How Many Different Poker Hands Are There

4/13/20224/13/2022

How Many Different Full House Poker Hands Are There
How Many Different Poker Hands Are There Now
Combinations In Poker

Algebra -> Probability-and-statistics-> SOLUTION: A poker hand consists of 5 cards dealt from an ordinary deck of 52 playing cards. How many different hands are there consisting of (a) four hearts and one spade? (b) three red card Log On

There are 13 card ranks in a deck. So if we exclude AA, A8 and A9, 10 different types of Ax hand remain that make top pair, each with 12 combinations. That’s a total of 120 (10. 12) different combinations of top pair on this texture. That’s still a relatively comfortable calculation. 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. The relative probability of being dealt a hand of each given shape is different.

I often read that there are 169 possible starting hands in Hold 'em, but my calculations show different. Bear with me If suits don't matter and AcKd = AdKs = any AKsuited then there are THIRTEEN hole cards with an ace say AA - A2 lets call AA A.1(the highest) and A2 A.13(the lowest).

Question 520629: A poker hand consists of 5 cards dealt from an ordinary deck of 52 playing cards. How many different hands are there consisting of (a) four hearts and one spade? (b) three red cards and two black cards? (c) at least 3 face cards? (d) 2 aces and 2 face cards?
Found 2 solutions by stanbon, edjones:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A poker hand consists of 5 cards dealt from an ordinary deck of 52 playing cards.
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How many different hands are there consisting of
(a) four hearts and one spade?
# of ways to get 4 hearts: 13C4 = 715
# of ways to get 1 spade: 13C1 = 13
---
Total # of ways = 715*13 = 9295
(b) three red cards and two black cards?
# of ways to get 3 red = 26C3 = 2600
# of ways to get 2 black = 26C2 = 325
---
Total # of ways = 845,000
(c) at least 3 face cards?
There are 12 face cards:
3 face cards: 12C3*40C2 = 220*780
+ 4 face cards: 12C4*40C1 = 495*40
+ 5 face cards: 12C5*40C0 = 792*1
---
Total # = 171600 + 19800 + 792 = 192,192
(d) 2 aces and 2 face cards?
# of ways to get 2 aces: 4C2 = 6
# of ways to get 2 face cards: 12C2 = 66
---
Total # of ways = 6*66 = 396
Cheers,
Stan H.

Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
a)13C4*13C1=715*13=9295
b)26C3*26C2=2600*325=845000
c)12C3*40C2+12C4*40C1+12C5=171600+19800+792=192192
d)4C2*12C2*36C1=156816
.
Ed

Sanderson M. Smith

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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.

How Many Different Full House Poker Hands Are There

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.

How Many Different Poker Hands Are There

TABLE 1 Are

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.

TABLE 2

HAND	N = NUMBER OF WAYS listed by McGervey	Computations and comments	Probability of HANDN/(2,598,960) and approximate odds.
Straight flush	40	There are four suits (spades, hearts, diamond, clubs). Using TABLE 1,4(10) = 40.	0.0000151 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.	0.000241 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.	0.001441 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.	0.0019651 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.	0.003921 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.	0.02111 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.	0.03251 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.	0.0006091 in 1,641

How Many Different Poker Hands Are There Now

'I must complain the cards are ill shuffled 'til Ihave a good hand.'

-Swift, Thoughts on Various Subjects

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