How Many Different Poker Starting Hands Are There

Sanderson M. Smith

  • Suited hands contain two cards of the same suit, like J♣9♣, A ♥ K ♥, K♠Q♠ and 9 ♦ 3 ♦. All other starting hands are in the offsuit category, like A♠8 ♦, 7♣5 ♥ and K ♥ 9 ♦.
  • However, the correct answer is quite different. There is a 27% chance that your opponent has QQ and a 73% chance that he has AK. The reason for this is that the two hands are each made up from a different amount of poker card combinations - called combos.

The only factors determining the strength of a starting hand are the ranks of the cards and whether the cards share the same suit. Of the 1,326 combinations, there are 169 distinct starting hands grouped into three shapes: 13 pocket pairs (paired hole cards), 13 × 12 ÷ 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 169. However, even though the hands are not identical from that perspective, they still form equivalent poker hands because each hand is an A-Q-8-7-3 high card hand. There are 7,462 distinct poker hands. There are 7,462 distinct poker hands.

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POKER PROBABILITIES (FIVECARD HANDS)

In many forms of poker, one is dealt 5 cards from astandard deck of 52 cards. The number of different 5 -card pokerhands is

52C5 = 2,598,960

A wonderful exercise involves having students verify probabilitiesthat appear in books relating to gambling. For instance, inProbabilities in Everyday Life, by John D. McGervey, one findsmany interesting tables containing probabilities for poker and othergames of chance.

This article and the tables below assume the reader is familiarwith the names for various poker hands. In the NUMBER OF WAYS columnof TABLE 2 are the numbers as they appear on page 132 in McGervey'sbook. I have done computations to verify McGervey's figures. Thiscould be an excellent exercise for students who are studyingprobability.

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

TABLE 1Starting
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 65
8 7 6 54
7 6 5 4 3
6 5 4 3 2
5 4 3 2 A

The following table displays computations to verify McGervey'snumbers. There are, of course , many other possible poker handcombinations. Those in the table are specifically listed inMcGervey's book. The computations I have indicated in the table doyield values that are in agreement with those that appear in thebook.

How many different poker starting hands are there every

How Many Different Poker Starting Hands Are There Every

TABLE 2
HAND

N = NUMBER OF WAYS listed by McGervey

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

Straight flush

40

There are four suits (spades, hearts, diamond, clubs). Using TABLE 1,4(10) = 40.

0.000015
1 in 64,974

Four of a kind

624

(13C1)(48C1) = 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 house

3,744

(13C1)(4C3)(12C1)(4C2) = 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

Flush

5,108

(4C1)(13C5) = 5,148.

Choose 1 suit, then choose 5 of the 13 cards in the suit. This figure includes all flushes. McGervey's figure does not include straight flushes (listed above). Note that 5,148 - 40 = 5,108.

0.001965
1 in 509

Straight

10,200

(4C1)5(10) = 45(10) = 10,240

Using TABLE 1, there are 10 possible sequences. Each denomination card can be 1 of 4 in the denomination. This figure includes all straights. 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 kind

54,912

(13C1)(4C3)(48C2) = 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

(4C2)(48C1)(44C1)(40C1)/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.

(4C2)(4C1)(44C1) = 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 Ihave a good hand.'

-Swift, Thoughts on Various Subjects

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