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When three cards are randomly selected at a time from a standard deck of 52 playing cards, what is the probability that all of these three cards are in the same suit (heart, diamond, spade, or club)?

I'm mortified to ask for help again about probability math. If you look at my profile you probably can see that I've asked several questions about probability. When I thought I can wrap my head around this kind of math, then I had this question and I knew I did not.

Can you please give me some hints on this question and advice me some useful tips for learning to solve this kind of math?

P/s : I already had the answer for this, but I don't know how to solve it!!!

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3 Answers 3

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Three cards are selected from a standard deck of $52$ cards. Disregarding the order in which they are drawn, the possible outcomes are $\binom{52}{3}$. Out of these, how many include all cards of the same suit (say hearts)? There are $\binom{13}{3}$ ways in which you can get all 13 heart cards.

Since there are 4 suits, there are $4\binom{13}{3}$ ways in which all cards drawn are of the same suit. Thus the probability is:

$$\frac{4\binom{13}{3}}{\binom{52}{3}}\approx 5.18\%$$

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    $\begingroup$ Please forgive me to ask this but what is this called (52 3) , obviously not division operation? $\endgroup$
    – f855a864
    Commented Oct 18, 2014 at 16:53
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    $\begingroup$ It is what is called a binomial coefficient In some schools they write is as $12C3$ instead of $\binom{12}{3}$ In the calculator it is the button that has a symbol $nCm$. $\binom{12}{3}$ is the number of subsets of three elements a set of $12$ elements has. It is also said to be the number of ways to select three things out of 12 options when the order "doesn't matter" $\endgroup$
    – Asinomás
    Commented Oct 18, 2014 at 16:57
  • $\begingroup$ Thanks alot Jorge! This is very informative! $\endgroup$
    – f855a864
    Commented Oct 18, 2014 at 16:58
  • $\begingroup$ you're very welcome, glad I could help $\endgroup$
    – Asinomás
    Commented Oct 18, 2014 at 16:59
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    $\begingroup$ There's a suggested edit--you want to switch ${12 \choose 3}$ to ${13 \choose 3}$. $\endgroup$
    – Caleb Stanford
    Commented Nov 29, 2015 at 21:50
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Here is another solution that might be easier to compute/reason with.

Your first card can be anything. So you have 52 choices out of 52 cards (because no matter what card you draw you can get a full hand of the same suite).

Your second card, has to be the same suit as your first card, so probability of that is $\frac{12}{51}$ because there are 13 of each suite and you have to subtract 1 for the one card you have drawn.

Your third card has to be the same suite as the first and the second, notice there are only 11 cards left of that suite, so selecting that specific card will be $\frac{11}{50}$

Giving a total probability of:

$$\frac{52}{52} \times \frac{12}{51} \times \frac{11}{50}$$

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$$4~\frac{^{13}C_3}{^{52}C_3}=0.0518 $$ $4$ suits per deck

$3$ cards required in a suit

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  • $\begingroup$ Use Latex please $\endgroup$
    – nonuser
    Commented Jul 7, 2019 at 14:45

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