Question

Compute the number of ways to deal each of the following five-card hands in poker. 1. Straight: the values of the cards form a sequence of consecutive integers. A jack has value 11, a queen 12, and a king 13. An ace may have a value of 1 or 14, so A 2 3 4 5 and 10 J Q K A are both straights, but K A 2 3 4 is not. Furthermore, the cards in a straight cannot all be of the same suit (a flush). 2. Flush: All five cards have the same suit (but not in addition a straight). 3. Straight flush: both a straight and a flush. Make sure that your counts for straights and flushes do not include the straight flushes. 4. Four of a kind. 5. Two distinct matching pairs (but not a full house). 6. Exactly one matching pair (but no three of a kind). 7. At least one card from each suit. At least one card from each suit, but no two values matching. 8. Three cards of one suit, and the other two of another suit, like three hearts and two spades.

Answer 1

Answer:

The number of ways to deal each hand of poker is

1) 10200 possibilities

2) 5108 possibilities

3) 40 possibilities

4) 624 possibilities

5) 123552 possibilities

6) 732160 possibilities

7) 308880 possibilities

8) 267696 possibilities

Step-by-step explanation:

Straigth:

The Straight can start from 10 different positions: from an A, from a 2, 3, 4, 5, 6, 7, 8, 9 or from a 10 (if it starts from a 10, it ends in an A).

Given one starting position, we have 4 posibilities depending on the suit for each number, but we need to substract the 4 possible straights with the same suit. Hence, for each starting position there are 4⁵ - 4 possibilities. This means that we have 10 * (4⁵-4) = 10200 possibilities for a straight.

Flush:

We have 4 suits; each suit has 13 cards, so for each suit we have as many flushes as combinations of 5 cards from their group of 13. This is equivalent to the total number of ways to select 5 elements from a set of 13, in other words, the combinatorial number of 13 with 5 ${13 \choose 5}$ .  However we need to remove any possible for a straight in a flush, thus, for each suit, we need to remove 10 possibilities (the 10 possible starting positions for a straight flush). Multiplying for the 4 suits this gives us

$4 * ( {13 \choose 5} -10) = 4* 1277 = 5108$

possibilities for a flush.

Straight Flush:

We have 4 suits and 10 possible ways for each suit to start a straight flush. The suit and the starting position determines the straight flush (for example, the straight flush starting in 3 of hearts is 3 of hearts, 4 of hearts, 5 of hearts, 6 of hearts and 7 of hearts. This gives us 4*10 = 40 possibilities for a straight flush.

4 of a kind:

We can identify a 4 of a kind with the number/letter that is 4 times and the remaining card. We have 13 ways to pick the number/letter, and 52-4 = 48 possibilities for the remaining card. That gives us 48*13 = 624 possibilities for a 4 of a kind.

Two distinct matching pairs:

We need to pick the pair of numbers that is repeated, so we are picking 2 numbers from 13 possible, in other words, ${13 \choose 2} = 78$ possibilities. For each number, we pick 2 suits, we have ${4 \choose 2} = 6$ possibilities to pick suits for each number. Last, we pick the remaining card, that can be anything but the 8 cards of those numbers. In short, we have 78*6*6*(52-8) = 123552 possibilities.  

Exactly one matching pair:

We choose the number that is matching from 13 possibilities, then we choose the 2 suits those numbers will have, from which we have $4 \choose 2$ possibilities. Then we choose the 3 remaining numbers from the 12 that are left ( $12 \choose 3 = 220$ ) , and for each of those numbers we pick 1 of the 4 suits available. As a result, we have

$13 * 4 * 220 * 4^3 = 732160$

possibilities

At least one card from each suit (no mathcing pairs):

Pick the suit that appears twice (we have 4 options, 1 for each suit). We pick 2 numbers for that suit of 13 possible ($13 \choose 2 = 78$ possibilities ), then we pick 1 number from the 11 remaining for the second suit, 1 number of the 10 remaining for the third suit and 1 number from the 9 remaining for the last suit. That gives us 4*78*11*10*9 = 308880 possibilities.

Three cards of one suit, and 2 of another suit:

We pick the suit that appears 3 times (4 possibilities), the one that appears twice (3 remaining possibilities). Foe the first suit we need 3 numbers from 13, and from the second one 2 numbers from 13 (It doesnt specify about matching here). This gives us

$4 * 13 \choose 3 * 3 * 13 \choose 2 = 4*286*3*78 = 267696$

possibilities.