Sanderson M. Smith

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POKER PROBABILITIES (FIVE CARD HANDS)In many forms of

poker, one is dealt 5 cards from a standard deck of 52 cards. The number of different 5 -card poker hands is_{52}C_{5}= 2,598,960A wonderful exercise involves having students verify probabilities that appear in books relating to gambling. For instance, in

Probabilities in Everyday Life, by John D. McGervey, one finds many interesting tables containing probabilities for poker and other games of chance.This article and the tables below assume the reader is familiar with the names for various poker hands. In the NUMBER OF WAYS column of TABLE 2 are the numbers as they appear on page 132 in McGervey's book. I have done computations to verify McGervey's figures. This could be an excellent exercise for students who are studying probability.

There are 13 denominations (A,K,Q,J,10,9,8,7,6,5,4,3,2) in the deck. One can think of J as 11, Q as 12, and K as 13. Since an ace can be "high" or "low", it can be thought of as 14 or 1. With this in mind, there are 10 five-card sequences of consecutive dominations. These are displayed in TABLE 1.

TABLE 1

A K Q J 10 K Q J 10 9 Q J 10 9 8 J 10 9 8 7 10 9 8 7 6 9 8 7 6 5 8 7 6 5 4 7 6 5 4 3 6 5 4 3 2 5 4 3 2 A The following table displays computations to verify McGervey's numbers. There are, of course , many other possible poker hand combinations. Those in the table are specifically listed in McGervey's book. The computations I have indicated in the table do yield values that are in agreement with those that appear in the book.

TABLE 2

HAND N = NUMBER OF WAYS listed by McGervey

Computations and commentsProbability of HAND N/(2,598,960) and approximate odds.

Straight flush40 There are four suits (spades, hearts, diamond, clubs). Using TABLE 1, 4(10) =

40.0.000015 1 in 64,974

Four of a kind624 (

_{13}C_{1})(_{48}C_{1}) =624.Choose 1 of 13 denominations to get four cards and combine with 1 card from the remaining 48.

0.00024 1 in 4,165

Full house3,744 (

_{13}C_{1})(_{4}C_{3})(_{12}C_{1})(_{4}C_{2}) =3,744.Choose 1 denominaiton, pick 3 of 4 from it, choose a second denomination, pick 2 of 4 from it.

0.00144 1 in 694

Flush5,108 (

_{4}C_{1})(_{13}C_{5}) = 5,148.Choose 1 suit, then choose 5 of the 13 cards in the suit. This figure includes

allflushes. McGervey's figure does not include straight flushes (listed above). Note that 5,148 - 40 =5,108.

0.001965 1 in 509

Straight10,200 (

_{4}C_{1})^{5}(10) = 4^{5}(10) = 10,240Using TABLE 1, there are 10 possible sequences. Each denomination card can be 1 of 4 in the denomination. This figure includes

allstraights. McGervey's figure does not include straight flushes (listed above). Note that 10,240 - 40 =10,200.0.00392 1 in 255

Three of a kind54,912 (

_{13}C_{1})(_{4}C_{3})(_{48}C_{2}) = 58,656.Choose 1 of 13 denominations, pick 3 of the four cards from it, then combine with 2 of the remaining 48 cards. This figure includes all full houses. McGervey's figure does not include full houses (listed above). Note that 54,912 - 3,744 =

54,912.0.0211 1 in 47

Exactly one pair, with the pair being aces.84,480 (

_{4}C_{2})(_{48}C_{1})(_{44}C_{1})(_{40}C_{1})/3! =84,480.Choose 2 of the four aces, pick 1 card from remaining 48 (and remove from consider other cards in that denomination), choose 1 card from remaining 44 (and remove other cards from that denomination), then chose 1 card from the remaining 40. The division by 3! = 6 is necessary to remove duplication in the choice of the last 3 cards. For instance, the process would allow for KQJ, but also KJQ, QKJ, QJK, JQK, and JKQ. These are the same sets of three cards, just chosen in a different order.

0.0325 1 in 31

Two pairs, with the pairs being 3's and 2's.1,584 McGervey's figure excludes a full house with 3's and 2's.

(

_{4}C_{2})(_{4}C_{1})(_{44}C_{1}) =1,584.Choose 2 of the 4 threes, 2 of the 4 twos, and one card from the 44 cards that are not 2's or 3's.

0.000609 1 in 1,641

"I must complain the cards are ill shuffled 'til I have a good hand."-Swift,

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