What are the probabilities of the following hands in poker a

What are the probabilities of the following hands in poker: (a) Two pair. That is, two sets of pairs, e.g. 3H, 3S, 9D, 9S, KD (b) Three of a kind. That is, three of one rank and two other mismatched cards (c) Full house. That is, a three of a kind and a pair.

Solution

(a) There are 52 cards in a deck of 4 suits. Each suit contains 13 contains. That means, there are 4 cards of same value.

Two cards of different values from 13 cards can be choosen in ¹³C₂ ways.

Now, once a card is choosen, there 4 cards having the same value.. so 2 cards from 4 can be choosen in ⁴C₂ ways. Thus, we have choosen 4 cards so far. After choosing 4 cards, we are left with 44 cards of different value. Out of these, we can choose one card in ⁴⁴C_{1 ways. Therefore, the number of ways of getting two pairs is} ¹³C₂X ⁴C₂X ⁴⁴C₁ =20592

and the number of possible hands in a pocker is ⁵²C₅=2598960

Then the probability of getting two pairs=20592/2598960=0.0079

(b) Three cards of suit can be choosen in C(13, 1)*C(4, 3) ways. Then we need to choose two different cards. We have 48 cards left so, one card can be choosen in C(48, 1) ways, Then we want the next card of different value which needs to be choosen from 44 cards and hence that can be done in C(44,1)

therefore, the total number of possible ways of getting 3 of a kind= C(13, 1)*C(4, 3)*C(48, 1)*C(44,1)=109824

And, the total possible number of hands of 5 cards=C(52, 5)=2598960

Then, the probability of getting three of a kind=109824/2598960=0.04226

(c) Possible ways of selecting three cards of the same suit=C(13, 1)*C(4, 3)=52

Then, we are left with 48 cards, so, a pair can be selected in C(12, 1)*C(4, 2)=72

therefore, the possible ways of getting a full house is =52*72=3744

And, the possible hands of 5 cards=C(52, 5)=2598960

Then, the probability of getting full house=3744/2598960=0.00144