• Support PF! Buy your school textbooks, materials and every day products Here!

Probability of a single pair poker hand

  • Thread starter hotvette
  • Start date
  • #1
hotvette
Homework Helper
988
3

Homework Statement


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

Homework Equations


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

3. The Attempt at a Solution
I've seen the following solution in several different places:
[tex]
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
[/tex] and every term makes perfect sense. What I don't understand though, is why the following:
[tex] n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1
[/tex] 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:

Answers and Replies

  • #2
MarneMath
Education Advisor
549
198
That's fine but you need to count the number of suites you can pick.
 
  • #3
hotvette
Homework Helper
988
3
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:
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


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

Homework Equations


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

The Attempt at a Solution


I've seen the following solution in several different places:
[tex]
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
[/tex] and every term makes perfect sense. What I don't understand though, is why the following:
[tex] n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1
[/tex] 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?
 
  • #5
hotvette
Homework Helper
988
3
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).
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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).
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!
 
  • #7
hotvette
Homework Helper
988
3
Thanks, I think you confirmed what I was asking.
 

Related Threads for: Probability of a single pair poker hand

  • Last Post
Replies
4
Views
648
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
2
Views
971
Replies
7
Views
21K
Replies
3
Views
947
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
Top