>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.