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Subject: Re: New and final solution of the Monty Hall Dilemma

Author: Uri Blass

Date: 07:18:19 09/27/02

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On September 27, 2002 at 10:04:42, Gerrit Reubold wrote:

>On September 27, 2002 at 09:45:47, Uri Blass wrote:
>
>>On September 27, 2002 at 08:50:27, Gerrit Reubold wrote:
>>>
>>>I don't care what that probability was _before_ he opened the door. The door
>>>with the goat is _now_ open.
>>
>>Suppose for the discussion that the candidate also guess door 1
>>and the host always open door 3 in case that door 1 has  a goat.
>
>Do you mean "... in case that door 3 has a goat" ?

You are right.
I should assume he always choose door 3.

I did a mistake here and I thought for a minute that if door 1 has a goat the
game is over but the rest of my claims hold.

>
>(I assume that the host _must_ open a door, he is not allowed to keep both doors
>closed.)
>
>I discuss this situation:
>- The candidate chooses door 1
>- The host chooses (say) door 3, and is lucky, there is a goat in door 3
>
>So there _is_ a game!
>
>Now the candidate should switch and double its winning chances.
>
>Do you agree?

No
>
>If you don't agree: Consider the game with 1.000.000 doors, the candidate
>chooses door 1. The host opens 999.998 doors, without knowing where the car is.
>By incredible luck all those doors have goats behind them. There is now door 1
>and door 432.102 closed. So again there is a game! Do you agree that the
>candidate should switch?

No

in 100000 cases you are going to have 999998 cases when there
is no game
1 case when there is a game when it is a good idea to switch and
1 case when there is a game when it is a bad idea to switch

It means that the probability to be right in switching
is the same as the probability to be wrong in switching.
>
>
>BTW, this seems all obvious to me, maybe we are discussing different situations?
>
>Greetings,
>Gerrit

I admit that I did a mistake in the start of my post.
The words
"in case that door 1 has  a goat" that were wrong.

Uri



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