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 July 22nd, 2015, 09:24 AM #1 Newbie   Joined: Jul 2015 From: murica Posts: 1 Thanks: 0 impossible probability problem??? the game is that there are 2 decks of standard 52-card decks. in one deck, you take 13 cards at random and set the rest aside. from the other deck, you randomly draw cards until you have the same cards that are in your hand. basically like bingo. the question is, what is an equation that shows the probability of drawing a bingo on any given card you draw? obviously, you need to draw at least 13 cards and you have a 100% chance of getting bingo on the 52nd card
 July 22nd, 2015, 12:37 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I would address this by finding the number of ways you could draw the chosen 13 cards as part of your $n$ cards, then dividing by the number of ways you could draw any $n$ cards. There's only one way to choose the 13 cards. You can order them in any way, giving 13! arrangements. There are $52-13\choose n-13$ ways to choose the other cards, which can be arranged in $(n-13)!$ ways. You can choose which card type (one of the special 13 or one of the others) goes in which slot in $n\choose13$ ways, for a grand total of $$1\cdot13!{52-13\choose n-13}(n-13)!{n\choose13}= 13!\frac{(52-13)!}{(n-13)!(52-n)!}(n-13)!\frac{n!}{13!(n-13)!}= \frac{39!n!}{(52-n)!(n-13)!}$$ ways for you to have a 'winning' hand with $n$ cards. Without any conditions, there are $\frac{52!}{(52-n)!}$ ways to get your $n$ cards, so the quotient $$\frac{39!n!}{(52-n)!(n-13)!}\cdot\frac{(52-n)!}{52!}= \frac{39!n!}{52!(n-13)!}$$ gives the chance of winning by the $n$th draw (for $13\le n\le52$ of course). Note that the probability of winning at $n=13$ is 13!*(52-13)!/52! = 1/635013559600, the probability of winning by $n=52$ is 1 = 100%, and the probability of winning by $n=51$ is 3/4.

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