Random Card Shuffling ProbabilitiesDate: 6/11/96 at 20:37:33 From: Anonymous Subject: Probability of two cards being.... I was wondering the other day what the probability of say at least two eights being next to each other in a random shuffling of a deck of cards. However a more interesting problem is the probability of at least two cards (2 eights or 2 queens etc.) being next to each other. The answer is 1-P(no similar cards are next to each other), but how do you count that? Date: 6/12/96 at 6:10:30 From: Doctor Anthony Subject: Re: Probability of two cards being.... If you want the probability that no 'eights' are next to each other, you can model the problem in the following way: Let 'eights' be represented by stars, *, and non-eights by strips |. Then if the cards are laid out in a row at random, you get something like this: |||*||||**||||....... and so on. You will notice that the condition for no two stars to be together is that a single star must occupy a gap between two strips. Suppose therefore that we first lay out the 48 strips (non-eights) in a row. Now there are 49 gaps available (49 because there is a space before the first strip and after the last one), and we must choose 4 out of the 49 in which to place a star. This can be done in 49_C_4 ways = 211876 ways. However the number of ways of laying out 48 strips and 4 stars in any order is 52!/(48!.4!) = 52_C_4 = 270725 The probability that no eights are together is therefore given by: (49_C_4)/(52_C_4) = 0.782624 The probability that at least two eights are together is therefore 1 - 0.782624 = 0.217376 These are also the a priori probabilities for any other kind of card, say kings or sixes. The fact that eights ARE together would however affect the number of gaps available when considering other face-value cards, so the probabilities are not independent, if, for example, you wanted probabilities that no eights and no queens are together. I will give the matter further thought, but I think your general request for the probability that NO two cards of the same face-value are together will prove very complex indeed. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 6/13/96 at 23:16:30 From: Doctor Peter Subject: Re: Probability of two cards being.... This problem involves counting to number of permutations, since we are dealing (no pun intended) with the order of the cards in the deck. There are 52! permutations of the deck, but the calculation of the number of permutations with at least two like cards together involves a fairly lengthy recursive formula. There is not enough space to give it here, but it involves the number of permutations of the 4 together plus 3 together plus 2 together all times the rest of the decks' permutations, which also needs to incorporate this. I made a spread- sheet model based on this, and got that the P(at least 2 together) is about equal to 96.35%. Contact me for additional explanation. -Doctor Peter, The Math Forum Check out our web site! |
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