# Probability that exactly three cards in at least one of the suits in an eight card hand from a standard deck.

so my solution was to...

a) find the number of hands where exactly one suit has exactly 3 cards b) find the number of hands where exactly two suits have exactly 3 cards c) add a) and b) then divide by c(52,8).

so starting with b) because it's slightly easier...

c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit... c(3,1) way to pick rest of suits and c(13,3) ways to pick cards in that suit... 2 cards left among 26, so just c(26,2).

so answer for b) would be - > 12*c(13,3)^2*c(26,2)

then going with a)

c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...

c(39,5) ways to pick rest of cards but must consider that we want to exclude possibility of having another 3 in another deck...

so what i did was count the ways in which we could have 3 cards in a suit among the 3 suits and 5 cards we need...

so c(3,1) ways of choosing a suit along with c(13,3) ways of selecting 3, and c(26,2) ways of selecting rest of the suits...

so c(39,5) - 3*c(13,3)*c(26,2) is my answer for a).

add them and divide by c(52,8) but i get a probability that is way too high...

Eight-card hands containing exactly three cards of one suit: Choose one of the four suits, three of the $13$ cards of that suit, and five of the $39$ cards from the remainder of the deck.
$$\binom{4}{1}\binom{13}{3}\binom{39}{5}$$

However, we have counted hands with exactly three cards of two suits twice, once for each way we could have designated one of those suits as being the suit from which exactly three cards are drawn. We only want to count such hands once, so we must subtract them from the total.

Eight-card hands containing exactly three cards of two suits: Choose two of the four suits, three cards of each selected suit, and two of the other $26$ cards in the deck. $$\binom{4}{2}\binom{13}{3}^2\binom{26}{2}$$

By the Inclusion-Exclusion Principle, there are $$\binom{4}{1}\binom{13}{3}\binom{39}{5} - \binom{4}{2}\binom{13}{3}^2\binom{26}{2}$$ eight-card hands with at least one suit from which exactly three cards are drawn. Since there are $\binom{52}{8}$ eight-card hands, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{13}{3}\dbinom{39}{5} - \dbinom{4}{2}\dbinom{13}{3}^2\dbinom{26}{2}}{\dbinom{52}{8}}$$

• thanks for the very clean solution. – Giancarlo Gatti Oct 31 '17 at 21:17

The numerator:

${13\choose 3}{13\choose 5}{13\choose 0}{13\choose 0}{4\choose 1,1,2} + {13\choose 3}{13\choose 4}{13\choose 1}{13\choose 0}{4\choose 1,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 2}{13\choose 0}{4\choose 2,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 1}{13\choose 1}{4\choose 2,2}+{13\choose 3}{13\choose 2}{13\choose 2}{13\choose 1}{4\choose 1,2,1} = 499163808$

$\frac {499163808}{{52\choose 8}} = 66.3\%$

• When I substitute your expression into a calculator (without the extra addition sign), I obtain $499163808$, which agrees with my answer. – N. F. Taussig Oct 31 '17 at 20:20
• @N.F.Taussig thanks, I see my error. – Doug M Oct 31 '17 at 20:23