How many 5-cards poker hands are there containing at least 3 of the 4 suits?

How many 5-cards poker hands are there containing at least 3 of the 4 suits?

I don't how know how to properly handle this question.

There 13 ranks of each card, and 4 suits

My Attempt:

Begin with the first 3: C(13,3) X 4 ways to choose suits

Then pick last 3 cards of anything:

C(49,2)

My solution: 1,345,344

Thanks,

• That technique will definitely count some hands multiple times. Can you see why? – Thomas Andrews Nov 23 '15 at 20:06
• Try the negative probability: the hands with just one or two suites. – Masacroso Nov 23 '15 at 20:10
• It's probably easier to count the number of hands with one or two suits, and then subtract them from the number of all hands. – Thomas Andrews Nov 23 '15 at 20:10
• One sanity check is that you should get a multiple of $13^3$. – Thomas Andrews Nov 23 '15 at 20:16
• How do i count the number of hands with one suit? and why one OR two? – chickenninja565 Nov 23 '15 at 20:19

EDIT ( thanks to Thomas ) ****

There are $\binom{52}{5}$ possible 5 card hands $$N_{tot} = \binom{52}{5}$$

for the number of hands containing just one suit there are 4 ways to choose the suit and $\binom{13}{5}$ ways to choose 5 cords from that suit

$$N_{1} = 4\binom{13}{5}$$

For the number of hands containing exactly 2 suits there are $\binom 42$ ways to choose the 2 suits
for each pair of suits there are $\binom{26}{5}$ 5 card hands, but that includes $\binom {13}{5}$ one suit hands from each of the 2 suits. $$N_{2} = \binom 42 \left(\binom{26}{5}-2\binom{13}{5} \right)$$ so $$N_{\text{at least 3}}=N_{tot}-N_1-N_2 = \binom {52}{5} - 4\binom{13}{5} - \binom 42 \left(\binom{26}{5}-2\binom{13}{5} \right)$$

P(all from one suit) $$= \frac {4\binom{13}{5}}{\binom{52}{5}}$$

P(all from two suits) $$= \frac {\binom 42 \binom{26}{5}}{\binom{52}{5}}$$

P(all from at least 3 suits) $$= 1-\frac {4\binom{13}{5}}{\binom{52}{5}} - \frac {\binom 42 \binom{26}{5}}{\binom{52}{5}}$$

• The second line over-counts - it counts some cases from the first line. You need $\binom{4}{2}\left(\binom{26}{5}-2\binom{13}{5}\right)$ in the numerator to not count single-suited hands. Also, OP was seeking a count, which easier to write - fractions are noisy. – Thomas Andrews Nov 23 '15 at 20:17
• sorry - I deleted my comment before I saw yours – WW1 Nov 23 '15 at 20:26
• Edit it so your answer is correct. Don't make readers read the wrong answer and then say, "oops, here's the correct answer." Put the correct answer at the top, and, if you absolutely must including the original wrong answer, put it after with text loudly saying "wrong answer." People don't come here for wrong answers. – Thomas Andrews Nov 23 '15 at 20:35

You can calculate the number of hands with at least three of the four suits directly as follows:

Case 1: All four suits are used.

One of the four suits must have receive cards. There are $\binom{13}{2}$ ways of selecting two of the thirteen cards from that suit. For each of the remaining suits, we can choose one of the $13$ cards of that suit in $\binom{13}{1}$ ways. Hence, the number of five card poker hands in which all of the suits are used is $$\binom{4}{1}\binom{13}{2}\binom{13}{1}^3$$

Case 2: Exactly three of the four suits are used.

There are two possibilities. Either there are three cards of one suit and one of each of the other two selected suits or there are two suits with two cards and one suit with one card.

1. There are three cards of one suit and one each of the other two selected suits. One of the four suits receives three cards. There are $\binom{13}{3}$ ways of selecting three of the thirteen cards from that suit. There are $\binom{3}{2}$ ways to select two of the three remaining suits. There are $\binom{13}{1}$ ways of selecting one card from each of these suits. Hence, there are $$\binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2$$ ways of selecting three different suits in which there are three cards of one suit and one card apiece of the other two selected suits.
2. There are two suits with two cards and one suit with one card. There are $\binom{4}{2}$ ways to select two of the four suits to receive two cards each. For each of those suits, there are $\binom{13}{2}$ ways of selecting two cards from that suit. There are $\binom{2}{1}$ ways to select the remaining suit. There are $\binom{13}{1}$ ways of selecting one card from that suit. Hence, there are $$\binom{4}{2}\binom{13}{2}^2\binom{2}{1}\binom{13}{1}$$ ways of selecting three different suits if two of those suits receive two cards apiece and the third suit receives one card.

Since the cases are mutually disjoint, the total number of poker hands that contain at least three different suits is $$\binom{4}{1}\binom{13}{2}\binom{13}{1}^3 + \binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2 + \binom{4}{2}\binom{13}{2}^2\binom{2}{1}\binom{13}{1}$$