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A random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit?

Choose one of the 4 suits, within that suit choose 13 available cards (without replacement,where order doesn't matter).

$4 \choose 1$ * $13 \choose 3$

Choose one of the remaining three suits, within that suit, choose 13 available cards.

$3 \choose 1$ * $13 \choose 3$

Same for the remaining two suits

$2 \choose 1$ * $13 \choose 3$

And the last suit available

$1 \choose 1$ * $13 \choose 3$

At this point I've picked $3*4=12$ cards. Hence I have one more left to pick. Which I pick from a pool of unpicked cards $52-12=40$

So the total number of ways in which I could get dealt a 13-card hand where I have at least 3 cards of each suit is:

$4 \choose 1$ * $13 \choose 3$ * $3 \choose 1$ * $13 \choose 3$ * $2 \choose 1$ * $13 \choose 3$ * $1 \choose 1$ * $13 \choose 3$ $ * 40$

Divide that by the total number of ways I can pick 13 cards from 52 ($52 \choose 13$)

$$ \frac{\binom{4}{1} * \binom{13}{3} * \binom{3}{1} * \binom{13}{3} * \binom{2}{1} * \binom{13}{3} * \binom{1}{1} * \binom{13}{3} * 40}{\binom{52}{13}} $$

What is wrong with my approach above?

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    $\begingroup$ As well as lulu's comment, you also should not have the 3*2*1 since the suits having 3 cards are all equivalent and so there is no "first", "second", "third" among them. $\endgroup$
    – David
    Commented Jan 13 at 10:40
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    $\begingroup$ I deleted my first comment as it only pointed out one error. To repeat that observation: you are overcounting by a factor of $4$ since there's no way to identify the "special" fourth card you added last. $\endgroup$
    – lulu
    Commented Jan 13 at 10:46
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    $\begingroup$ I see, that's helpful and makes sense. So the $\binom{4}{1}$, $\binom{3}{1}$, $\binom{2}{1}$ are redundant because I would have to pick all the suits regardless. Hence the answer would be $ \frac{\binom{13}{3}^4 * 40}{\binom{52}{13}} $ $\endgroup$
    – IGottaLearnMath
    Commented Jan 13 at 11:05
  • $\begingroup$ You may write an answer and accept that answer. $\endgroup$
    – Benjamin Wang
    Commented Jan 13 at 11:12
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    $\begingroup$ @IGottaLearnMath Your comment-answer still overcounts. (Think about how you're counting the suit in which $4$ cards occur.) $\endgroup$
    – paw88789
    Commented Jan 13 at 11:21

1 Answer 1

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Your denominator is correct. For the numerator, choose the suit from which four cards are drawn, select four cards from that suit, and select three cards each from each of the other suits, which can be done in $$\binom{4}{1}\binom{13}{4}\binom{13}{3}\binom{13}{3}\binom{13}{3}$$ distinguishable ways.

As David pointed out in the comments, the order in which the suits are selected does not matter, so there is no need to multiply by the $4!$ ways of selecting the suits. However, it does matter from which suit four cards are selected. There are four ways to select this suit. Also, as lulu pointed out in the comments, designating a card as the fourth card from a suit overcounts by a factor of $4$ since any of the four cards in that suit could be your additional card. Notice that selecting $5\spadesuit, 9\spadesuit, J\spadesuit$ as your three spades and $2\spadesuit$ as your additional card results in the same hand as selecting $2\spadesuit, 5\spadesuit, 9\spadesuit$ as your three cards and $J\spadesuit$ as your additional card.

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  • $\begingroup$ Is the adjustment by a factor of 4 already done in the binomial you presented above for the numerator? I’m also not sure why it matters which suits we pick 4 cards from because once we have picked 3 suits with 3 cards in the same suit, the 4th suit should follow/be implied. E.g. if I picked 3 hearts, spades, and diamonds, I have to fill the rest of my hand with at least 3 clubs, so why do I have to choose from that suit as the one that will have 4 cards? $\endgroup$
    – IGottaLearnMath
    Commented Jan 13 at 12:50
  • $\begingroup$ The suit from which four cards are selected is distinguished by its size. Therefore, we need to select which of the four suits will be the one from which four cards are drawn, which is why I included the factor $\binom{4}{1}$ in the numerator. Notice that there are $\binom{4}{3} = \binom{4}{1}$ ways to select the three suits from which exactly three cards are drawn. Once you select them, the remaining suit becomes the one from which four cards are drawn. In either approach, you have to distinguish which of the four suits is the one from which four cards are drawn. $\endgroup$
    – N. F. Taussig
    Commented Jan 13 at 13:24

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