Re: Rephrasing the question
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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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Of course, the two questions you've posed are different. A question more related to the first situation goes like this:
Assume at least one 2 is on the board. What is the probability that a 2 exists at slot i, but not any j < i, 0 < j <= 5.
Notice here that the probabilites here will decrease from 1 to 5 since any time a 2 comes at the first card, it will always be the first 2 on the board. However, if a 2 comes as the 2nd card, there is a slight chance that the other 2 already came at the first card. So the probability that you will be beaten by the 2nd card is slightly smaller than the first. This is the reason the probabilities decrease in my first answer.
However, suppose there is only 1 card in the deck which can make the underdog lose.
P(hitting the card in the 1st card) = (1/48)
P(2nd card) = (47/48) * (1/47) = (1/48)
P(3rd card) = (47/48) * (46/47) * (1/46) = (1/48).
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So we come to the riveting result that a particular card is equally likely to show up on each street. [img]/images/graemlins/smirk.gif[/img]
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