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 March 8th, 2014, 03:38 AM #1 Newbie   Joined: Dec 2010 Posts: 8 Thanks: 0 Card combinatorics and probabilities Suppose there is a number K of cards in a deck, a majority L of which are black and a minority (K-L) of which are red. Suppose that we draw one card at a time and after we've recorded the result, we put the card back into the deck and re-shuffle it. Furthermore, suppose that one team, Team Black, wins if a majority of the drawn cards are black, and that the other team, Team Red, wins if a majority of the drawn cards are red. Draws are not allowed so if the two teams draw the same number of cards, we flip a coin to determine who wins. Suppose we draw N cards. It is obvious that the greater K is, the greater is the chance that Team Black will win, and that N approaches 1 in the limit. The formula (where N is odd) is given on p. 264 of this paper: http://www.socsci.uci.edu/~bgrofman/69% ... 0truth.pdf But suppose now that we instead use an alternative procedure. According to this procedure, if we encounter a card that already has been drawn (say the ace of spades or the ace of hearts), we don't count that card (though we do put it back in the deck). My conjecture is that this will not change the probability that Team Black will win (it holds for N=3, if I've calculated correctly). However, I can't prove this generally. Any help with this would be much appreciated. Please ask if anything's unclear.
March 8th, 2014, 04:46 AM   #2
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Re: Card combinatorics and probabilities

Sorry it should not say

Quote:
 Suppose we draw N cards. It is obvious that the greater K is, the greater is the chance that Team Black will win, and that N approaches 1 in the limit.
Rather: "It is obvious that the greater N is, the greater is the chance that Team Black will win, and that this probability approaches 1 in the limit."

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