[Math] the probability that a $5$ card poker hand has at least one pair

Question

What is the probability that a $5$ card poker hand has at least one pair (possibly two pair, three of a kind, full house, or four of a kind)?

I need to use the inclusion-exclusion principle.

My thinking thus far:$$P(0pair) = \frac{\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{10}{1}\cdot\binom{9}{1}}{\binom{52}{5}} \approx 0.059$$

Since we don't really need to worry about straights or flushes since the no pair probability includes those random choices, we now just need to subtract the no pair probability from $1$ to arrive at our conclusion:$$P(>=1pair) = 1 – P(0pair) = 1 – 0.059 = 0.941$$

Is my line of thinking correct here, or am I missing something?

Answer 1

The number of (five card) poker hands (from a standard 52-card deck) (where order of cards doesn't matter) which do not have any repeated ranks (i.e. no pairs, but possibly straight or flush) can be found via multiplication principle:

• pick the five distinct ranks used in the hand

  $\binom{13}{5}$

• pick the suit used for each rank

  $4^5$

The number of no-pair hands is then:

$\binom{13}{5}4^5$

Implying that the probability is:

$\binom{13}{5}4^5/\binom{52}{5}\approx 0.50708$

The probability then of having at least one pair is one minus the probability of no pairs and is then:

$1-\binom{13}{5}4^5/\binom{52}{5}\approx 0.49292$

Note: this could be found via the multiplication principle of probability instead of the multiplication principle of counting.

The first card can be picked to be anything: $\frac{52}{52}$. The next card would need to avoid all others of the same rank as the first, the probability of doing so would be $\frac{48}{51}$. The third card would need to avoid both of the first two with probability $\frac{44}{50}$. Continuing, we see $Pr(0pair)=\frac{52\cdot 48\cdot 44\cdot 40\cdot 36}{52\cdot 51\cdot 50\cdot 49\cdot 48}\approx 0.50708$. Continue as before by subtracting away from one.