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In the popular game of bridge, each of four players is dealt a hand of 13 cards from a well shuffled deck of 52 standard playing cards. Find the probability that any randomly dealt hand of 13 cards contains all three face cards of the same suit, where a face card is a Jack, a queen or a king.

My attempt:

  • There are $\binom{52}{13} = 635,013,559,600$ possible hands.
  • Because there are 13 cards from a well shuffled deck of 52 cards, to obtain a random dealt hand of 13 cards containing all three face cards of the same suit, we need to have $[\binom{3}{3}* \binom{49}{10}]/\binom{52}{13}$

Am I doing the right thing?

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  • $\begingroup$ Unfortunately not. That would be the probability that a hand contains all three face cards of a specific suit, e.g. spades. You need to use inclusion-exclusion for this. $\endgroup$ – JMoravitz Aug 26 '18 at 22:26
  • $\begingroup$ I am not sure how to use that here. Any help with that? $\endgroup$ – Lady Aug 26 '18 at 22:26
  • $\begingroup$ What do you already know about inclusion-exclusion principle? We dont' care which suit contains all three of its face cards, it could be clubs with all three, it could be spades, it could be hearts, it could be all of those at the same time, etc... What do you suppose might be useful events over which to run inclusion-exclusion over? $\endgroup$ – JMoravitz Aug 26 '18 at 22:29
  • $\begingroup$ What about using combinations? Won't that help with this question? $\endgroup$ – Lady Aug 26 '18 at 22:31
  • $\begingroup$ What about "what about using combinations?" Of course we're going to use combinations., but that isn't the only tool we're using for this problem. $\endgroup$ – JMoravitz Aug 26 '18 at 22:31
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There are indeed $\binom{52}{13}$ different 13-card hands and this will indeed be the size of our sample space and thus our denominator when we finish our calculations.

For the numerator, we need to pause for a moment and understand what the problem is actually asking, since this appears to be where you got stuck.

We are asked to find the probability that in our hand of thirteen cards, there is at least one suit for which we have all three face cards. For example $(A\spadesuit,2\spadesuit,3\spadesuit,\dots,10\spadesuit,J\spadesuit,Q\spadesuit,K\spadesuit)$ has all three of the face cards for spades. Similarly if all those cards happened to be hearts instead it would also count since we would have all of the face cards for hearts. Similarly still, a hand like $(J\spadesuit,Q\spadesuit,K\spadesuit,J\heartsuit,Q\heartsuit,K\heartsuit,J\diamondsuit,Q\diamondsuit,K\diamondsuit,\dots)$ would count since we have all of the face cards from spades (we also happen to have all of the face cards from hearts and diamonds too).

Let $\spadesuit$ represent the event that we have have all of the face cards from spades. Similarly, let $\diamondsuit, \heartsuit, \clubsuit$ represent the event that we have all of the face cards from diamonds, hearts, and clubs respectively.

You are asked to find $Pr(\spadesuit\cup\diamondsuit\cup \clubsuit\cup \heartsuit)$


To do this, let us apply inclusion exclusion. We expand the above as:

$Pr(\spadesuit\cup \diamondsuit\cup\clubsuit\cup\heartsuit) = Pr(\spadesuit)+Pr(\diamondsuit)+\dots-Pr(\spadesuit\cap \diamondsuit)-Pr(\spadesuit\cap \clubsuit)-\dots+Pr(\spadesuit\cap \diamondsuit\cap \clubsuit)+\dots-Pr(\spadesuit\cap\diamondsuit\cap \clubsuit\cap \heartsuit)$

Now, let us calculate each individual term in the expansion.

The calculation you did before is relevant. Indeed, we calculate $Pr(\spadesuit)=\dfrac{\binom{3}{3}\binom{49}{10}}{\binom{52}{13}}$. This is again merely the probability that we have all of the face cards from the spades and is not the final probability that we were tasked with calculating.

We continue and calculate more terms:

For example $Pr(\spadesuit\cap \diamondsuit)=\dfrac{\binom{6}{6}\binom{46}{7}}{\binom{52}{13}}$

We then notice what symmetry there is in the terms and can simplify some. Finally, we write the final expression for our final answer (and get an exact number only if actually requested or required, usually opting to leave the answer in terms of binomial coefficients without additional simplification).

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  • $\begingroup$ thank you so much. $\endgroup$ – Lady Aug 26 '18 at 23:34

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