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How many poker hands have two pairs?
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I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $binom132$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $binom132cdot binom42cdot binom42cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $binom133$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $binom133cdot binom42cdot binom42cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......
combinatorics poker
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
asked Jul 26 at 18:22
Stargazer
311
add a comment |Â
up vote
3
down vote
favorite
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $binom132$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $binom132cdot binom42cdot binom42cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $binom133$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $binom133cdot binom42cdot binom42cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......
combinatorics poker
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
asked Jul 26 at 18:22
Stargazer
311
2
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
2
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $binom132$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $binom132cdot binom42cdot binom42cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $binom133$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $binom133cdot binom42cdot binom42cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......
combinatorics poker
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
asked Jul 26 at 18:22
Stargazer
311
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $binom132$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $binom132cdot binom42cdot binom42cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $binom133$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $binom42$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $binom133cdot binom42cdot binom42cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......
combinatorics poker
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
asked Jul 26 at 18:22
Stargazer
311
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
edited Jul 27 at 9:07
N. F. Taussig
38.1k93053
38.1k93053
asked Jul 26 at 18:22
Stargazer
311
asked Jul 26 at 18:22
Stargazer
311
asked Jul 26 at 18:22
Stargazer
311
311
2
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
2
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12
add a comment |Â
2
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
2
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12
2
2
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
2
2
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12
add a comment |Â
3 Answers
3
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oldest
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up vote
0
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After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are $3 choose 2$ and $3 choose 1$, respectively, and of course each equals $3$.
answered Jul 26 at 20:10
John
21.9k32346
add a comment |Â
up vote
0
down vote
Assume your choose three ranks R1, R2, R3.
In the first solution:
You choose (R1, R2) first $binom132$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $binom132$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $binom133$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
answered Jul 26 at 19:59
tdluong
1315
add a comment |Â
up vote
0
down vote
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). Â That is okay. $$binom132binom 42^2binom441
\binom132binom 42^2binom 111binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?  Ah, that is better.$$binom 133binom 42^2binom 41binom 32$$
...and of course $binom133binom 32=frac13!3!10!frac3!2!1!=frac13!2!11!frac11!10!1!=binom 132binom111$
answered Jul 27 at 9:34
Graham Kemp
80k43275
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are $3 choose 2$ and $3 choose 1$, respectively, and of course each equals $3$.
answered Jul 26 at 20:10
John
21.9k32346
add a comment |Â
up vote
0
down vote
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are $3 choose 2$ and $3 choose 1$, respectively, and of course each equals $3$.
answered Jul 26 at 20:10
John
21.9k32346
add a comment |Â
up vote
0
down vote
up vote
0
down vote
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are $3 choose 2$ and $3 choose 1$, respectively, and of course each equals $3$.
answered Jul 26 at 20:10
John
21.9k32346
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are $3 choose 2$ and $3 choose 1$, respectively, and of course each equals $3$.
answered Jul 26 at 20:10
John
21.9k32346
answered Jul 26 at 20:10
John
21.9k32346
answered Jul 26 at 20:10
John
21.9k32346
answered Jul 26 at 20:10
John
21.9k32346
21.9k32346
add a comment |Â
add a comment |Â
up vote
0
down vote
Assume your choose three ranks R1, R2, R3.
In the first solution:
You choose (R1, R2) first $binom132$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $binom132$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $binom133$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
answered Jul 26 at 19:59
tdluong
1315
add a comment |Â
up vote
0
down vote
Assume your choose three ranks R1, R2, R3.
In the first solution:
You choose (R1, R2) first $binom132$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $binom132$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $binom133$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
answered Jul 26 at 19:59
tdluong
1315
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume your choose three ranks R1, R2, R3.
In the first solution:
You choose (R1, R2) first $binom132$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $binom132$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $binom133$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
answered Jul 26 at 19:59
tdluong
1315
Assume your choose three ranks R1, R2, R3.
In the first solution:
You choose (R1, R2) first $binom132$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $binom132$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $binom133$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
answered Jul 26 at 19:59
tdluong
1315
edited Jul 26 at 21:00
answered Jul 26 at 19:59
tdluong
1315
answered Jul 26 at 19:59
tdluong
1315
answered Jul 26 at 19:59
tdluong
1315
1315
add a comment |Â
add a comment |Â
up vote
0
down vote
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). Â That is okay. $$binom132binom 42^2binom441
\binom132binom 42^2binom 111binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?  Ah, that is better.$$binom 133binom 42^2binom 41binom 32$$
...and of course $binom133binom 32=frac13!3!10!frac3!2!1!=frac13!2!11!frac11!10!1!=binom 132binom111$
answered Jul 27 at 9:34
Graham Kemp
80k43275
add a comment |Â
up vote
0
down vote
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). Â That is okay. $$binom132binom 42^2binom441
\binom132binom 42^2binom 111binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?  Ah, that is better.$$binom 133binom 42^2binom 41binom 32$$
...and of course $binom133binom 32=frac13!3!10!frac3!2!1!=frac13!2!11!frac11!10!1!=binom 132binom111$
answered Jul 27 at 9:34
Graham Kemp
80k43275
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). Â That is okay. $$binom132binom 42^2binom441
\binom132binom 42^2binom 111binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?  Ah, that is better.$$binom 133binom 42^2binom 41binom 32$$
...and of course $binom133binom 32=frac13!3!10!frac3!2!1!=frac13!2!11!frac11!10!1!=binom 132binom111$
answered Jul 27 at 9:34
Graham Kemp
80k43275
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). Â That is okay. $$binom132binom 42^2binom441
\binom132binom 42^2binom 111binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?  Ah, that is better.$$binom 133binom 42^2binom 41binom 32$$
...and of course $binom133binom 32=frac13!3!10!frac3!2!1!=frac13!2!11!frac11!10!1!=binom 132binom111$
answered Jul 27 at 9:34
Graham Kemp
80k43275
answered Jul 27 at 9:34
Graham Kemp
80k43275
answered Jul 27 at 9:34
Graham Kemp
80k43275
answered Jul 27 at 9:34
Graham Kemp
80k43275
80k43275
add a comment |Â
add a comment |Â
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2
In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3.
– Lord Shark the Unknown
Jul 26 at 18:23
2
@LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked.
– Ross Millikan
Jul 26 at 18:35
Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too.
– amWhy
Jul 26 at 18:56
Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton?
– Stargazer
Jul 26 at 19:02
Never mind, I understand it now that I thought some more about it.
– Stargazer
Jul 26 at 19:12