Probability for #suits in 5 card poker hand

Question

Given a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits. All one suit is straight-forward - $\frac{\binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$- pick five different ranks, each from the same suit.

Likewise, 4 seems fairly simple: $\frac{\binom{4}{1}\binom{13}{2}\binom{13}{1}^3}{\binom{52}{5}}$ - pick one suit to grab two cards from, then pick one card from each other suit.

Its on 2 and 3 that I get kind of stuck - I'm not sure how to set them up! I don't see why something along the lines of

$\frac{\binom{4}{2}*\binom{26}{6} - \binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$ doesn't work for 2 suits; i.e. picking 2 suits, choose 5 cards, subtracting off the ways in which you could end up with one suit. Similarly, for 3 I would expect

$\frac{\binom{4}{3}*\binom{39}{5}-\binom{4}{2}*\binom{26}{5}}{\binom{52}{5}}$ to give the answer (picking 5 cards from the group containing 3 suits, subtracting off those hands with fewer than 3 suits), but if I sum the probabilities it comes out incorrectly.

Thank you very much for your help!

Answer 1

The "exclusion-based" formulation you're trying to use for exactly two suits works fine; you just have a minor error. You can choose the two suits in $4\choose 2$ ways, and then choose a five-card hand from the $26$ cards in those two suits in $26\choose 5$ ways. This counts a number of single-suit hands as well, which you want to exclude. In fact, each of the $4\times{13\choose 5}$ single-suit hands is included exactly $3$ times in your original count (once with each possible "partner suit"). After excluding these with the correct multiplicity, you have $$ {4\choose 2}{26\choose 5}-3{4\choose 1}{13\choose 5}=379236 $$ hands containing exactly two suits. As a double-check, you can calculate from the other direction ("inclusion-based"). A hand with two suits has either four cards from one suit and one from the other, or three cards from one suit and two from the other. There are $12$ ways to choose the major and minor suits. For each (ordered) pair of suits, there are ${13\choose 4}{13\choose 1} + {13\choose 3}{13 \choose 2}=31603$ ways to make a hand; the result is $12\times 31603 = 379236$ again.

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With the two-suit result in hand, you can finish the problem. The number of single-suit hands, as you argued, is $4\times{13\choose 5}=5148.$ The number of four-suit hands is $4\times 13^3\times{13\choose 2}=685464.$ Therefore, the number of three-suit hands (the only remaining possibility) is ${52\choose 5}-5148-379236-685464=1529112.$ To double-check this, note that a three-suit hand has $(1,2,2,0)$ or $(3,1,1,0)$ cards per suit. There are $12$ ways to choose the "loner" (first) and "excluded" (last) suits. For each (ordered) pair of suits, there are $13\times{13\choose 2}^2 + {13\choose 3}\times 13^2=127426$ ways to make a hand; the result is $12\times 127426 = 1529112$ again. The associated probabilities are: $$ \begin{eqnarray} P_1 &=& \frac{5148}{2598960} &=& 0.198\% \\ P_2 &=& \frac{379236}{2598960} &=& 14.592\% \\ P_3 &=& \frac{1529112}{2598960} &=& 58.836\% \\ P_4 &=& \frac{685464}{2598960} &=& 26.375\% \\ \end{eqnarray} $$

Answer 2

As an example of another technique, which is useful for counting poker hands, particularly two pairs, we count the $3$-suit hands.

There are two types of $3$-suit hand: $3$-$1$-$1$ and $2$-$2$-$1$.

We count the $3$-$1$-$1$ hands. The suit we have $3$ of can be chosen in $\dbinom{4}{1}$ ways. For each such way, the actual cards can be chosen in $\dbinom{13}{3}$ ways. Now the suits we have $1$ of can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{1}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2.$$

Next we count the $2$-$2$-$1$ hands. The suit in which we have a singleton can be chosen in $\dbinom{4}{1}$ ways. For each such choice, the actual card can be chosen in $\dbinom{13}{1}$ ways. Now the suits we have doubletons in can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{2}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{1}\binom{3}{2}\binom{13}{2}^2.$$