in reply to The Monty Hall Trap Simulator

Assuming monty *always* opens a wrong door

You have a critical flaw in your logic. You assume that your original choice has anything to do with the chance of you winnning.. And it doesn't. Your first choice is irrelevant. When monty asks you to switch you always have the option of picking from one of 2 doors. One that will win, and one what will loose.

Agreed, when you remove 1 door, you still have a .333 chance that your ORIGINAL decision is correct. HOWEVER, once one door is removed, you are left with a new choice between 2 doors. One being the car, and one being the goat. You now have a .5 chance of walking away with a goat, but a .333 chance that your original guess is correct.

The probablilty of you being right on your first choice is ALWAYS .333, HOWERVER the probability of you winning is ALWAYS .5

"If you're ever on the show this means you should change your door!"

Sure, go ahead... You still have the same shot at winning.

Rick

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Re: Re: The Monty Hall Trap Simulator
by legLess (Hermit) on Jul 17, 2002 at 22:26 UTC
    Not to be a pedant, but you're wrong. The problem is not quite as you stated it. The problem is, "Given three doors, and the following cirumstances, what's your chance to win if you switch?" Your restatement is, "Given two doors, what's your chance to win?" Your answer is correct for your problem, but your problem is different from what we're discussing here.
    --
    man with no legs, inc.
[reply]
      Yes.. The original poster was absolutely correct. As are you.

      Here is the logic I used to realize my mistake (that and 3 pieces of paper ;) )

      If you pick a loosing door, and monty removes a loosing door, 100% of the time you will win if you switch.

      Since you have a .66 chance of picking a loosing door, you have a .66 chance of winning when switching.

      Rick
[reply]

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