# Probability of a single pair poker hand

Homework Helper

## Homework Statement

How many one-pair 5-card poker hands are there in a standard 52-card deck?

## Homework Equations

$${}^nC_k = \frac{n!}{(n-k)!}$$

3. The Attempt at a Solution
I've seen the following solution in several different places:
$$n = {}^{13}C_1 \cdot {}^{4}C_2 \cdot {}^{12}C_3 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^{4}C_1 = 1098240$$ and every term makes perfect sense. What I don't understand though, is why the following:
$$n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1$$ underestimates the answer by a factor of 4. To me 13 choose 4 makes just as much sense as 13 choose one followed by 12 choose three, but of course one is correct and the other isn't. Can someone explain?

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
MarneMath
That's fine but you need to count the number of suites you can pick.

Homework Helper
Hmm, doesn't the 4C2 take care of suits? After thinking about this for a while, it seems to me the factor of 4 comes from the fact that {xx yzw} can be arranged 4 ways given that the order of yzw doesn't matter. Is this the correct way to look at it?

I think so, because I tried the same logic with 3 of a kind {xxx yz}. One solution is 13C1 x 4C3 x 12C2 x 4C1 x 4C1 = 54912 but a second way is 13C3 x 4C3 x 4C1 x 4C1 x 3C1 where the last 3C1 is the number of ways to arrange 3 values in {xxx yz} given the order of yz doesn't matter.

I also tried it with 4 of a kind {xxxx y}. One solution is 13C1 x 4C4 x 12C1 x 4C1 = 624 but a second way is 13C2 x 4C4 x 4C1 x 2C1 where the last 2C1 is the number of says to arrange 2 values in {xxx y}.

If my logic is correct, this is all (finally) making sense.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

How many one-pair 5-card poker hands are there in a standard 52-card deck?

## Homework Equations

$${}^nC_k = \frac{n!}{(n-k)!}$$[/B]

## The Attempt at a Solution

I've seen the following solution in several different places:
$$n = {}^{13}C_1 \cdot {}^{4}C_2 \cdot {}^{12}C_3 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^{4}C_1 = 1098240$$ and every term makes perfect sense. What I don't understand though, is why the following:
$$n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1$$ underestimates the answer by a factor of 4. To me 13 choose 4 makes just as much sense as 13 choose one followed by 12 choose three, but of course one is correct and the other isn't. Can someone explain?[/B]
Pleas turn off the bold font; it makes it look like you are yelling at us, and is very distracting.

Anyway, I do not understand where your ${13 \choose 4} \equiv {}^{13}C_4$ comes from. What is the logic behind that way of writing things?

Homework Helper
Strange, I never turned bold on in the first place, but clearly it was bold. Fixed now. Anyway, my logic was that for a single pair, I'm ultimately choosing 4 different cards from a pool of 13, one for the pair plus the remaining three. The math works if I consider that there are 4 different ways to group the 4 cards when using that scenario (each of the 4 cards could be the pair and the order of the remaining three does not matter). The logic seems to work for all of the scenarios I tried (one pair, two pair, four of a kind, 4 of a kind).

Ray Vickson
After you have chosen 4 different cards from the 13 ranks, you need to pick ONE of those 4 ranks for duplication across suits, so there are $C^4_1 = 4$ ways of choosing the rank to be paired, then $C^4_2$ ways of choosing the pair from the 4 suits at that rank. Then, of course, for each of the other three ranks there are $C^4_1$ ways of choosing the suit. That gives the total number as $C^{13}_4\, C^4_1 \,C^4_2 \, C^4_1 \, C^4_1 \, C^4_1 = 1098240$. THIS is the same as the other computation you mentioned!