Poker

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Resources

Poker probability - http://en.wikipedia.org/wiki/Poker_probability

Poker Math

Sample space

13 kinds - Ace, King, etc.

4 suits - Clubs, Hearts, Diamonds, Spades

|S| = C(52,5) different hands

Event Space

For a simple example, the hands containing the Ace of Diamonds are:

  1. No choice on Ace of Diamonds, only 1 card possible
  2. Remaining  = C(51,4) to pick 4 cards out of the remaining 51

C(51,4) possible hands containing the Ace of Diamonds.

For another simple example, the hands containing the Ace and King of Diamonds are:

  1. No choice on the Ace and King of Diamonds, only two cards possible
  2. Remaining  = C(50,3) to pick 3 cards out of the remaining 50

C(50,3) possible hands containing the Ace and King of Diamonds.

General method

To determine the number of possible hands, need to account for the following:

  1. Number of kinds
  2. Number of suits
  3. Remaining cards
Example 1

How many hands with 3 Aces?

Number of kinds = 1 since picking Aces as the kind

Number of ways to pick 3 Aces from 4 Aces = C(4,3)

Number of ways to pick the 2 cards from the remaining 48 = C(48,2)

Reason only 48 is the fourth Ace must be excluded to prevent picking 4 Aces.

C(4,3)*C(48,2) = 4512 hands with 3 Aces

 

Example 2

How many hands with 5 of the same suit?

Number of ways to pick 5 different kinds from 13 = C(13,5)

Number of ways to pick 1 suit from 4 suits = C(4,1)

C(4,1)*C(13,5) = 5148 hands with 5 of same suit.

 

Example 3

How many hands with 3 of the same suit?

3 kinds from 13 = C(13,3)

1 suit from 4 suits = C(4,1)

2 remaining cards from 39, removing the 13 suit cards of the same kind, C(39,2)

C(13,3)*C(4,1)*C(48,2) = 847,704 with 3 of the same suit

 

Example 4

How many hands with 2 pairs?

2 kinds from 13 = C(13,2)

2 cards from 4 suits for first pair = C(4,2)

2 cards from 4 suits for second pair = C(4,2)

1 remaining card from 44, 52-8=44 by removing the 8 cards of the 2 kinds, C(44,1)

C(13,2)*C(4,2)*C(4,2)*C(44,1) = 123552 with 2 pair

 

Example 5

How many hands with 3 of the same kind? Ignore full houses.

1 kind from 13 = C(13,1)

3 suits from 4 suits = C(4,3)

2 remaining from 48, 52-4=48 by removing the 4 cards of the same kind, C(48,2)

C(13,1)*C(4,3)*C(48,2) = 58656 with 3 of a kind

There are 3744 full houses, which normally are excluded.

 

Example 6

How many hands with a full house?

Here the order of the kinds matter because 3 Queens/2 Jacks is different than 3 Jacks/2 Queens

2 kinds from 13 = P(13,2)

3 cards from 4 suits = C(3,2)

2 cards from 4 suits = C(4,2)

P(13,2)*C(4,3)*C(4,2) = 3744