(Note that the probability calculated in this example assumes no information is known about the cards in the other players' hands; however, experienced poker players may consider how the other players place their bets (check, call, raise, or fold) in considering the probability for each scenario. The mean of the hypergeometric distribution is nk / N, and the variance (square of the standard deviation) is nk (N − k) (N − n)/ N2 (N − 1). N 2 N [5]. The classical application of the hypergeometric distribution is sampling without replacement. . {\displaystyle k} = still unseen. balls and colouring them red first. 1 a N {\displaystyle k=0,n=2,K=9} For example, if a problem is present in 5 of 100 precincts, a 3% sample has 86% probability that k = 0 so the problem would not be noticed, and only 14% probability of the problem appearing in the sample (positive k): The sample would need 45 precincts in order to have probability under 5% that k = 0 in the sample, and thus have probability over 95% of finding the problem: In hold'em poker players make the best hand they can combining the two cards in their hand with the 5 cards (community cards) eventually turned up on the table. th 5 The mean is given by: μ = E(x) = np = na / N and, variance σ2 = E(x2) + E(x)2 = na(N − a)(N − n) N2(N2 − 1) = npq[N − n N − 1] where q = 1 − p = (N − a) / N. I want the step by step procedure to derive the mean and variance. This has the same relationship to the multinomial distribution that the hypergeometric distribution has to the binomial distribution—the multinomial distribution is the "with-replacement" distribution and the multivariate hypergeometric is the "without-replacement" distribution. K 0000003543 00000 n K N 0000003619 00000 n If we replace M N by p, then we get E(X) = np and V(X) = N n N 1 np(1 p). and x na x np N μ == = (2) The variance of the hypergeometric distribution can be computed from the generic formula that 222 2 N 2 {\displaystyle \Phi } Then, the number of marbles with both colors on them (that is, the number of marbles that have been drawn twice) has the hypergeometric distribution. In a test for under-representation, the p-value is the probability of randomly drawing N 0000005749 00000 n 2 0000007297 00000 n 0000008316 00000 n ( 0000002463 00000 n This identity can be shown by expressing the binomial coefficients in terms of factorials and rearranging the latter, but it − In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of k The player would like to know the probability of one of the next 2 cards to be shown being a club to complete the flush. 9 and successes in , ( Hypergeometric − successes. K ( n k / 2 K 20 30 ⋅ , K N . K {\displaystyle N} Φ ) {\displaystyle n} K The pmf is positive when Intuitively we would expect it to be even more unlikely that all 5 green marbles will be among the 10 drawn. The following table describes four distributions related to the number of successes in a sequence of draws: The model of an urn with green and red marbles can be extended to the case where there are more than two colors of marbles. 0000001178 00000 n n K ", "Calculation for Fisher's Exact Test: An interactive calculation tool for Fisher's exact probability test for 2 x 2 tables (interactive page)", Learn how and when to remove this template message, "HyperQuick algorithm for discrete hypergeometric distribution", Binomial Approximation to a Hypergeometric Random Variable, https://en.wikipedia.org/w/index.php?title=Hypergeometric_distribution&oldid=989602957, Articles lacking in-text citations from August 2011, Creative Commons Attribution-ShareAlike License, The result of each draw (the elements of the population being sampled) can be classified into one of, The probability of a success changes on each draw, as each draw decreases the population (, If the probabilities of drawing a green or red marble are not equal (e.g.