[ShaneP]

Well, I might not be understanding something totally, but I'll put things in a different way to show why the 50/50 versus 90/10. I'm not sure why your friend is fixated on a specific card, so I might be missing something. Actually it IS 90% that ANY card is in the big pile...which would include the ace or any other card. Then it's 10% that any particular card is the one you picked initially, and 10% chance that it was in the 9 pile and survived the random cuts. (if your friend knew it, I'd say then you just Bayesian update...but that probably doesn't help)

I'll assume you're trying to pick the ace.

The first case can be seen as: pick one card. OK, pick another card (that's the one of the pile of nine that wasn't chosen). Now, here's all the cards you didn't pick. Gee, there's an Ace and a 5 (or a 6 and a 3, or whatever) that haven't been seen. You've got no new information about which card is which. It's just you've picked the two cards in a slightly different way than above, but which results in exactly the same distribution. So if you know the Ace and 5 are still out there, you don't have any way of differentiating between the cards.

The Monte Hall problem is : pick one card. OK, now I'll look through the cards you didn't pick, and show you eight of the nine. Now, do you want to switch. Since we can't reveal the ace, given there was a 90% chance it was in the pile of nine, and it can't be revealed, you now have a 90% chance that the ace is the card that was left after the other cards were revealed to be not aces.

One other thing to remember, is that yes, the ace has a 90% chance of being in the pile of 9, but if it is, it has a 89% chance of being revealed, so that really cuts down the probability of the card being in the pile of nine and being the card left unrevealed at the end.

I think the difference can be summed up as follows: you don't have any information (since they are essentially chosen in the same way) to differentiate between the two cards in the first case, but in the Monte Hall version, the cards are chosen quite differently...one was chosen by the participant, and one was chosen by Monte by taking away bad cards.

A final comment: in the DonD version with 10 cards, there's only a 20% that the ace isn't revealed until there are only two cards left...it probably doesn't affect the understanding of the problem, but in the monte hall problem there is a 100% chance the ace isn't revealed until the final two. That should be another indication that the final two cards are chosen differently.

Hope this helps...as I said, I dont' think I'm quite parsing the initial question totally, but I think this is on the right path...

Shane