Problem 3.1 Easy Free 5 variations

Four of a Kind and Poker Hands

A poker hand consists of five cards dealt from a standard $52$ -card deck. Let $p$ be the probability that the hand is a four-of-a-kind (four cards of one rank, plus one other card). Find $1/p$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

All $\binom{52}{5}=2\,598\,960$ hands are equally likely. To build a four-of-a-kind, first choose the rank of the quadruple: there are $13$ ways. That rank contributes all four of its cards, so there is no suit choice. The fifth card can then be any of the remaining $48$ cards. Each such hand arises exactly once, so there are $13\cdot 48=624$ four-of-a-kind hands. Hence

p=\frac{624}{2\,598\,960}=\frac{1}{4165}, \qquad \boxed{\;\frac1p=4165\;}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 3.1a

A poker hand consists of five cards dealt from a standard $52$ -card deck. Find the probability that it is a full house: three cards of one rank and two cards of another rank.

Work it out, then compare with the answer.

Solution

The triple and the pair play different roles, so their ranks are chosen in order.

Ranks: $13$ choices for the triple, then $12$ for the pair.
Suits: $\binom43=4$ for the triple and $\binom42=6$ for the pair.

This gives $13\cdot12\cdot4\cdot6=3744$ hands, and

\boxed{\frac{3744}{\binom{52}{5}}=\frac{6}{4165}\approx 0.00144}

Variation 3.1b

A poker hand consists of five cards dealt from a standard $52$ -card deck. Find the probability that it is two pair: two cards of one rank, two cards of a different rank, and one card of a third rank.

Work it out, then compare with the answer.

Solution

The two pairs play the same role, so their ranks are chosen as an unordered set.

Pair ranks: $\binom{13}{2}=78$ ways.
Suits: $\binom42^2=36$ for the two pairs.
Kicker: any of the $44$ cards of the other $11$ ranks.

This gives $78\cdot36\cdot44=123\,552$ hands, and

\boxed{\frac{123\,552}{\binom{52}{5}}=\frac{198}{4165}\approx 0.0475}

Variation 3.1c

A poker hand consists of five cards dealt from a standard $52$ -card deck. Find the probability that it contains exactly three cards of one rank, with the remaining two cards of two different ranks (not a full house).

Work it out, then compare with the answer.

Solution

Triple: $13$ choices of rank and $\binom43=4$ choices of suits.
Other two cards: their ranks are interchangeable, so choose them as a set in $\binom{12}{2}=66$ ways, then a suit for each in $4^2=16$ ways.

This gives $13\cdot4\cdot66\cdot16=54\,912$ hands, and

\boxed{\frac{54\,912}{\binom{52}{5}}=\frac{88}{4165}\approx 0.0211}

Variation 3.1d

A poker hand consists of five cards dealt from a standard $52$ -card deck. Find the probability that it contains exactly one pair: two cards of one rank and three cards of three other, mutually distinct ranks.

Work it out, then compare with the answer.

Solution

Pair: $13$ choices of rank and $\binom42=6$ choices of suits.
Other three cards: choose their ranks as a set in $\binom{12}{3}=220$ ways, then a suit for each in $4^3=64$ ways.

This gives $13\cdot6\cdot220\cdot64=1\,098\,240$ hands, and

\boxed{\frac{1\,098\,240}{\binom{52}{5}}=\frac{352}{833}\approx 0.423}

Variation 3.1e

You are dealt five cards from a standard $52$ -card deck and told that at least three of them share the same rank. Given this information, find the probability that all four cards of that rank are in your hand.

Work it out, then compare with the answer.

Solution

Five cards cannot contain three cards of each of two ranks, so the rank appearing at least three times is unique. The two cases are therefore disjoint.

Exactly three of the rank: $13$ ranks, $\binom43=4$ suit choices, and $\binom{48}{2}=1128$ ways to pick the other two cards from the other ranks. This gives $58\,656$ hands, full houses included.
All four of the rank: $13\cdot48=624$ hands.

Hence

\boxed{\frac{624}{624+58\,656}=\frac{624}{59\,280}=\frac{1}{95}\approx 0.0105}

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