Re: Why is Deal or No Deal not the Monte Carlo problem?
here's one stumbling block in the attempt to clear this up:
In the Monte Carlo version, it's 90% chance that one particular card that you choose to "shoot for" is in the bigger pile, rather than the smaller pile.
There is a 90% chance that the Ace is in the big pile.... and, independently, it's a 90% chance that the 5 is in the big pile.
When you get down to two cards, how is it that you can determine that it is the Ace, not the 5, that is 90% likely to be the card you haven't chosen?
I think this argument changes the conditions of the Monte Carlo problem, but I can't explain it properly
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