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How many different bridge hands have exactly $2$ six-card suits?

I think the answer is :

$C(13,6)$$C(4,1)$$C(13,6)$$C(4,1)$$C(26,1)$

First I choose $6$ numbers from $13$ numbers, and choose $1$ suit for one $6$ card suit. Since there are $2$ six-card suits, I need to multiply the same thing again. And the remaining $1$ card can be chosen from $26$ cards.

Does that sound reasonable?

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2 Answers 2

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In the game of bridge, we have thirteen card hands. The phrase "exactly two six-card suits" implies exactly what it says, that there is a suit with six cards, a second different suit with six cards, and a third suit different than both previous with one card.

For example: $$\begin{array}~ \heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$

Break it up via multiplication principle.

  • Choose which two suits are each represented by six cards: (we must choose both of these simultaneously since otherwise we cannot distinguish between having chosen $\spadesuit$ and then $\heartsuit$ versus $\heartsuit$ and then $\spadesuit$) (Thus there are $\binom{4}{2}$ number of choices for this step)
  • Choose which suit is represented by a single card: ($\binom{2}{1}$ number of choices for this)
  • Choose which numbers are used in each suit: (at this point we have already chosen which suit is which and can distinguish between them. For each of the suits with six cards there will be $\binom{13}{6}$ choices and for the suit with one card there will be $\binom{13}{1}$ choice)

Thus, there are:

$$\binom{4}{2}\binom{2}{1}\binom{13}{6}\binom{13}{6}\binom{13}{1}$$ different bridge card hands satisfying these conditions.

(note: You could have opted to use $\binom{26}{1}$ to represent choosing the one card that wasn't one of the two suits with six-cards. $\binom{26}{1}=\binom{2}{1}\binom{13}{1}$)

(note further: If you used $\binom{4}{1}$ for the first suit, and then $\binom{3}{1}$ for the second suit, you have double-counted. The sequence of choices "choose hearts, choose spades, choose diamonds, choose A23456, choose A23456, choose A" yields the same result as "choose spades, choose hearts, choose diamonds, choose A23456, choose A23456, choose A", thus we must pick both suits simultaneously and use $\binom{4}{2}$ instead)

$$\begin{array}~ \heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$

vs

$$\begin{array}~ \spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$

See the related poker question: Probability of getting two pair in poker

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I think it should be C(13,6)C(4,2)C(13,6)C(26,1), since the second suit must be different from the first. Other than that, it looks fine to me.

I think your tutors answer doesn't require the suits form the sextuples to be different, and it doens't require the extra card to be form a different suit than the sextuples. However, than the correct answer is:

$$C(13,12)C(4,1)C(39,1)+ C(13,13)C(4,1)+ C(4,2)C(13,6)C(13,6)C(26,1)+ C(4,1)C(3,1)C(13,7)C(13,6)$$

So you should ask your tutor about the exact conditions for this problem.

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  • $\begingroup$ But one of my tutor give me such an answer:C(4,1)*C(13,12)*C(40,1)+C(4,2)*C(13,6)*C(13,6)*C(40,1) $\endgroup$
    – JimfyWinsy
    Commented May 10, 2015 at 17:03
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    $\begingroup$ @wythagoras the condition seems quite clear. A hand with one twelve-card suit is definitely not a hand with exactly two six-card suits. Also, read my answer as to why you should not use $\binom{3}{1}$ for selecting the second suit. Instead you should have selected both sextuple suits simultaneously. You have as a result doublecounted. $\endgroup$
    – JMoravitz
    Commented May 10, 2015 at 17:42
  • $\begingroup$ @JMoravitz Indeed you are right, I edited it, but it doesn't explain why the tutor got the answer thet JimfyWinsy gave. The 40 implies that there is chosen form the remaining 52-12=40 cards, so including those form the same suit. $\endgroup$
    – wythagoras
    Commented May 10, 2015 at 17:58
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    $\begingroup$ Teachers and tutors alike are prone to the same human mistakes as the rest of us. The tutor seems to have been confused as to the specific interpretation of the question or the specifics of answering a question like this in general. Even with his (flawed) presumed interpretation of the problem, he has overcounted, in which case your alternate answer seems to (almost) fix his mistakes (you still have a $\binom{4}{1}\binom{3}{1}$ in the third summand that should be a $\binom{4}{2}$). I would again reiterate that in problems like these "six-card suit" implies "exactly six-card" not "at least." $\endgroup$
    – JMoravitz
    Commented May 10, 2015 at 18:04
  • $\begingroup$ Yeah, you are right. +1 to your answer. $\endgroup$
    – wythagoras
    Commented May 10, 2015 at 18:19

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