Re: Rephrasing the question
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Each street is equally likely. You assume one and only one of the 2's will be on the board of 5 cards. This problem is equivalent to having 5 cards in your hand 1 of which is a 2. Shuffle them out and turn them over face up. The 2 has an equal probability of being in each of the 5 streets.
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Yes, this looks quite obvious and simple. However, It clearly contradicts your first answer, in which you showed (and rigtly so) that the probability of the 2 falling on the first card is greater than on the second, and on the second it's greater than on the third , and so on (this is if we don't KNOW the 2 already hit and player B outrew player A...).
How do you settle this?
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