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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?

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5  
Hint: Try calculating the probability of not getting a pair. –  Glen The Udderboat Mar 3 '13 at 17:45
    
I'm curious as to whether there is a "smart" way to exactly calculate the supplemental question though. +1 –  Glen The Udderboat Mar 3 '13 at 20:02
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3 Answers

up vote 5 down vote accepted

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 $$\frac{3796}{22100} \approx 16.9\%.$$

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Depends on whether being dealt three of a kind counts as also being dealt a pair. You might have to subtract a little more. –  Thomas Andrews Mar 3 '13 at 17:52
    
true, thank you. I'll edit my post accordingly. –  azimut Mar 3 '13 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? –  JackOfAll Mar 3 '13 at 22:11
    
Yes, exactly. You're welcome! –  azimut Mar 3 '13 at 22:19
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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.

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  • Number of total combinations of 3 cards = $52\choose3$
  • Combinations of 3 cards with only a pair = $13\times {4\choose2}\times 48$ (There are 13 possible ranks of cards (from Ace to King) that could form the pair, 4C2 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)
  • Combinations of 3 cards with a three-of-a-kind = $13\times {4\choose3}$ (Similar reasoning to the above)

Thus, the probability of getting only a pair (and not a three-of-a-kind) is $p_{pair}=\frac{72}{425}$

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

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

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The other card can be any of 48 cards –  Thomas Andrews Mar 3 '13 at 17:49
    
thank you, I've corrected that. would my answer be correct now? –  Vincent Tjeng Mar 3 '13 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. –  azimut Mar 3 '13 at 18:08
    
Azimut is right. –  Thomas Andrews Mar 3 '13 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. –  Vincent Tjeng Mar 4 '13 at 1:57
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