In the last blog post, I provided an intro to poker combinatorics. Now I’d like to take it a step further and deep dive into the math to determine the probability of winning hands like two pairs and royal flushes. We’re assuming a 52-card deck for this math.

The probability of gaining a pair of Queens in a 5-card hand includes 2 Queens (with a

as their is one Queen per 13 possible hand values from Ace to King multiply times _{13} C _{1}

as we are getting 2 queens out _{4} C _{2} `of 4`

) and 3 other card ranks. Since we’re down a Queen, the other cards in the deck would be 12 (one less 13) and we are choosing 3 completely different card ranks. So this equates to

multiply times 4 ways to get let’s say a 2 multiply times 4 ways to get let’s say a 10 and multiply times 4 ways to get let’s say a Jack. The math looks like this:_{12} C _{2}

We’re going to use this “combination without repetition” equation again.

(13! / (13 – 1) ! 1!) = 13

(4! / (4 – 2)! 2! = 6

(12! / (12 – 3)! 3! = 220

4 x 4 x 4 = 64

13 x 6 x 220 x 64 = 1.098 million ways to get a pair

1.1 million ways to get a pair / 2.59 million possible hands = 0.42 or 42% chance of getting a pair in 5-hand poker.

The probability of gaining a two pair so let’s say a pair of Kings and a pair of Aces in a 5-card hand includes 2 Kings (with a

as their is one King per 13 possible hand values from Ace to King multiply times _{13} C _{1}

as we’re getting 2 Kings out of 4) and 2 Aces (with a _{4} C _{2}

as their is one Ace per 13 possible hand values minus the King multiply times _{12} C _{1}

as we’re getting 2 Aces out of 4) and 1 other card rank. Since we’re down a King and an Ace, the other cards in the deck would be 11 (two less 13) and we are choosing 1 completely different card ranks. So this equates to _{4} C _{2}

multiply times 4 ways to get let’s say a 9. The math looks like this:_{11} C _{1}

Same “combination without repetition” equation.

(13! / (13 – 2) ! 2!) = 78

(4! / (4 – 2)! 2! = 6

(4! / (4 – 2)! 2! = 6

(11! / (11 – 1)! 1! = 11

Multiply times 4

78 x 6 x 6 x 11 x 4 = 123,552 ways to get a two pair

123,552 ways to get a two pair / 2.59 million possible hands = 0.048 or 4.8% chance of getting a two pair in 5-hand poker.

The probability of gaining 3 of a king so let’s say a three 5s includes 2 Kings (with a

as their is one 5 per 13 possible hand values from Ace to King multiply times _{13} C _{1}

as we are getting three 5s out of 4) and 2 other card ranks. Since we’re down one card, the other cards in the deck would be 12 (one less 13) and we are choosing 2 completely different card ranks. So this equates to _{4} C _{3}

multiply times 4 ways to get let’s say a 8 multiply times 4 ways to get let’s say a 9. The math looks like this:_{12} C _{1}

Same “combination without repetition” equation.

(13! / (13 – 1) ! 1!) = 13

(4! / (4 – 3)! 3! = 4

(12! / (12 – 2)! 2! = 66

Multiply times 4 x 4

13 x 4 x 66 x 4 x 4 = 54,912 ways to get three of a kind

54,912 ways to get three of a kind / 2.59 million possible hands = 0.021 or 2.1% chance of getting 3 of a kind in 5-hand poker.

The probability of getting a full house (3 of a kind plus a two pair so let’s say a pair of Kings and three Aces in a 5-card hand) includes 2 Kings (with a

as their is one King and one Ace per 13 possible hand values from Ace to King multiply times _{13} C _{2}

as we’re getting 2 Kings out of 4 multiply times _{4} C _{2}

as we’re getting 3 Aces out of 4). Since we’re 5 cards, there are no remaining other cards left in your hand. The math looks like this:_{4} C _{3}

Same “combination without repetition” equation.

(13! / (13 – 1) ! 1!) = 13

(4! / (4 – 2)! 2! = 6

(12! / (12 – 1)! 1! = 12

(4! / (4 – 3)! 3! = 4

13 x 6 x 12 x 4 = 3,744 ways to get a full house

3,744 ways to get a full house / 2.59 million possible hands = 0.0014 or 0.14% chance of getting a full house in 5-hand poker.

The probability of getting 4 of a kind includes four 10 (with a

as there is one 10 per 13 possible hand values from Ace to King multiply times _{13} C _{1}

as we are getting four tens) and one other card. Since we’re down a 10, the other cards in the deck would be 12 (one less 13) and we are choosing 1 completely different card ranks let’s say an Ace. So this equates to _{4} C _{4}

multiply times _{12} C _{1}

. The math looks like this:_{4} C _{1}

Same “combination without repetition” equation.

(13! / (13 – 1) ! 1!) = 13

(4! / (4 – 4)! 4! = 1

(12! / (12 – 1)! 1! = 12

(4! / (4 – 1)! 1! = 4

13 x 1 x 12 x 4 = 624 ways to get a four a kind

624 ways to get a four a kind / 2.59 million possible hands = 0.00024 or 0.024% chance of getting a four of a kind in 5-hand poker.

Finally, there are only 4 ways to get a royal flush. Not very good odds and after playing years of poker and countless hours, I have never seen anyone ever get a royal flush (or a straight flush for that matter).