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Four cards are drawn from a standard 52-card deck without replacement. Find the probability that exactly 3 cards are of the same suit and the remaining card is of a different suit.

What I did: (4C1)(13C3)(3C1)(13C1) divided by (52C4)

But the final right answer is 0.375. Can you please tell me how to solve it and why mine is wrong? Thank you.

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  • $\begingroup$ Sorry that was a typo. I am actually dividing by (52C4) and still getting the wrong answer. $\endgroup$
    – user54379
    Oct 12, 2014 at 16:33
  • $\begingroup$ Your expression is now correct. $\endgroup$ Oct 12, 2014 at 16:37
  • $\begingroup$ I don't get 0,375 either $\endgroup$ Oct 12, 2014 at 17:12
  • $\begingroup$ Why do you think $0.375$ is the correct answer? $\endgroup$
    – David
    Oct 12, 2014 at 17:16
  • $\begingroup$ $0.375$ (which is $3$ out of $8$) seems too high to me even if I didn't know how to solve this problem cuz remember there are $4$ suits so you could get $1$ from each suit and you could also easily get $2$ max from one suit. If I had to guess I would say no more than maybe $20$% "winners" based on your criteria. $\endgroup$
    – David
    Oct 12, 2014 at 17:32

3 Answers 3

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There are 4 ways to choose the suit we want three of. There are then 3 ways to choose the remaining suit (we have one of). Then there are ${13 \choose 3}$ ways to choose the three card from the first suit, and $13$ ways to choose the card from the final one.

We divide this by ${52 \choose 4}$ ways to pick four cards, so we have as our probability:

$$\frac{4 \times 3 \times {13 \choose 3} \times 13}{{52 \choose 4}} \approx 0.1648$$

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Consider that there are 13 cards per suit. Then, to fulfill the stated conditions, we need

Then $P$($3$ cards same suit and $1$ card other suit| $4$ suits) = $4$ * $\dfrac{\binom{13}{3}\binom{39}{1}}{\binom{52}{4}}$ = $4$ * $\dfrac{286*39}{270,725}$ = $0.1648$

@Henno, @David, and @calculus: You are correct, there are four ways (suits) for this occur.

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  • $\begingroup$ This seems wrong. What about the 4 possible suits available? $\endgroup$
    – David
    Oct 12, 2014 at 17:10
  • $\begingroup$ I aggree with David. $\endgroup$ Oct 12, 2014 at 17:16
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Another approach is to draw the cards one by one.

Draw 3 cards of the same suit and 1 card of another suit:

$yyyx$

In total there are 4 ways.

And the same kind of drawings can be made with the other 3 suits. Thus we have in total $4\cdot (1+3)=16$ ways.

The probability of drawing cards one way is $\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}\cdot \frac{39}{49}$

All together the probabilitiy is $16\cdot \frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}\cdot \frac{39}{49}\approx 0.1648$

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