# Dealt 3 cards. Odds of being dealt any pair?

This is not to aid a gambling habit. I am simply curious how to do this math.
You get dealt 3 cards. What are the odds of having any pair? (We can exclude 3 of a kind)

Total number of hands = $\begin{pmatrix}52 \\ 3\end{pmatrix}$ = 22100

What do I do next? (Added from response below)

How many ways can I get a pair of 2's, for example? $\begin{pmatrix}4 \\ 2\end{pmatrix}$ = 6 And there are 13 types of pairs I can get. So, 13x6 = 72. So, there is only a 72/22100 chance of being dealt a pair?

Supplemental: If there are 5 players, what are the odds at least 1 person is holding a pair?

• Hint: Try calculating the probability of not getting a pair. Mar 3, 2013 at 17:45
• I'm curious as to whether there is a "smart" way to exactly calculate the supplemental question though. +1 Mar 3, 2013 at 20:02

Let's count the hands where you don't get a pair: For the first card, there are $52$ possibilities, for the second there are $48$ (since the cards of the first type are forbidden now) and for the third hand there are $44$ possibilities. Now for the hand, the order doesn't matter, so there are $$\frac{52\cdot 48\cdot 44}{6} = 18304$$ hands without a pair. Now the number of hands with a pair is $$\binom{52}{3} - 18304 = 3796.$$ Thus, the probability of getting a hand with a pair is $$\frac{3796}{22100} \approx 17.2\%.$$

EDIT: The above solution counts three of a kind as a hand with a pair. In the case that $3$ of a kind are forbidden, we have to subtract the $$\frac{52\cdot 3\cdot 2}{6} = 52$$ hands with $3$ of a kind. Now there are $$3796 - 52 = 3744$$ "good" hands, so the probability of getting a pair, but not three of a kind, is $$\frac{3744}{22100} \approx 16.9\%.$$

• Depends on whether being dealt three of a kind counts as also being dealt a pair. You might have to subtract a little more. Mar 3, 2013 at 17:52
• true, thank you. I'll edit my post accordingly. Mar 3, 2013 at 17:53
• Thanks for the explanation!! Last thing, dividing by 6 in the first calculation is to remove the 3! ways of ordering 3 cards? Mar 3, 2013 at 22:11
• Yes, exactly. You're welcome! Mar 3, 2013 at 22:19
• @thotpolice Thanks for correcting the typo. Apr 3, 2015 at 13:34

Assume you are dealt 3 cards in a row (so no other cards go missing from the deck between the cards you receive). The chances of getting a pair are simply 1 - (the chances of getting no pair). The first card is some card with probability 1. There are now 3 cards in the deck, which if dealt to you, would give you a pair, so the chances of not getting a pair on the second card are 48/51. Now there are 6 cards in the deck which will give you a pair if you receive them, so the chances of not getting a pair on the third card is 44/50. Overall chances of getting no pair is 1*(48/51)*(44/50), so the chances of getting a pair is 1 minus that product.

Here's another way, there are 2 ways to get a pair from 3 drawn cards:

1) Second card drawn matches first, third does not match first or second.

$$\frac{3}{51} \cdot \frac{48}{50} = \frac{144}{2550}$$

2) Second card drawn doesn't match first, third matches first or second.

$$\frac{48}{51} \cdot \frac{6}{50} = \frac{288}{2550}$$

Add the two probabilities to get:

$$\frac{144 + 288}{2550} = \frac{432}{2550} \text{ or } 16.94\%$$

• Number of total combinations of 3 cards = $$52\choose3$$
• Combinations of 3 cards with only a pair (and not a three of a kind) = $$13\times {4\choose2}\times 48$$ (There are 13 possible ranks of cards (from Ace to King) that could form the pair, $${4\choose2}$$ possible ways in which that pair could be formed by different suits, and 48 possible cards for the other card - all except the card that forms the pair)

Thus, the probability of getting only a pair (and not a three-of-a-kind) is $$p_{pair}=\frac{13\times {4\choose2}\times 48}{52\choose3} = \frac{72}{425}$$

For the extension question, @Azimut pointed out that my approach below is incorrect. I've left it here for posterity.

For $$n$$ people in general, the probability that at least one person has a pair is equal 1 subtracted by the probability that no person has a pair. The probability that each person does not have a pair is $$(1-p_{pair})$$, and for $$n$$ people not to have a pair it would be $$(1-p_{pair})^n$$

For your given example, the required probability would be

$$1-(1-p_{pair})^5\approx0.6047$$

• The other card can be any of 48 cards Mar 3, 2013 at 17:49
• thank you, I've corrected that. would my answer be correct now? Mar 3, 2013 at 17:54
• The probability for 5 players is not correct. You can't compute it in the way you did, since the events are not independent. However, I guess your 60% is not a bad estimate. Mar 3, 2013 at 18:08
• Azimut is right. Mar 3, 2013 at 18:24
• thank you! that teaches me not to post answers too quickly :) will think about it and post the edited one when I can. Mar 4, 2013 at 1:57

The probability of getting any pair is 13 times the probability of getting a specific pair. Let's look at a pair of aces.

There are three ways it can happen AAX AXA XAA where X is any card except an ace, so the probability of getting a pair of aces is the sum of these three possibilities.

$$\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{48}{50} + \frac{4}{52} \cdot \frac{48}{51} \cdot \frac{3}{50} + \frac{48}{52} \cdot \frac{4}{51} \cdot \frac{3}{50} \approx 0.013$$

Now we just multiply by 13 (there are 13 possible pairs) and 100 to get the percentage.

$0.013 \cdot 13 \cdot 100=16.9 \%$