Will you fold?

Probability Level 5

In most five-card poker games, a hand is an unordered set of five cards. The following is the ranking of poker hands. In case an hand belongs to both possible classification of hands, only the higher rank is considered .

$1$ . Royal Flush: The set of best cards from 10 to A all in the same suit.

A♦ K♦ Q♦ J♦ 10♦

$2.$ Straight Flush: Five Cards of Consecutive ranks of the same suit.

Q♣ J♣ 10♣ 9♣ 8♣

$3.$ Four of a Kind: Four Cards of the same rank.

9♣ 9♠ 9♦ 9♥ J♥

$4.$ Full House: Two Cards of the same rank and three other cards of a similar rank.

6♠ 6♥ 6♦ A♠ A♣

$5.$ Flush: Five Cards all in the same suit

K♥ Q♥ 9♥ 5♥ 4♥

$6.$ Straight: Five Cards of consecutive ranks

10♣ 9♦ 8♥ 7♣ 6♠.

$7.$ Three of a kind: Three cards of the same rank

2♦ 2♠ 2♣ K♠ 6♥

$8.$ Two Pairs: Two pairs of cards each consisting of a pair of matching ranked cards.

10♠ 10♣ 4♠ 4♥ 8♥

$9.$ One Pair: A pair of cards of the same rank

4♥ 4♠ K♠ 10♦ 5♠

If a hand does not match any of the above, it can be called a high-card or a no-pair .

How many such high-card hands exist provided you're choosing from a standard deck?

The answer is 1302540.

1 solution

Brian Charlesworth
Aug 13, 2015

There are three conditions that need to be met:

(i) the denomination of each card must be distinct from all the others,
(ii) the five denominations chosen cannot be arranged to form a straight, and
(iii) the suits for the five cards chosen cannot all be the same.

Disregarding suits for the moment, there are $\dbinom{13}{5}$ ways to choose five denominations, but $10$ of these involve denominations that can be arranged to form a straight, (namely Ace through $5$ up to $10$ through Ace).

Now bringing in the factor of suits, we have $4$ suit choices for each of the $5$ denominations, i.e., $4^{5}$ ways of choosing the suits, but $4$ of these will involve all denominations having the same suit.

Thus the number of strictly high-card hands is

$\left(\dbinom{13}{5} - 10\right)(4^{5} - 4) = \boxed{1302540}.$

did not expect such a huge number!

Avinash Kamath - 5 years, 9 months ago

It is large, and it represents just over 50% of all possible hands. The probability of getting a single pair is

$\dfrac{\dbinom{13}{1} \dbinom{4}{2} \dbinom{12}{3} \dbinom{4}{1}^{3}}{\dbinom{52}{5}} = 0.422569...,$

so all the hands other than high card and single pair constitute just around 7.7% of all possible hands.

Brian Charlesworth - 5 years, 9 months ago

Will you fold?

The answer is 1302540.

1 solution

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