Associated Topics || Dr. Math Home || Search Dr. Math

### Poker Hand Probabilities

```
Date: 02/11/99 at 13:32:34
From: Jason
Subject: Poker Probabilty

I have to figure out the probability of different poker hands. I know
all of the hands but need help figuring out the probability of each.
I would like to see if my answers are correct.

Thank you,
Jason Charpentier
```

```
Date: 02/11/99 at 13:51:01
From: Doctor Anthony
Subject: Re: Poker Probabilty

a. No pairs [5 different face values, not in sequence, not all cards
in the same suit]

There are   C(13,5) x 4^5  ways with all cards of different face value.
There are C(4,1) x C(13,5) ways where cards are all of the same suit,
and 10 x 4^5 - 10 x 4 ways where cards form a straight.

The number of ways that a straight can be formed is as follows:

10 is the number of ways of choosing the first face value of the straight.
4^5 is the number of ways of choosing the suit for each of the 5 cards.
10 x 4 is number of ways of choosing a straight all of one suit.

So Prob(5 different face values, not in sequence, not in same suit)

C(13,5) x 4^5 - C(4,1) x C(13,5) - (10 x 4^5 - 10 x 4)
= ------------------------------------------------------ = 0.501177
C(52,5)

----------------------------------------------------------------------

b. One pair [two of one face value, and 3 cards of different face
values, no matching the pair]

13 x 4C2 x 12C3 x 4^3
Prob(1 pair) = ---------------------- = 0.422569
52C5

----------------------------------------------------------------------

c. Two pairs [one pair of each two different face values and a card of
a third face value]

13C2 x 4C2 x 4C2 x 11 x 4
Prob(2 pairs) =  -------------------------  = .0475390
52C5

The various terms in above expression arise as follows.

13C2 = (13 choose 2) is the number of ways of choosing the two
different face values for the two pairs.

4C2 = (4 choose 2) is the number of ways of choosing the two suits for
one pair, and the second 4C2 is the number of ways of choosing the two
suits for the second pair.

11 is the number of ways of choosing the face value for the 5th card.
It must not be the same face value as either of the pairs.

4 is the number of ways of choosing the suit for the 5th card.

52C5 = (52 choose 5) is the number of unrestricted ways that 5 cards
can be selected from the 52 in the pack.

----------------------------------------------------------------------

d. Three of a Kind [exactly 3 cards of one face value and 2 different
cards]

13C1 x 4C3 x 12C2 x 4^2
Prob(Three of a kind) =  -------------------------  = .02112845
52C5

13C1 is number of ways of choosing the face value for the three cards.

4C3 is the number of ways of choosing the suits for these three cards.

12C2 is the number of ways of choosing the face values for the other
two cards. (order does NOT matter)

4^2 is the way the suits for these last two cards can be chosen.

----------------------------------------------------------------------

e. Straight [5 cards in sequence, but not all of same suit.  Ace high
or low]

10 x 4^5 - 10 x 4
Prob(straight) =  --------------------  =  .0039246
52C5

----------------------------------------------------------------------

f. Flush [5 cards of the same suit but not in sequence, not including
the straight flush and royal flush below]

4C1 x 13C5 - 4 x 10
Prob(Flush) = --------------------  = .0019654
52C5

----------------------------------------------------------------------

g. Full house [3 cards of one face value and 2 cards of another face
value]

13P2 x 4C3 x 4C2
Prob(Full house) = ------------------  =  .00144058
52C5

----------------------------------------------------------------------

h. four of a kind [four cards of one face value and one other card]

13C1 x 12 x 4
Prob(Four of a kind) =  --------------  =  .00024009
52C5

----------------------------------------------------------------------

i. straight flush [five cards in sequence and of the same suit, but
not ace king queen jack ten]

9 x 4C1
Prob(Straight flush) =  ---------- =  .00001385
52C5

----------------------------------------------------------------------

j. Royal Flush [ace, king, queen, jack, ten in the same suit]

1 x 4C1
Prob(Royal flush) =  ----------  =  .000001539
52C5

For another source, see Ivars Peterson's MathLand, September 7, 1996,
in Science News Online:

http://www.sciencenews.org/sn_arch/9_7_96/mathland.htm

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search