UTC)Hummm ...
Date: 2007-11-08 11:23 pm (UTC)"... the chance of winning is now "concentrated" in the other curtain ..."
If you flip a coin, the chances of heads/tails is always 50-50, the prior landings don't affect the next flip.
So if you have two curtains, i don't see how it is not a 50-50 chance regardless of what went before which doesn't affect the next choice-except in one's mind.
Re: Hummm ...
Date: 2007-11-09 07:15 am (UTC)Re: Hummm ...
Date: 2007-11-09 02:12 pm (UTC)But let me see if I can make this clearer. Let me change the rules of the game (in such a way as to not affect the outcome). First, I'll have the contestant, whom we'll call Alice, pick one of the three curtains.
1. Wouldn't you agree that having done so, there is a 1/3 probability that Alice has picked the prize?
2. Wouldn't you agree that, if the host makes no offers of any kind, that her probability of winning will remain 1/3?
Now, without opening any of the other curtains, let's have the host offer to let Alice change her mind and take what is behind both other curtains.
3. Wouldn't you agree that if Alice does not change her mind, her probability of winning will remain 1/3?
4. Wouldn't you agree that if Alice does change her mind, and agrees to accept what is behind the two remaining curtains, the probability of her winning the prize is now 2/3?
Being a sharp cookie, Alice knows that if she has won, there is nothing behind one of the curtains. (Of course, if she has lost, there will be nothing behind either curtain, but that doesn't affect the general argument.)
5. Does knowing that there is nothing behind (at least) one of the two other curtains change her probability of winning if she changes her mind (that is, change it from 2/3)?
Now let's assume that, having changed her mind, the host now reveals that one of the two curtains she agreed to pick has nothing behind it.
6. Does seeing that there is nothing behind one of the two other curtains change her probability of winning after having changed her mind (that is, change it from 2/3)?
Did you try the mind experiment with a million curtains? I still think it's quite convincing.
Cheers...
Re: Hummm ...
Date: 2007-11-09 09:05 pm (UTC)When i tried the mind experiment with a million curtains i was still left with just two at the end and still, it was a 50-50 chance which curtain had something behind it ... the previous 999,998 empty curtains does not influence me in any way about the two left.
Leastwise that's how i think about it ... while i've never watched the aforementioned TV show, i'd guess if one of the three curtains reveal the not-wanted-prize, getting the contestant to consider changing their choice has more to do with hype then actual winning or losing and the audience gets to join in.
Re: Hummm ...
Date: 2007-11-10 12:40 am (UTC)The thing that made the million-curtain variation of the problem so compelling for me is knowing that, in randomly picking one curtain out of a million, Alice is almost certain to lose. (Her probability of winning is 10.)
Now we know that, of the remaining 999,999 curtains, 999,998 are certain losers. If the host shows us which curtains those are, that means the one remaining, closed curtain is almost certainly (99.9999%) a winner, and Alice should switch.
Cheers...