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Suppose I draw 4 cards from a fair deck. What is the probability that I will end up with a hand with 4 different suits? What about 2 distinct suits or 3 distinct suits or all the same suit?

For 4 different suits, my solution is (13^4)∕52C4 which is about 0.105

[13 possible cards from each suit]

For 3 distinct suits, my solution is (4C3)(13C2)(13^2)/(52C4) which is about 0.195

[4C3 is me choosing 3 suits out of the 4, 13C2 is me choosing 2 cards from a specific suit]

For 2, my solution is (4C2)(13C3)(13)/(52C4)+ (4C2)(13C2)(13C2)/(52C4) which is about 0.217

[Case 1: 3 of the same suit and 1 of a different suit, Case 2: 2 of the same suit then another of the same suit]

For all the same suit, my solution is 4(13C4)/(52C4) which is about 0.011

[4 suits to pick from, then choose 4 cards from the suit]

As far as I know, the total probability is supposed to add up to 1 (in this situation I can only draw either all of the same suit, 2 distinct suits, 3 distinct suits, or all different suits), but when I add up my solutions, it doesn't even reach 1. Obviously, I did something wrong, but I can't figure out where. Help?

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2 Answers 2

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1 suit $\dfrac {{4\choose 1}{13\choose 4}}{{52\choose4}} \approx 0.011 $

2 suits - consider 3,1 and 2,2 splits separately

3,1 split $\dfrac {{4\choose 1}{13\choose 3}\cdot {3\choose 1}{13\choose 1}}{{52\choose4}}\approx 0.135$

2,2 split $\dfrac {{4\choose 2}{13\choose 2}^2}{{52\choose4}}\approx 0.165$

3 suits $\dfrac {{4\choose 1}{13\choose 2}{3\choose 2}{13\choose 1}^2}{{52\choose4}}\approx 0.584$

4 suits $\dfrac {{4\choose 4}{13\choose 1}^4}{{52\choose4}}\approx 0.105$

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For 2 suits, case 1, you need to specify which suit you will have three of, and which suit you will only have one of. That makes it into $$\frac{4\cdot 3\cdot\binom{13}{3}\cdot 13}{\binom{52}{4}}$$ Same thing with three suits: pick one suit to have two of, and then two suits to only have singles:$$\frac{4\cdot\binom32\binom{13}2\cdot13^2}{\binom{52}{4}}$$

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  • $\begingroup$ For 2 suites case 2, does it also turn into (4)(3)(13C2)(13C2)/(52C4), as I choose which suite gets a pair and which suite gets the other pair? $\endgroup$
    – J. Jo
    Feb 22, 2017 at 17:51
  • $\begingroup$ @J.Jo No, for case 2, it's still $\binom42$, since the roles of the two suits are the same. $\endgroup$
    – Arthur
    Feb 22, 2017 at 17:52
  • $\begingroup$ got it thanks!! $\endgroup$
    – J. Jo
    Feb 22, 2017 at 18:03

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