Why is Deal or No Deal not the Monte Carlo problem?

[ShaneP]

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What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

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If you're exposing losers only -- you're adding information.

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I know, but saying that isn't going to convince them in the discussion

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Here's one approach (how to solve it mathematically)...chance you pick the A to begin with: 1/10.

Chance it is in the remaining pile: 9/10. But we have to pick 8 of the 9 AT RANDOM that aren't the ace to get to the final situation, so the chances of that are (9/10) * (1/9) = 1/10. So there's a 1/10 chance that the ace was in the pile of 9 and we didn't pick it. So the probability either of the two remaining cards is the ace is 50/50, since each occurs 10% of the time.

BTW, the 1/9 can be seen easily if you note that you're flipping all but one, so you could actually just pick the one you aren't going to flip instead of trying to sum the probabilities for each individual flip.

The difference between that is the cards in this case are picked at random, without knowledge. In the Monte Hall problem, if we always have to show 8 of the 9 cards (and none can be the ace) then the last 1/9 probability becomes a probability of 1 (since we can always flip over 8 cards without an ace turning over if we know which one is the ace). Thus if the 'host' flips over 8 of the 9 cards and tells us he isn't flipping an ace, then the probability the unflipped card is an ace is (9/10) * 1 = 9/10, and we should switch.

Shane