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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?

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    $\begingroup$ This is the correct line of thinking, however your value for $P(0pair)$ is incorrect. Break it up via multiplication principle: pick the five ranks (simultaneously to avoid overcounting). Then, pick which suit is used for each rank. $\endgroup$
    – JMoravitz
    Commented Mar 30, 2016 at 2:45
  • $\begingroup$ Do I pick what suit is used for each rank to avoid a flush? If this is the case, what about a straight? $\endgroup$
    – Jodo1992
    Commented Mar 30, 2016 at 2:45
  • $\begingroup$ Do not concern yourself with flushes or straights. If there are two cards of the same rank in the hand, it is impossible for it to be a flush or a straight (since the pair would use different suits it can't be a flush, since the pair uses the same rank it can't be a straight) $\endgroup$
    – JMoravitz
    Commented Mar 30, 2016 at 2:47
  • $\begingroup$ True, okay I will update my question once I work that out. $\endgroup$
    – Jodo1992
    Commented Mar 30, 2016 at 2:47
  • $\begingroup$ @Jodo1992 I think what JMoravitz is saying is that your evaluation gives the number of unique "numbers" ("numbers" including jack, queen, king, and ace). However there are various different permutations of, for instance a) Clubs: 2,3,4,5,6; b) Diamonds, 2,3,4,5,6; c) Clubs 2,3 Diamonds 4,5,6; d) Hearts 2, Clubs 3, Spades 4, Diamonds 5, Clubs 6; etc. This makes sense because the idea that a pair is 94% likely is unintuitive. $\endgroup$
    – Jared
    Commented Mar 30, 2016 at 2:52

2 Answers 2

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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.

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  • $\begingroup$ I like the explanation of the multiplication principle. I (think I) understand the other argument (the ranks), but I would like a little more explanation. To me "rank" is a bad descriptor--I'm not sure what's better, but perhaps add that "rank" means a collection of 5 distinct "faces", hence there are $\binom{13}{5}$ distinct collections of distinct "faces". $\endgroup$
    – Jared
    Commented Mar 30, 2016 at 3:01
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    $\begingroup$ @Jared, "rank" is the term used for the number or letter on the card. There are thirteen ranks in the standard 52-card deck (A,2,3,...,10,J,Q,K). $\endgroup$
    – JMoravitz
    Commented Mar 30, 2016 at 3:03
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I get. $P($no pair$) = \dfrac{52*48*44*40*36}{52*51*50*49*48}$

You could also say choose 5 of 13 numbers, each of 4 suits over 5 of 52 cards.

$\dfrac {4^5 {13\choose {5}}}{52\choose5}$

This "no pair accounting allows for the possibility of a flush or a straight.

Easiest to do each of them on their own... and since all straight and flush hands overlap the no-pair hands you can just subtract.

flush: $\dfrac {4 {13\choose {5}}}{52\choose5}$ straight: $\dfrac {10*4^5}{52\choose5}$ and each of those include the straight flush: $\dfrac {4*10}{52\choose5}$

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