Probabilities of Poker Hands
The following is a list of possible winning poker hands ranked from best to worst hand:
Royal Flush (10, J, Q, K, A of the same suit)
Straight Flush (5 in order of the same suit)
Four of a Kind (4 of the same number)
Full House (2 of a kind and 3 of a kind)
Flush (5 cards of the same suit)
Straight (5 in order of varying suits)
Three of a kind (3 of the same number)
Two pairs (2 of one number and 2 of another number)
Two of a kind (2 of the same number)
Find the number of hands of 5 cards that you could be dealt (there are 52 cards in a deck).
Find the number of possible hands of each type and the probability of getting each of the
above hands.
Solution:
52 52!
Total possible hands:   
 2598960
 5  48!5!
There are 52 cards in the deck, and we are choosing 5 of these cards.
5 4
 J, Q, K, A of the same suit):      1 4  4
Royal Flush (10,
5 1
There are 5 cards to choose from, and we need all of them. Then, we must choose 1
of the 4 suits that they will be.
4
P(E) 
 0.0000015

2598960
9 4
Straight Flush (5 in order of the same suit):      9  4  36
1 1
There are 9 possible orders you could have (2, 3, 4, 5, 6; 3, 4, 5, 6, 7; 4, 5, 6, 7, 8; etc.).

Then, we must choose one suit for this order.
36
P(E) 
 .00001385

2598960
13 4 48
Four of a Kind (4 of the same number):        131 48  624
1  4  1 
We must first choose the rank of our 4 of a kind. Then, we want all 4 suits of this

rank. Lastly, we must choose our final card from the remaining 48 in the deck.
624
P(E) 
 .00024

2598960
13 4 12 4
Full House (2 of a kind and 3 of a kind):          13 4 12  6  3744
1  3 1  2


First, let’s choose our 3 of a kind. There are 13 ranks (or numbers) to choose from.
The rank we chose then has four suits, we must choose 3 of these 4 suits. Now, we
can choose our two of a kind. There are only 12 remaining ranks to choose from.
After choosing our rank, we must then pick two of the 4 suits for this rank.
3744
P(E) 
 .00144
2598960
4 13
Flush (5 cards of the same suit):    =5148
1 5 
There are 4 suits; we will choose one of them. Then, we are going to choose any 5

cards from the 13 cards of that chosen suit.
5148
P(E) 
 .00198
2598960
Straight (5 in order of varying suits): (9)(45)= 9,216
There are 9 possible orders you could have (2, 3, 4, 5, 6; 3, 4, 5, 6, 7; 4, 5, 6, 7, 8; etc.).
Each of the 5 cards you have can be 4 different suits, which gives us 45.

9216
P(E) 
 .0035
2598960
13 4 49
49!
Three of a kind (3 of the same number):        13 4 
 61152
47!2!
1  3  2 
Choose a rank, then choose 3 suits for this number to be. Next, we have to choose

two of the remaining 49 cards.
61152
P(E) 
 .024 
2598960
13 4 12 4 48
Two pairs (2 of a #, 2 of another #):            13 6 12  6  48  269568
1  2 1  2  1 
First we choose the rank of the first pair, then choose 2 suits of that rank. Then, we

choose the rank of our second pair from the remaining 12.then, we choose the suit
of these cards. Lastly, choose a card from the remaining 48 cards.

269568
P(E) 
 .1037
2598960
13 4 50
50!
Two of a kind (2 of the same number):        13 6 
 1528800
47!3!
1  2  3 
Choose the rank of the pair. Then, choose two of the four suits. Lastly, choose 3

cards from the remaining 50 cards.
1528800
P(E) 
 .588
2598960
