>>I believe if I used the cards or ran Christof's program, both of which seem to be for a 3 door problem, I would duplicate your results. It took me quite a few messages to other people for me to realize what the difficulty is: it is really only a 2 door problem, so I believe that the cards and program do not apply here. I believe it can be solved most simply by logic:
>>
>>1. There are exactly 2 doors remaining.
>>2. Exactly one of them contains a boat, the other contains a goat.
>>3. The probability that either door contains the boat is 0.5, or 50%.
>>4. We already "own" one of the 2 doors.
>>5. Therefore, the probability we already "own" the boat is 50%, the same as for the other, remaining door.
>>
>>It's too late on a Sunday evening for me to try to fathom why the other approaches might give a different answer. Is there any flaw in my logic that would invalidate my analysis? If not, I believe we have to accept conclusion #5.
>
>Ok. So, if we are truly dealing with a 3-door problem, then we are in agreement. So the question is are we dealing with a 2 or a 3 door problem. Prior to showing a door is it a 3 door problem? If I was to tell you, I am going to show you are a goat and allow you to switch, but you must decide to switch now prior to me showing you a door, would that be a 2 door problem or a 3 door problem?
>
>Again the key question is what are my tactics going into the game. Prior to going in we know that we are wrong 66% of the time in the 3 door problem. We also know in the 3 door problem that if we switch, we would be right 66% of the time.
>
>Perhaps the problem is the way it is presented. I believe that if I go into this situation with the 3 door tactics presented regardless of what door is originally chosen or which door is shown that I will get the boat 66% of the time.

Dan, Jim, Christof, Rick, and everyone else I've argued with about this:

Colour me red. Humble pie for me for a week. I finally figured out what you guys knew all along:

Initial pick: Goat1, Goat2, or Boat. All are equally likely.

Monty ALWAYS exposes a goat.

If I chose Goat1, he exposes Goat2, making the "other" door the Boat.
If I chose Goat2, he exposes Goat1, making the "other" door the Boat.
If I chose the Boat, he exposes either Goat, leaving the "other" door the other Goat.

So, clearly, in 2 of the 3 equally likely scenarios, the "other" door is the Boat. In only 1 of the 3 scenarios is the door I've originally chosen, the Boat.

The moral to the story: it's better to switch than fight.

My two-door argument falls down in that my initial choice is clearly the Boat only 1/3 of the time, not 1/2 as you would expect in a true two-door problem.

My apologies, gentlemen, and thank you for being more sporting than I was over this.