What is the probability that the cards are of the same suit? This specific type of question deals with the use of combinations and permutations. I know that because the cards aren't being replaced after being taken out, order does matter and therefore permutations are used. So I started off with this:

(13)P(2) since at first there are 13 cards of the same suit, then 12 after one of the suit is picked out and not replaced. The total came out to be 156 ways.

For the sample space I did (52)P(2) since at first there are 52 cards to choose from, and then there are 51 cards to choose from after the first card is picked and not replaced. The total came out to be 2652 ways.

So then to find the probability of the question:

(13)P(2)/(52)P(2)= 156/2652= 1/17

What is the probability that the cards are both face cards?

I started off with this:

There are 12 face cards in total, and because cards aren't being replaced, order does matter. So since there are 12 face cards as choices in the first try, then 11 in the second I did (12)P(2) as the numerator. The total came out to be 132.

For the denominator, which is the sample space, I did (52)P(2), since you are choosing 2 out of 52 cards and got 2652 ways.

So now I do :

(12)P(2)/(52)P(2)= 132/2652= 11/221

What is the probability that the cards are a diamond and a spade?

I was a little confused on this one, but I tried. Even though I said order matters and to use permutations, I felt like for the numerator I was supposed to combinations since choosing a diamond and spade didn't matter. Since there are 13 spades and 13 diamonds, for the numerator I did (13)C(1)*(13)C(1) which equals to 169 ways of choosing a diamond and a spade.

For the sample space I stayed with (52)P(2) since order still matter on this one which equals to 2652 ways.

So the final step is :

[(13)C(1)*(13)C(1)]/(52)P(2)= 169/2652= 13/204

I would like to know if I did these questions right, because i'm a little confused between using combinations and permutations.

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    $\begingroup$ You haven't accounted for the fact that there are four suits available. This number would be correct for the probability that both cards are spades, but not that both cards are same (unspecified) suit. $\endgroup$ – JMoravitz Oct 8 '15 at 1:20
  • $\begingroup$ Why would order matter? Would A-K be different than K-A? But if you are using permutations for the numerator and denominator, then the errors (IMO) will cancel out. $\endgroup$ – turkeyhundt Oct 8 '15 at 1:29
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    $\begingroup$ Here is a simpler way. Imagine the cards are drawn one at a time. Whatever card was drawn first, the probability the next one matches it in suit is $12/51$, which if you like (I don't) simplifies to $4/17$. $\endgroup$ – André Nicolas Oct 8 '15 at 1:31
  • $\begingroup$ Your second is not quite right. For the bottom and top, we need to use "combinations" in both, or "permutations" in both. Your $\binom{13}{1}\binom{13}{1}$ counts the number of hands that have a diamond and a spade. So the denominator should be $\binom{52}{2}$. If you want permutations at the bottom, for the top you need to count diamond then spade, and spade then diamond, $2\binom{13}{1}^2$. $\endgroup$ – André Nicolas Oct 8 '15 at 1:39
  • $\begingroup$ So you are saying to do: [(13)P(1)*(13)P(1)+(13)P(1)*(13)P(1)]/(52)P(2)= 138/2652 = 13/102? $\endgroup$ – user273609 Oct 8 '15 at 2:03

Almost.   You have the probability that two cards come from a particular suit.   You also have to consider that there are four suits from which those cards can be selected.

Further, we typically use combination rather than permutation for this, because order of selection in a card hand is not important.   (Hands are "heaps" rather than "lists").   (Though often the common factor cancels.)

We count ways to select a suit, and two cards from that suit. Divide by all ways to select two cards of any suit.

$$\mathsf P(\text{the two cards selected are of same suit}) = \frac{{^4\mathrm C_1}{^{13}\mathrm C_2}}{{^{52}\mathrm C_2}} \times\frac{2!}{2!} = \frac{{^4\mathrm C_1}{^{13}\mathrm P_2}}{{^{52}\mathrm P_2}} = \frac{4}{17} $$

There is an alternative.   Whatever card is drawn 'first', there are $12$ of the $51$ cards remaining that match the 'first' card's suit. $12/51=4/17$.

  • $\begingroup$ Thank you so much! I always thought since you weren't doing replacement in a problem it automatically meant permutation since the order does matter. $\endgroup$ – user273609 Oct 8 '15 at 1:57
  • $\begingroup$ @user273609 The main criteria is that you measure the favoured space and the total space in the same way. As David K said, you need consistency in forming the numerator and denominator. $$\frac{{^{4}\mathrm{C}_{1}\;{^{13}\mathrm{P}_{2}}}}{{^{52}\mathrm{P}_{2}}} = \frac{{^{4}\mathrm{C}_{1}\;{^{13}\mathrm{C}_{2}}}}{{^{52}\mathrm{C}_{2}}}\frac{2!}{2!}$$ $\endgroup$ – Graham Kemp Oct 8 '15 at 2:17
  • $\begingroup$ Since you say that we typically use combinations, shouldn't the 2!/2! term appear on the other side of the equality? i.e. it is intended to show that comb/comb = perm/perm * 2!/2!. Having said that, said equality would be clearer if it were on a separate line below after some explanation. (I am not referring to the comment but to the main answer) $\endgroup$ – PatrickT Nov 4 '17 at 16:41

You have calculated the probability of drawing two cards from one particular suit, but there are 4 suits to consider. Then there are actually $4*_{13}P_2$ ways to select two cards of the same suit.


For this particular question it does not matter whether you do permutations or combinations, as long as you are consistent about using the same counting method in the numerator and the denominator.

Your method with permutations is correct (up to the point where you forgot to account for the other three suits), exactly as you showed it.

If you used combinations instead, the calculation would be

$$ \frac{\binom{13}{2}}{\binom{52}{2}} = \frac{78}{1326} = \frac{1}{17}. $$

This is not a coincidence. It is because although you draw the two cards out one at a time, you do not really care which came first. The 4 of clubs and 6 of clubs can be considered the same event as 6 of clubs and 4 of clubs, as long as you also consider the 4 of clubs and 6 of diamonds to be the same event as 6 of diamonds and 4 of clubs. But there is no reason not to consider the order, and if you do, you have twice as many "same suit" events and twice as many events overall, so the ratio stays the same.

The probability $\frac{1}{17}$ then is the correct probability for drawing both cards from the same pre-selected suit; for example, the probability to draw two cards from clubs. Now add the probability to draw two diamonds, or two hearts or two spades (all of which are mutually exclusive events).


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