Poker game probability

Question

From a deck of 52 cards, pick 3 cards randomly, what is the probability that there is a pair among the three cards? (A pair means that two of the cards are the same number) I guess it allows to have all three cards the same number.

Answer 1

As Alex commented, finding the complementary probability is easier.

Firstly, choose your card values. In this case it is $\binom{13}{3}$ as you want three different valued cards.

Now for each card, the number of ways for them to have their respective values are $\binom{4}{1}$, or simply $4$, as there are four suits. There are three cards in your hand.

Your total probability will be $$1-\frac{\binom{13}{3}\times4^{3}}{\binom{52}{3}}$$ Does this make sense?

Answer 2

let's say first card we draw has some value. now, for drawing 2 more cards, there are two ways to get at least one pair among them. you must have either of the cards match the first, or the second two match each other. since the probability of neither of the cards matching the first is $(48/51)*(47/50)$, the probability of either matching is $1-(48/51)*(47/50)$. now, consider when the last two are the same. since this only happens when the third card matches the second, the probability is $3/50$. therefore, since the probability of neither of the cards matching the first nor matching each other is $(48/51)*(47/50)^2$, the probability of a pair would be $1-(48/51)*(47/50)^2$, or around 16.8%

Answer 3

The direct computation is $$ \underbrace{\frac{\binom{13}{1}\binom{4}{2}\binom{48}{1}}{\binom{52}{3}}}_{\substack{\text{probability of drawing a pair}\\ \text{without three of a kind}}} + \underbrace{\frac{\binom{13}{1}\binom{4}{3}}{\binom{52}{3}}}_{\substack{\text{probability of drawing}\\ \text{three of a kind}}} = \frac{73}{425} $$

Answer 4

It's 1 - 48/51 * 44/50 ≈ 17.176%. One minus the probability of all different ranks. First can be anything, then one of 48 of a different rank, then one of 44 of a 3rd rank.

Answer 5

You could compute the probability that the first and second cards have the same rank (this is $3/51=1/17$), then multiply this by three to account for the equally likely first-and-third-card and second-and-third-card pairs. However, this triple-counts the case where all three cards have the same rank, which occurs with probability $(3/51) \cdot (2/50) = 1/425$. Subtracting the excess gives $$ p = 3\cdot\frac{1}{17} - 2\cdot \frac{1}{425}=\frac{73}{425}\approx 0.1718. $$