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Calculating the Probability of Poker Hands 📂Sabermetrics

Calculating the Probability of Poker Hands

Definition of Suits and Ranks

Before getting into probabilities, it is recommended to look up poker hands if you are not familiar with poker itself. Before calculating probabilities, let’s define two things:

  • Suits: Element of the set {♠,◇,♤,♣}
  • Ranks: Element of the set {A,2,3,4,5,6,7,8,9,10,J,Q,K}

If two or more hands are satisfied at the same time, follow the higher one. The probabilities below assume that you draw 5 cards and consider the highest hand possible at that time.

One Pair

  • Choose one rank for which there exists a pair of cards of the same rank $( _{ 13 }{ { C }_{ 1 } }=13)$
  • Choose 2 out of 4 suits for the pair $( _{ 4 }{ { C }_{ 2 } })$
  • Choose three different ranks different from the first chosen pair $( _{ 12 }{ { C }_{ 3 } })$
  • Choose suits for the three selected cards $( _{ 4 }{ { C }_{ 1 } }\times _{ 4 }{ { C }_{ 1 } }\times _{ 4 }{ { C }_{ 1 } }={ 4 }^{ 3 })$ $$ \therefore \frac { 13\times _{ 4 }{ { C }_{ 2 }\times _{ 12 }{ { C }_{ 3 }\times { 4 }^{ 3 } } } }{ _{ 52 }{ { C }_{ 5 } } } $$

Note that when choosing three different numbers other than the first chosen pair, you should not use 48x47x46 because it is a permutation, which considers the order of drawing three cards from the remaining 48 cards.

Two Pair

  • Choose two ranks for which there exist two pairs of cards of the same rank $( _{ 13 }{ { C }_{ 2 } })$
  • Choose suits for the selected pairs $( _{ 4 }{ { C }_{ 2 } }\times _{ 4 }{ { C }_{ 2 } }={ 6 }^{ 2 })$
  • Choose one different rank from the initially chosen pairs $( _{ 11 }{ { C }_{ 1 } }=11)$
  • Choose a suit for the selected card $( _{ 4 }{ { C }_{ 1 } }=4)$ $$ \therefore \frac { _{ 13 }{ { C }_{ 2 } }\times { 6 }^{ 2 } \times 11\times 4 }{ _{ 52 }{ { C }_{ 5 } } } $$

Just calculating like one pair could lead to mistakes by incorrectly choosing ranks. Calculating the first pair and the second pair separately creates an order between the pairs.

Three of a Kind

  • Choose a rank for which there exist three cards of the same rank $( _{ 13 }{ { C }_{ 1 } }=13)$
  • Choose suits for the selected rank $( _{ 4 }{ { C }_{ 3 } }=4)$
  • Choose two different ranks from the initially chosen rank $( _{ 12 }{ { C }_{ 2 } })$
  • Choose suits for the two selected ranks $(4\times 4= { 4 }^{ 2 })$ $$ \therefore \frac { 13\times 4\times _{ 12 }{ { C }_{ 2 } }\times { 4 }^{ 2 } }{ _{ 52 }{ { C }_{ 5 } } } $$ Not too difficult.

Straight

  • Assuming A=1, J=11, Q=12, K=13, either have five consecutive ranks or A,K,Q,J,10, from A~5 to 9~K, total of 9 possibilities, or A,K,Q,J,10 $(9+1=10)$ Choose suits for the five cards $(4\times 4\times 4\times 4\times 4={ 4 }^{ 5 })$
  • Except cases higher than a straight flush $(4\times 10=40)$ $$ \therefore \frac { 10\times { 4 }^{ 5 } -40 }{ _{ 52 }{ { C }_{ 5 } } } $$

A bit tricky to express, but just having five consecutive ranks fills the requirement.

Flush

  • All five cards are of the same suit Choose one out of the four suits $( _{ 4 }{ { C }_{ 1 } }=4)$
  • Choose five cards of the same suit $( _{ 13 }{ { C }_{ 5 } })$
  • Except cases higher than a straight flush $(4\times 10=40)$ $$ \therefore \frac { 4\times _{ 13 }{ { C }_{ 5 }-40 } }{ _{ 52 }{ { C }_{ 5 } } } $$

It’s important to subtract 40 because of the straight flush.

Full House

  • Being one pair and three of a kind at the same time for two different ranks Choose a rank for the pair $( _{ 13 }{ { C }_{ 1 } }=13)$
  • Choose suits for the pair $( _{ 4 }{ { C }_{ 2 } })$
  • Choose a rank for the three of a kind $( _{ 12 }{ { C }_{ 1 } }=12)$
  • Choose suits for the three of a kind $( _{ 4 }{ { C }_{ 3 } })$ $$ \therefore \frac { 13\times _{ 4 }{ { C }_{ 2 } }\times 12\times _{ 4 }{ { C }_{ 3 } } }{ _{ 52 }{ { C }_{ 5 } } } $$

Simply multiplying the cases of getting a pair by the cases of getting a three of a kind and calling it a full house is not correct.

Four of a Kind

  • Choose a rank for which there exist four cards of the same rank $( _{ 13 }{ { C }_{ 1 } }=13)$
  • Choose the remaining card $( _{ 48 }{ { C }_{ 1 } }=48)$ $$ \therefore \frac { 13\times 48 }{ _{ 52 }{ { C }_{ 5 } } } $$

It’s as if a one pair occurred with the same rank twice.

Straight Flush

  • Being both a straight and a flush $(10\times 4=40)$
  • Except for the royal straight flush $(4)$ $$ \therefore \frac { 40-4 }{ _{ 52 }{ { C }_{ 5 } } } $$

Royal Straight Flush

  • Exists A,K,Q,J,10 and is a flush, thus only four possibilities. $$ \therefore \frac { 4 }{ _{ 52 }{ { C }_{ 5 } } } $$