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Monty Hall problem

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Monty Hall problem

Postby john_titor » Tue Oct 23, 2012 6:12 pm

Can someone explain why the odds of picking the car are not 50/50 with two doors left? I tried to read the wiki entry but got bored, stupid math.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

"Ask Marilyn's" response was that the contestant should switch to the other door. If the car is initially equally likely to be behind each door, a player who picks door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks door 1 and does switch has a 2 in 3 chance, because the host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.

http://en.wikipedia.org/wiki/Monty_Hall_problem
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Re: Monty Hall problem

Postby ArturoBandini » Tue Oct 23, 2012 7:58 pm

I love this problem. The 50/50 solution is seductively intuitive. I find this explanation from the wikipedia page (under "Other simple solutions") to be the most satisfying:
An intuitive explanation is to reason that a player whose strategy is to switch loses if and only if they initially pick the car; that happens with probability 1/3, so switching must win with probability 2/3
I think that a big part of the confusion is that the player actually makes two choices, but only thinks about the second choice when considering the overall probability of winning the car. If you think about the first choice only, you either picked the car (33%) or you didn't (67%). If you bring along this consideration in your second choice, it's clear to see that you should choose to switch doors, because you probably didn't guess correctly first, and if that's so, you will certainly win the car when you switch.
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Re: Monty Hall problem

Postby Igor » Tue Oct 23, 2012 8:04 pm

Yeah, the problem lies mostly in explaining the answer. It kind of comes down to "If you picked goats, they are going to show you where the other goats are too."
Last edited by Igor on Tue Oct 23, 2012 8:26 pm, edited 1 time in total.
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Re: Monty Hall problem

Postby Kenneth Burns » Tue Oct 23, 2012 8:10 pm

The way things are going, I think I'd prefer the goat.
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Re: Monty Hall problem

Postby kurt_w » Tue Oct 23, 2012 8:13 pm

I agree with Arturo that the Wikipedia excerpt he quotes makes it very clear what's going on there.

Your first pick has only a one in three chance of being right. So switching (once you know which of the two to switch to) moves you over to the two-thirds side of those odds.
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Re: Monty Hall problem

Postby Sandi » Tue Oct 23, 2012 8:38 pm

Ah the The Monty Hall Paradox.

We had this discussion way back in 2005 on one of my daily blogs. We found the problem has it's own web page.

It seems the best choice is to switch and win 2/3 of the time. Switching will give you a win 1/3 of the time.

EDIT: Oops edit to correct, had it bass ackwards. Reading too fast can be clumsy.
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Re: Monty Hall problem

Postby pjbogart » Tue Oct 23, 2012 10:30 pm

Sandi wrote:Ah the The Monty Hall Paradox.

We had this discussion way back in 2005 on one of my daily blogs. We found the problem has it's own web page.

It seems the best choice is to switch and win 2/3 of the time. Switching will give you a win 1/3 of the time.

EDIT: Oops edit to correct, had it bass ackwards. Reading too fast can be clumsy.


Thanks for that correction. Oy.
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Re: Monty Hall problem

Postby john_titor » Wed Oct 24, 2012 5:05 pm

If I flip a coin 5,386 times and all of them have come up heads, the odds of it coming up tails on the next toss is still 50/50, right? How does eliminating all choices but two, result in anything different than 50% chance? Maybe I'll just have to take a Wiggum on this as my wookie is bent.
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Re: Monty Hall problem

Postby massimo » Wed Oct 24, 2012 5:20 pm

Your simple odds of choosing the winner are 1/3 if there are three doors. And 2/3 that you will choose a loser. But with the switching strategy, picking a loser at first guarantees a win.

If door C is the winner,

Pick A, B is revealed, switch to C. WIN.
Pick B, A is revealed, switch to C. WIN.
Pick C, A or B is revealed, switch. LOSE.

Your odds are 2/3 with the switching strategy.

After all these years it still blows my mind, but it's true.
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Re: Monty Hall problem

Postby pjbogart » Wed Oct 24, 2012 6:06 pm

massimo wrote:If door C is the winner,

Pick A, B is revealed, switch to C. WIN.
Pick B, A is revealed, switch to C. WIN.
Pick C, A or B is revealed, switch. LOSE.


Pick A, B is revealed, switch to C. WIN.
Pick B, A is revealed, switch to C. WIN.
Pick C, B is revealed, switch. LOSE.
Pick C, A is revealed, switch. LOSE.

It's still 50/50
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Re: Monty Hall problem

Postby green union terrace chair » Wed Oct 24, 2012 6:12 pm

The problem is that the question is wrong.
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Re: Monty Hall problem

Postby ArturoBandini » Wed Oct 24, 2012 6:19 pm

pjbogart wrote:Pick A, B is revealed, switch to C. WIN.
Pick B, A is revealed, switch to C. WIN.
Pick C, B is revealed, switch. LOSE.
Pick C, A is revealed, switch. LOSE.

It's still 50/50
Interesting take. I have little formal education in probability, but here is the problem I see with this model.

For the first two lines (Pick A, Pick B), the exact outcome of the game is determined 100% after the initial choice, assuming the player has committed to switching. This means that it is 100% certain which door will be opened, and which door will be switched to after the initial choice has been made. In the latter two cases, there is a 50% probability of the game resolving to either of the two possible cases (assuming the host chooses from the two doors randomly), so that makes me think that these cases should be weighted less than the fully determinate starting choices.

It might also help if the doors are not labeled as A B C, but rather as Goat, Goat, Car. In that case, the two doors containing Goat are degenerate and indistinguishable, which again boils down the problem into only three possible paths, each fully determined by the initial door choice.
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Re: Monty Hall problem

Postby john_titor » Wed Oct 24, 2012 6:45 pm

There is an table on the wiki page that does just that. But...
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Re: Monty Hall problem

Postby snoqueen » Wed Oct 24, 2012 6:46 pm

My before-peeking solution was different. I thought the guy who picked A and was shown the goats in C was right the first time: A has the car.

Reason: If the host wants to keep the expenses down for his game show, he wants to give the contestant the goats. If the contestant picked a goat the first time, the host would just open the door and give him the goat and the show would move on.

If the contestant picked the car the first time, it's in the host's interest to make the contestant change his pick. So the contestant, if he's smart, stands pat.

It's more of a psychology problem than a logic problem, but that's real life, right?
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Re: Monty Hall problem

Postby kurt_w » Wed Oct 24, 2012 6:55 pm

What you say in the two paragraphs following the word "Reason:" is a bit hard to follow.

But, no, it's entirely a logic problem, not a psychology problem -- well, except that one aspect of human psychology is that people frequently behave illogically.

If every contestant understood the logic of it, they would always switch, it would become boring, and the game-show host would lose a lot of cars.
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