When you pick three cards, without replacement, from a standard 52 card deck, what are the probabilities of:

  • only one suit in your three cards
  • two different suits in your three cards
  • three different suits in your three cards

For the first I have the probability of $4 \cdot \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} = \frac{22}{425} $

But I cannot think of a way to determine the possibilities you have two or three different suits in the three chosen cards. Any help is appreciated.


Our denominator is the number of ways to select three cards from a deck of $52$ cards is $$\binom{52}{3}$$

Another way to calculate the probability of selecting exactly one suit is to choose the suit, then choose three cards from that suit.

$$p(\text{all of the same suit}) = \frac{\dbinom{4}{1}\dbinom{13}{3}}{\dbinom{52}{3}}$$

For exactly two suits, we must choose the suit from which two cards are drawn, draw two cards of that suit, choose one of the other three suits, then choose one card from that suit.

$$p(\text{exactly two suits}) = \frac{\dbinom{4}{1}\dbinom{13}{2}\dbinom{3}{1}\dbinom{13}{1}}{\dbinom{52}{3}}$$

For three different suits, we must choose the three suits from which a card is drawn and select one card from each of those suits.

$$p(\text{exactly three suits}) = \frac{\dbinom{4}{3}\dbinom{13}{1}^3}{\dbinom{52}{3}}$$


For two different suits:

We need two different suits out of 4, so $^4C_2$ ways.

We need 2 cards from one of the selected suits and one from the other suit so $^{13}C_2.^{13}C_1.2!$ ways.

So favourable no. of cases: $ ^4C_2.^{13}C_2.^{13}C_1.2!$

Total no. of cases: $^{52}C_1.^{51}C_1.^{50}C_1$

So, probability = $$\frac{^4C_2.^{13}C_2.^{13}C_1.2!}{^{52}C_1.^{51}C_1.^{50}C_1}$$

For three different suits:

Following the same steps we have favourable cases as $^4C_3.^{13}C_1.^{13}C_1.^{13}C_1$

Total no. of cases remains same as before.

Hence, probability = $$\frac{^4C_3.^{13}C_1.^{13}C_1.^{13}C_1}{^{52}C_1.^{51}C_1.^{50}C_1}$$


The easiest way is to calculate next the probability of getting three different suits in your three cardst. Let denote the different suits as $a,b,c$ and $d$.

There are 4 combinations: $abc, acd, abd, bcd$

Each combination can be picked in $3!=6$ ways. Therefore the probability is

$24\cdot \frac{13}{52}\cdot\frac{13}{51}\cdot\frac{13}{50}$

Let $X$ be the amount of different cards. Notice that $P(X=1)+P(X=2)+P(X=3)=1$

The only remaining probability is $P(X=2)$, which is the probability of your second question. You have to solve the equation for $P(X=2)$.


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