Question

In a five card poker hand, find the probability of getting two pairs where one pair is a face card (Jack, Queen, King) and the other pair is not a face card (Ace, two, ... , ten)

Answer 1

Deck Of Cards

A deck of 52 cards has 4 suits, Spades, Clubs, Diamonds and Hearts

26 are Red cards (Hearts and Diamonds) and 26 are Black (Spades and Clubs)

There are 13 cards in each suit, from 2 till 10, the 3 face cards J, Q, K and finally Ace.

Therefore there are 4 cards of each type i.e 4 kings, 4 queens etc

____________________________________________

Probability = Favorable Outcomes / Total Outcomes

Favorable Outcomes = To get a 2 pair with constraints

We have 3 face cards and 10 other types

Let us choose a face card first = 3C1 = 3 ways

From these (each have 4) we choose 2 = 4C2 = 6 ways

Now we choose one pair for the second pair from the remaining 10 cards = 10C1 = 10

From this chosen card, we choose 2 in 4C2 = 6 ways

Finally the last card, should neither be the face card we have chosen nor the normal cards, so we choose one from the remaining 11 types of cards (one face and one normal gone from the 13 types of cards) = 11C1

From this one type chosen, we need to choose 1 card amongst 4 different suits = 4C1

Therefore Total Number of ways = 3 * 6 * 10 * 6 * 11 * 4 = 47520

Total Outcomes = choose 5 cards out of 52 in 52C5 ways = 2598960

The required probability = 47520 / 2598960 = 0.0183