I am working on creating a magic trick and it got me wondering about this related probability question. Ten cards are chosen at random from a standard $52$ card deck. You may use only these ten cards to form a five card poker hand. What is the probability that you are able to form a hand that is at least as good as a three of a kind?
By "at least as good as a three of a kind", I specifically mean that we need to be able to create one of the following:
- Three of a Kind (Three cards with matching values, eg 7H, 7C, 7D, JH, 2S)
- Straight (Five consecutive card values, eg 6H, 7C, 8D, 9H, 10D)
- Flush (Five cards of the same suit, eg 6D, KD, 2D, 4D, JD)
- Full House (A three of a kind in one value AND a pair of another value, eg 9H, 9S, 9D, 3C, 3H)
- Four of a Kind (Four cards with matching values, eg 10C, 10H, 10S, 10D, 7H)
- Straight Flush (Five consecutive card values that also have matching suits, eg 6D, 7D, 8D, 9D, 10D)
- Royal Flush (A Straight flush with values 10, J, Q, K, A, eg 10H, JH, QH, KH, AH)
Best Answer
It'll be a substantial amount of work, but it's doable. My intuition says the probability should be pretty high - let's see how that works out.
Final result: Approximately $57.8\%$. Somewhat lower than I would have initially guessed, but still quite large.
And now, the calculation. I'll sort by hand type.
Many of a kind. We count all of the ten-card stacks in which it's possible to make 4 of a kind, full house, or 3 of a kind.
To get four of a kind, we choose the denomination and then choose six other cards, for $13\cdot \binom{48}{6}$ ways. Some of these double up with two denominations each having four; that happens in $\binom{13}{2}\cdot \binom{44}{2}$ ways, which we subtract off for their double-counting.
We won't separately track full house hands; every hand that could be a full house is already three of a kind, so these add nothing to the count.
To get three of a kind, choose a denomination, then choose three cards from that denomination and seven cards of other denominations. That's $13\cdot \binom{4}{3}\cdot\binom{48}{7}$ ways. We can also have three of two kinds in $\binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}$ ways, or three of three kinds in $\binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1}$ ways. Counting by inclusion-exclusion, we subtract the "three of two kinds" options and add back the "three of three kinds" options. It's also possible to get three and four in $13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3}$ ways, or three, three, and four in $13\cdot \binom{12}{2}\cdot\binom{4}{3}^2$ ways. We subtract the former since they're already counted as four of a kind hands, and add the latter back by inclusion-exclusion.
So far, with just counting hands of four of a kind, full house, or three of a kind, we're at \begin{align*}M &= 13\cdot \binom{48}{6} - \binom{13}{2}\cdot \binom{44}{2} + 13\cdot \binom{4}{3}\cdot\binom{48}{7} - \binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}\\ &\quad + \binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1} - 13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3} + 13\cdot \binom{12}{2}\cdot\binom{4}{3}^2\end{align*} ways out of the $N=\binom{52}{10}$ ways to choose ten cards. Computing it, $N\approx 1.582\cdot 10^{10}$ and $M\approx 3.811\cdot 10^9$, for a probability $\frac MN$ of about $24.1\%$.
Straights. Our count will exclude all "many of a kind" hands.
To get at straights, we look at the list of denominations among the cards we've drawn. We can have one, two, three, or four cards per denomination - but we've already counted everything with three or more of a kind, so we'll just look at ones and twos. The options:
Adding these up, straights that aren't also 3 or more of a kind account for \begin{align*}S &= 9\cdot 6^5 + (9\cdot 8-8)\cdot\binom{6}{2}\cdot 6^4\cdot 4^2 + \left(9\cdot \binom{8}{2}-8\cdot 7\right)\cdot \binom{7}{4}\cdot 6^3\cdot 4^4\\ &\quad +\left(9\cdot\binom{8}{3}-8\cdot\binom{7}{2}\right)\cdot \binom{8}{6}\cdot 6^2\cdot 4^6 + \left(9\cdot\binom{8}{4}-8\cdot\binom{7}{3}\right)\cdot \binom{9}{8}\cdot 6\cdot 4^8\\ &\quad +\left(9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6\right)\cdot 4^{10}\end{align*} Computing that, it's about $3.253\cdot 10^9$, for an added probability $\frac{S}{N}\approx 20.6\%$. We're up to nearly half.
Since straight flushes, including the royal variety, are already counted as both straights and flushes, we won't be treating them separately. Instead, we'll look at stacks that contain both a straight and a flush - even if those aren't the same five-card hand. That can wait, though.
Flushes. We sort by the length of the long suit, then account for the elements outside that suit. The elements off-suit may cause us to double-count with three or more of a kind, so we'll subtract them.
Adding these up, flushes that aren't also 3 or more of a kind account for \begin{align*}F &= 4\binom{13}{9}\cdot 39 + 4\binom{13}{8}\left(\binom{39}{2}-8\binom{3}{2}\right) + 4\binom{13}{7}\left(\binom{39}{3}-13-7\binom{3}{2}36\right)\\ & +4\binom{13}{10}+ 4\binom{13}{6}\left(\binom{39}{4}-13\binom{36}{1}-6 \binom{3}{2} \binom{36}{2}+\binom{6}{2}\binom{3}{2}^2\right) - \binom{4}{2}\binom{13}{5}^2\\ & + 4\binom{13}{5}\left(\binom{39}{5}-13\binom{36}{2}-5\binom{3}{2}\binom{36}{3}+5\binom{3}{2} 12+\binom{5}{2}\binom{3}{2}^2 33\right) \end{align*} Computing that, it's about $2.922\cdot 10^9$, for an added probability of about $18.5\%$. We're up past $60\%$ overall.
Straights and flushes. Unfortunately, we still have some overcounting to deal with. It's possible to have both a straight and a flush in the same stack, and it's possible with any level of overlap. We'll work with the straight count from earlier to find the doubles. Sorting in the same way:
Adding these up, stacks with both straights and flushes but not three or more of a kind come to \begin{align*}X &= 9\cdot \left(4\cdot 3^5 - \binom{4}{2}\right) + (9\cdot 8-8)\cdot \binom{6}{2}\cdot 4\left(5\cdot 3^4 + 2\cdot 3^5 - 3\right)\\ & + \left(9\cdot 28-8\cdot 7\right)\cdot\binom{7}{4}\cdot 4\left(7\cdot 3^3+16\cdot 3^4+6\cdot 3^5 - 9\right)\\ & + \left(9\cdot 56-8\cdot 28\right)\cdot\binom{8}{6}\cdot 4\left(4\cdot 3^2+24\cdot 3^3+45\cdot 3^4+20\cdot 3^5-30\right)\\ & + \left(9\cdot 70-8\cdot 35\right)\cdot\binom{9}{8}\cdot 4\left(2\cdot 3+16\cdot 3^2+56\cdot 3^3+112\cdot 3^4+70\cdot 3^5 - 105\right)\\ & + \left(9\binom{8}{5} - 8\binom{7}{4} - 6\right) 4 \left(1+10\cdot 3+45\cdot 3^2+120\cdot 3^3+210\cdot 3^4+252\cdot 3^5 - 378\right)\end{align*} Computing that, it's about $0.848\cdot 10^9$, for a probability of about $5.4\%$.
The overall probability we seek is $\dfrac{M+S+F-X}{N} = \dfrac{9139533260}{15820024220}\approx 57.8\%$.