Walter,

Al (and others) are right, it makes no difference if you change your guess. Deciding NOT to change doors is also a choice, therefore your chance is always 1/x where x is the number of remaining doors. When Monty Hall opens one of the wrong doors, 'x' is reduced by one. When he gives you a chance to choose again, your chances are 1/x whether you change your guess or not. Changing does not improve your chances.

Note to Dan Cummings: the name of the show was Let's Make a Deal.

Rick

>Al,
>
>>I'm amazed that both you and Jim Booth have it all wrong! :-) You've both fallen into the trap of making something more complicated than it is. Consider these incontrovertible facts:
>
>Actually they are right.
>
>>1. There are 2 remaining doors.
>>2. Exactly one of them has a goat, the other a boat.
>
>Imagine there are 1000 doors, 999 goats one boat. The host let you choose one. The propability you choose a goat is .999 . The trick sits in the fact that host HAS (in .999 of the cases) to open to 998 remaining goat doors (IOW in 99,8% he's got no choice of choosing which door to open) So now you faced with 2 doors. One you choose, and one remained. The chances you were right is .001 the chances that the boat is after the remaining door is .999
>
>Note that using 3 or 1000 doors make no difference. In the cases of 3 doors the host has a propability of 2/3 that he must open the remaining goat (as you initially choose the other) leaving a 2/3 propability that the boat is after the other remaining door.
>
>Walter,