Question

Find the probabilities for the following poker hands. They are arranged in decreasing order of probability.

(e) Straight. (Five cards in a sequence. Does not include a straight flush. Ace can be high or low.)

(f) Three of a kind. (Three cards of one face value. Does not include four of a kind or full house.)

(g) Two pair. (Does not include four of a kind or full house.)

(h) One pair. (Does not include any of the aforementioned conditions.)

Answer 1

Number of ways of selecting 5 cards out of 52 cards is

(e)

Staright can be start from 10 cards start from ace, 1, 2,3, 4...10. Each denomination can be choosed in C(4,1)= 4 ways. So number of possible straights is

From the above, we need to remove flush. A straight flush means all 5 cards are from same suit and they form a straight. Since number of possible straight from each suit is 10 and number of ways of selecting one suit out of 4 is C(4,1) =4 so possible number of straight flush is 10*4 = 40.

Therefore number of straight without being a flush is 10240 - 40 = 10200

So the probabilty of a straight without being a flush is

10200 / 2598960 = 0.0039

(f)

Number of ways of selecting 1 denominations out of 13 is C(13,1). Number of ways of selecting 3 cards out of 4 cards of selected denomination is C(4,3). And then select two denominations out of remaining 12 denominations is C(12,2) and then 1 card from each selected denominations is C(4,1)C(4,1). So number of ways are there to draw a 5 card poker hand that contains 3 a kind is

C(13,1)C(4,3)C(12,2)C(4,1)C(4,1) = 54912 ways

So probability of getting three of a kind:

(g)

There are total 13 denominations and each denomination has 4 cards. So number of ways of selecting 2 denominations and then 2 cards out of 4 is

And since we need exactly 2 pairs so remaining 1 card must come from different denomination so number of ways of selecting 1 denominations out of remaining 11 denominations and then 1 card from selected denomination is

So number of ways of selecting 2 pairs is :

So required probability is

(g)

There are total 13 denominations and each denomination has 4 cards. So number of ways of selecting 1 denomination and then 2 cards out of 4 is

And since we need exactly 1 pair so remaining 3 cards must come from different denominations so number of ways of selecting 3 denominations out of remaining 12 denominations and then 1 card from each selected denominations is

So number of ways of selecting 1 pair is :

So required probability is