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I draw a hand of 13 from a deck of 52 standard playing cards. What is the probability that I do not have a card from every suit?

I count the number of ways I can draw 13 from 3 suits

$$\frac{{4\choose3}{39\choose13}}{52\choose13}$$

but I mind the intersection. Each possible pair of suits that I may have drawn from only is counted twice. And in the possibility that I pick from only one suit: each possibility is counted three times.

$$\frac{{4\choose3}{39\choose13}-{4\choose2}{26\choose13}}{52\choose13}$$

This removes the overcounted iterations from the pair of suits I could have drawn from, but now I'm not considering the possibility that I drew from only one suit, so:

$$\frac{{4\choose3}{39\choose13}-{4\choose2}{26\choose13}+{4\choose1}{13\choose13}}{52\choose13}$$

which is the probability I'm looking for.

Is my reasoning sound? Have I made any mistakes? Is there a better solution?

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    $\begingroup$ Good use of Inclusion/Exclusion. $\endgroup$ – André Nicolas May 5 '15 at 1:17
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    $\begingroup$ @ whorl There are other ways of solving this, but they will be less elegant. Your way works perfectly fine. An example of a direct approach might have been: (For a simplification being a hand of 5 cards instead of 13). Break this into cases: distribution of suits (denoted most common to least) could be one of: (5,0,0,0), (4,1,0,0), (3,2,0,0), (3,1,1,0), or (2,2,1,0). Then count how many hands fall into each of those cases. Generally, casework like this should be avoided if at all possible as it can be very tedious. The approach for 13 card hands is similar but much longer. $\endgroup$ – JMoravitz May 5 '15 at 1:37
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    $\begingroup$ I kind of take it back. You really should not have divided by $\binom{52}{13}$ until the end. There should have been a count of the "favourables" (your final numerator), then the division. You call for example $\binom{4}{3}\binom{39}{13}/\binom{52}{13}$ a number of ways, but it isn't. And even $\binom{4}{3}\binom{39}{13}$ is not a count of anything, it is a step toward the final count. $\endgroup$ – André Nicolas May 5 '15 at 1:40
  • $\begingroup$ @AndréNicolas You're right, that mess is an artifact of my first two attempted solutions before I realized I wasn't paying attention to intersections. $\endgroup$ – whorl May 5 '15 at 1:44
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$\boxed{\color{green}{\checkmark}}$ Yes; by the Principle of Inclusion and Exclusion, the ways of not drawing from all four suits is:

  • Include: the $\binom{4}{1}\binom{52-13}{13}$ ways to pick one suit and not draw from this, then
  • Exclude: the $\binom{4}{2}\binom{52-26}{13}$ ways to pick two suits and not draw from both, then
  • Include: the $\binom{4}{3}\binom{52-39}{13}$ ways to pick three suits and not draw from these.

So the probability of doing this is, as you obtained:

$$\dfrac{\dbinom{4}{1}\dbinom{39}{13}-\dbinom{4}{2}\dbinom{26}{13}+\dbinom{4}{3}\dbinom{13}{13}}{\dbinom{52}{13}}$$


NB $\binom{4}{1}=\binom{4}{3}$

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