Question

Assume there is a new version of porker, still using a standard deck of 52 cards. It requires hands of six cards. There are some types of hands to get in the Porker. How  many different hands are possible for each type?

1. There are 4 cards of one common value and 2 cards of another common value.

2. All six cards are from the same suit, with exactly one ace and with no king.

3. There are 3 cards of one common value and 3 cards of another common value.

Answer 1

1) We can select any of the 6 cards in 52C6 ways

Now, there are 13 cards of same value, out of that we want any 1 whose we will take all 4 suits. This is possible in 13C1*4C4 ways = 13

Next, we want two cards out of the 12 cards left of same value. Fir these 2 we can have any two suits out of the four (clubs, spades, heart and diamond). This is possible in 12C1*4C2 ways = 72

So required Probability = 13*72/52C6 = 0.000004

2) There are 4 suits so we can select anyone of the suit in 4C1 ways. Next, we want one Ace which has 1 way. We want any of the 5 cards from the suit without the king. Each suit has 13 cards, we already have the ace and without the king, there are 11 cards left, so we have 11C5 ways to select. This can be done in: 4C1*1*11C5 = 1848

So required Probability= 1848/52C6 = 0.000091

3) Let's select anyone of the cards of common value from the 13 available. From that we can have any 3 of the 4 suits available. This is done is 13C1*4C3 ways = 52

Again for the next 3 cards to be of the same value we have 12 left and any 3 out of 4 suits can be selected again: 12C1*4C3 = 48

So required Probability= 52*48/52C4= 0.00922