1. Poker Solutions

2. How Many Hands are Possible?
All of these problems will have a numerator which is the number of ways to form the hand of interest.
The denominator will always be the number of ways to form any five card hand
Order does not matter, so we use combination
We are choosing 5 things from a universe of 52, so evaluate 52C5 to get the denominator which turns out to be 2,598,960

3. One Pair
How many ways can you form exactly one pair from a deck of cards?
Choose one of 13 denominations to pair up
Within that chosen denomination, choose 2 of the 4 available cards
Of the 12 remaining denominations, choose 3 to get the other cards in the hand
This ensures that we don’t accidentally get two pairs or some higher hand
Within each of those 3 chosen denominations, choose one of the 4 available cards
So the numerator is: 13C1 x 4C2 x 12C3 x 4C1 x 4C1 x 4C1
Dividing by 52C5 gives P(one pair) = .423

4. Two Pairs Choose two of 13 denominations that will have the pairs.
Choose two of the four card in each chosen denomination to form the pairs.
Choose one of 11 denominations and one of its four cards to finish the hand.
Numerator: 13C2 x 4C2 x 4C2 x 11C1 x 4C1
Divide to get P(two pairs) = .047

5. Three of a Kind
Choose one of 13 denominations and choose 3 of its 4 cards
Choose two of the remaining 12 denominations and choose one of four cards from each
Numerator: 13C1 x 4C3 x 12C2 x 4C1 x 4C1
Divide to get P(3 of a kind) = .021

6. Straight
The order of the cards (low to high) is J-Q-K-A
The low card of any straight cannot be higher than a 10
So there are 9 denominations possible to start a straight
Choose one of the 9 denominations and choose one of its 4 cards
Choose one of 4 cards from the next higher denomination
Choose one of 4 cards from each of the next 3 higher denominations
Numerator: 9C1 x 4C1 x 4C1 x 4C1 x 4C1 x 4C1
Divide to get P(straight) = .0035

7. Flush Choose one of four suits for the flush Choose 5 of its 13 cards
Numerator: 4C1 x 13C5
Divide to get P(flush) = .0020

8. Full House
Is a full house of three 5’s and two 6’s the same as two 5’s and three 6’s?
No! The hand with the higher three of a kind would win.
That means order matters, so we start with permutation
Choose two of the 13 denominations to form the full house.
Within the denomination that will use 3 of a kind, choose 3 of its 4 cards (and order does not matter)
Within the denomination that will use 2 of a kind, choose 2 of its 4 cards.
Numerator: 13P2 x 4C3 x 4C2
Divide to get P(full house) = .0014

9. Four of a Kind Choose one of 13 denominations and use all 4 cards
Choose one of 12 remaining denominations and choose one of its 4 cards
Numerator: 13C1 x 4C4 x 12C1 x 4C1
Divide to get P(4 of a kind) = .0002

10. Straight Flush Choose one of the 4 suits for the flush
Choose one of the 9 possible bottom cards to start the straight
Numerator: 4C1 x 9C1
Divide to get P(straight flush) =
Note: Some people distinguish between a straight flush and a “royal flush”, which is just a straight flush that goes 10-J-Q-K-A
There are only 4 ways to choose the suit, so the probability is 4 / 52C5 = 1.54 x 10-6
This was actually included in the calculation of P(straight flush) above

11. What About None of the Above?
To find the probability of getting a 5 card hand that contains none of the named groupings, just subtract from 1 all of the probabilities we have calculated.
The result is that there are 1,179,936 ways to get nothing, and the probability is .454

12. Summary Hand # of ways Probability One Pair 1,098,240 .4226 Two Pair
123,552
.0475
Three of a Kind
54,912
.0211
Straight
9,216
.0035
Flush
5,148
.0020
Full House
3,744
.0014
Four of a Kind
624
.0002
Straight Flush 36
.00001
Garbage!
1,179,936
.4540