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Given a deck of 40 cards, with 4 suits of 10 cards each and cards numbered from 1 to 10, we draw 4 cards. Let the random variable N be the number of different suits. I am asked to compute $P(N=2)$.

The solution provided by using Laplace rule is:

Number of possible draws: ${40 \choose 4}=91390$

Number of positive outcomes: $4{10 \choose 3}*3*10+{4 \choose 2}{10 \choose 2}{10 \choose 2}=26550$

So $P(N=2)\approx0.29$.

The reasoning for this I understand. But I appraoched the problem differently and the result I get is different and therefore wrong. Here it is:

I first computed the probability of drawing, say, hearts, hearts, spades and spades, which would be $P(HHSS) = \frac{10}{40}\frac{9}{39}\frac{10}{38}\frac{9}{37}$. We want to take different orderings into account, so we multiply by ${4 \choose 2}$. And also we don't only want hearts and spade, but all suit combinations, so I multiply by ${4 \choose 2}$ again.

This is for having 2 of each suit, but we can also have 1 and 3 of each, so similarly:

$P(HHHS)=\frac{10}{40}\frac{9}{39}\frac{8}{38}\frac{10}{37}$. We also take ordering into account so multiply by ${4 \choose 1}$ and same as above, I account for the different suit combinations with ${4 \choose 2}$. The final expression is:

$P(N=2)={4 \choose 2}{4 \choose 2}\frac{10}{40}\frac{9}{39}\frac{10}{38}\frac{9}{37}+{4 \choose 2}{4 \choose 1}\frac{10}{40}\frac{9}{39}\frac{8}{38}\frac{10}{37}$

but this results in a number close 0.21.

Could someone please help me figure out what is wrong?

Thanks in advance

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    $\begingroup$ For the $2\times 2$ case you are correct to think that $(H,S)$ is the same as $(S,H)$ but that's not true for the $3\times 1$ case. $\endgroup$
    – lulu
    Commented Aug 27, 2023 at 13:48
  • $\begingroup$ Now I see, thank you very much. Feel free to add it as an answer (maybe with more detail) and I will accept it. Othersie I will add the answer myself. $\endgroup$
    – ERed
    Commented Aug 27, 2023 at 14:57
  • $\begingroup$ Feel free to write it up yourself. Your computation was $99\%$ complete...you just need to double your last case to finish. $\endgroup$
    – lulu
    Commented Aug 27, 2023 at 15:30
  • $\begingroup$ Also, in your calculation, you wrote $\binom{4}{2}$ rather than $\binom{4}{1}$ for the three of one suit and one of a different suit case. $\endgroup$
    – N. F. Taussig
    Commented Aug 27, 2023 at 18:19

1 Answer 1

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The reason my calculation was wrong is due to the fact that I was only counting the number of ways in which I can combine HHHS (3 hearts 1 spade) by changing the position of S with the term ${4 \choose 1}$ but I also need the number of ways in which SSSH (3 spades and 1 heart) can occur, as pointed out. So by multiplying the second term by 2 (due to summetry) the result is now correct.

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