The monty Hall problem

[Kent]

I just read about this problem. i think it is ridiculous because so many people got it wrong.


The problem is this: You are in a game show. In front of you are three doors( A grand prize is behind one of the doors). We can label each door with "a", "b", and "c". Let us suppose you picked door a, and the host of the show open one of the other two doors, and reveals that there is no prize behind it. The host asked weather you want to switch to the other closed door or stay with your original choice? what should you do to maximize your chances of winning ?


show solution

hint: I solved it by lising each possibilities...............

[Bo]

I just read about this problem. i think it is ridiculous because so many people got it wrong.

i think it is a briljant problem, because so many people get it wrong

[pgrmdave]

It simply becomes a 50/50 chance, and choosing either A or the other door doesn't matter - sticking with A is a choice, just as much as switching.

[Tormod]

Before you started, A had a 1/3 chance of having the prize. Now we know that one of the other doors has a 50% chance of hiding the prize. Your odds are better if you switch.

[C1ay]

Originally Posted by Kent

i think it is ridiculous because so many people got it wrong.

what should you do to maximize your chances of winning ?

That's what makes it a great problem, not a ridiculous one. It demostrates how intuition can fail you.

You should change doors to maximize your chances of winning. The odds do not change as intuition would have you believe. I'll leave the explanation of how the odds change until later so as not to spoil the puzzle for others.

[pgrmdave]

I disagree - I think that, you now have to make a completely new choice - either your original choice, or the new choice. Either one has a chance of hiding the prize. Sticking with your original is still making a choice.

[TeleMad]

Originally Posted by pgrmdave

I disagree - I think that, you now have to make a completely new choice - either your original choice, or the new choice. Either one has a chance of hiding the prize. Sticking with your original is still making a choice.

No, they're right...you should switch.

I too didn't feel comfortable with the "solution" so I modeled it programmatically, running through the scenario one million times using the pseudo-random number generator of the language. It showed you'd win 66.7% of the time if you switch.

Then it was easier for me to accept the following.

When you started, A had 1/3 chance of being correct and the other two doors combined had 2/3 chance of being correct. When the host reveals that one of those other two doors does not have the prize, that pair's total 2/3 chance collapses onto the only remaining door of that pair: finding someting out about one of those two doors doesn't affect what you know about A. Thus, at this point, there is a 1/3 chance of A being correct and a 2/3 chance of the other door being correct. You should switch from A to the other one.

[Qfwfq]

Originally Posted by TeleMad

Thus, at this point, there is a 1/3 chance of A being correct and a 2/3 chance of the other door being correct. You should switch from A to the other one.

This is the usual solution and it is correct, providing one thing:

Originally Posted by TeleMad

When the host reveals that one of those other two doors does not have the prize, that pair's total 2/3 chance collapses onto the only remaining door of that pair:

Although not so obviously, this implicitly supposes that the host knew where the prize is, before opening one of the doors. If, otoh, you were sure the door had been opened casually, there would be no reason to switch. If you're not sure whether or not the host knew then the estimate of the probability is intermediate.

finding someting out about one of those two doors doesn't affect what you know about A.

???