>>
>>You have three cards. One ace. Two Jacks. You are trying to get the ace. There are nine possible scenarios based on which card you pick but I will only cover three.
>>
>>1.
>> 1 2 3
>> A J J
>>
>>2.
>> 1 2 3
>> J A J
>>
>>3.
>> 1 2 3
>> J J A
>>
>>We will automatically switch each time.
>>I will assume you take card #1 without looking at it.
>>
>>In the first example, I show you #2, you pick #3. You get a J and lose.
>>In the second example, I show you #3, you pick #2. You get an A and win.
>>In the third example, I show you #2, you pick #3. You get an A and win.
>> _
>>That's 2/3 or 66.6. Try it yourself with cards, if you don't believe it. I know it seems like it should be a 50/50 shot, but it really is 2/3.
>
>Sorry, Dan, you're wrong. See my latest reply to Walter Meester. Where you guys are all falling down is in including the result of the already known door in your analysis. The analysis is of 2 doors, not 3; one contains a boat, the other a goat. You own one of the two doors. Therefore the door you already own has a 50% chance of containing the boat. There is no advantage or disadvantage to changing your choice to the other unknown door.

You have to include the other door in the analysis. If it was really 50/50, then the above should give you a 50/50 result. The above gives you a 66.6 result. We are only concerned with the best possible results and what advice to give the player when they choose a door. If you're right then you would tell the player it doesn't make any difference whether you change or not. I say the above example proves that the results will give you a 66.6 result. Take the above example and prove to me that it returns a 50/50 result for the three possibilities.