## Poker probabilities

I got this problem from a friend, and although the answer can be found with a web search I will reproduce it here for future reference.

What are the probabilities of the different poker hands? First let me be specific in that I’m talking about the probability of getting one of the poker hands at the first draw from a full deck of cards. So we have 52 cards (no wild cards) and we draw 5, or we a number of players and we give 5 cards to each.

Both of the above situations are equivalent. There is no difference in the way you calculate the probability if you draw the first 5 cards or you draw the 2nd, 6th, 10th, 14th and 18th card from the deck.

#### Royal Flush

Royal flush is ace, king, queen, jack, ten in the same suit. The probability is:

$P_{royal\_flush} = \frac{4}{C(52, 5)} \approx 0.0001539\%$

There is only 1 possible combination for each suit that can give you a royal flush, multiplied with the number of suits.

#### Straight Flush

Straight flush is 5 cards in rank sequence and of the same suit. The probability is (including royal flush):

$P_{straight\_flush} = \frac{4 \cdot 10}{C(52, 5)} \approx 0.001539\%$

There are 10 ways of getting 5 cards from 14 in rank sequence (because we can have ace-high or an ace-low straight flush) for each suit, multiplied with the number of suits.

#### Four of a kind

Four of a kind means 4 cards of one rank, and an unmatched card of another rank. The probability is:

$P_{4\_of\_kind} = \frac{13 \cdot 48}{C(52, 5)} \approx 0.024\%$

There are 13 possible ways to get 4 cards of the same rank (that’s the number of ranks available) and there are (52 – 4) possible ways to get the remaining 5th card.

#### Full house

Full house means 3 matching cards of one rank, and 2 matching cards of another rank. The probability is:

$P_{full\_house} = \frac{P(13, 2) \cdot C(4, 3) \cdot C(4, 2)}{C(52, 5)} \approx 0.144\%$

There are $P(52, 5)$ possible ways to get 2 suits from the 13 available. This is because in this case it matters if for example we have 3 cards of clubs and 2 cards of spades or 3 cards of spades and 2 cards of clubs. There are $C(4, 3)$ and $C(4, 2)$ possible ways of getting 3 and respectively 2 cards from the 4 cards of the same rank.

#### Flush

Flush means 5 cards of the same suit, not in rank sequence, excluding the straight flush and royal flush. The probability is:

$P_{flush} = \frac{4 \cdot C(13, 5) - 40}{C(52, 5)} \approx 0.19654\%$

There are $C(13, 5)$ possible ways to get 5 cards from each suit and we have to subtract the 40 straight flushes (see above) from that number.

#### Straight

Straight means 5 cards in rank sequence but in more than one suit. The probability is:

$P_{straight} = \frac{10 \cdot 4^5 - 40}{C(52, 5)} \approx 0.39246468\%$

There are 10 ways of getting 5 cards in rank sequence and $4^5$ combinations available for those 5 cards. As before we have to subtract the 40 straight flushes from that number.

#### Three of a kind

Three of a kind means 3 cards of the same rank, plus 2 unmatched cards. The probability is:

$P_{3\_of\_kind} = \frac{13 \cdot C(4, 3) \cdot C(12, 2) \cdot 4^2}{C(52, 5)} \approx 2.1129\%$

OK, so there are $C(4, 3)$ possible ways of selecting 3 cards from 4 of the same rank, and we have to multiply this with the number of ranks. For the remaining 2 cards there are $C(12, 2)$ possible ways of selecting the rank so it would not match the rank of the first 3 cards, and $4^2$ combinations available for those 2 cards.

#### Two pair

Two pair means two cards of the same rank, plus two cards of another rank (that match each other but not the first pair), plus one unmatched card. The probability is:

$P_{2\_pair} = \frac{C(13, 2) \cdot C(4, 2) \cdot C(4, 2) \cdot 11 \cdot 4}{C(52, 5)} \approx 4.7539\%$

We have $C(13, 2)$ possible ways of selecting 2 different ranks from the 13 available, multiplied twice with $C(4, 2)$ which is the number of possible ways of selecting 3 cards from 4 of the same rank. For the remaining card we have only 11 suits available and 4 possible cards per suit.

#### One pair

One pair means two cards of the same rank, plus three other unmatched cards. The probability is:

$P_{2\_pair} = \frac{13 \cdot C(4, 2) \cdot C(12, 3) \cdot 4^3}{C(52, 5)} \approx 42.2569\%$

We have $C(4, 2)$ possible ways of selecting 2 cards from 4 of the same rank multiplied with the number of ranks. For the remaining 3 cards we have $C(12, 3)$ possible ways of selecting a different rank and $4^3$ combinations available for those 3 cards.

For a final overview, the below pie chart shows all the probabilities discussed previously (all of them are in the chart, just that some of them are really small).