Combinatorics of two pairs in poker
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I've read the answer on counting the number of possible two pair hands in poker and I understand it. However, I don't know why my initial approach was wrong. Can someone point to the flaw in my approach.
Here's my logic. I'm going to assume a hand in the particular order XXYYZ, and then divide by 5! to account for the fact that the order of cards can be in any order.
The first card X is any one of the 52 cards.
The second card X is any one of the remaining 3 cards that would make the initial pair XX.
The third card Y is any one of the 48 cards remaining in the deck that's not X.
The fourth card Y is any one of the remaining 3 cards that would make the second pair YY.
The fifth card is any one of the 44 cards remaining in the deck that's not X nor Y.
So my answer is: (52*3*48*3*44)/(5!)
This doesn't even end up as an integer. So where did my logic go wrong?
probability combinatorics combinations
asked Dec 17 '18 at 16:25
smasma
162
$endgroup$
add a comment |
$begingroup$
I've read the answer on counting the number of possible two pair hands in poker and I understand it. However, I don't know why my initial approach was wrong. Can someone point to the flaw in my approach.
Here's my logic. I'm going to assume a hand in the particular order XXYYZ, and then divide by 5! to account for the fact that the order of cards can be in any order.
The first card X is any one of the 52 cards.
The second card X is any one of the remaining 3 cards that would make the initial pair XX.
The third card Y is any one of the 48 cards remaining in the deck that's not X.
The fourth card Y is any one of the remaining 3 cards that would make the second pair YY.
The fifth card is any one of the 44 cards remaining in the deck that's not X nor Y.
So my answer is: (52*3*48*3*44)/(5!)
This doesn't even end up as an integer. So where did my logic go wrong?
probability combinatorics combinations
asked Dec 17 '18 at 16:25
smasma
162
$endgroup$
$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
1
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55
add a comment |
$begingroup$
I've read the answer on counting the number of possible two pair hands in poker and I understand it. However, I don't know why my initial approach was wrong. Can someone point to the flaw in my approach.
Here's my logic. I'm going to assume a hand in the particular order XXYYZ, and then divide by 5! to account for the fact that the order of cards can be in any order.
The first card X is any one of the 52 cards.
The second card X is any one of the remaining 3 cards that would make the initial pair XX.
The third card Y is any one of the 48 cards remaining in the deck that's not X.
The fourth card Y is any one of the remaining 3 cards that would make the second pair YY.
The fifth card is any one of the 44 cards remaining in the deck that's not X nor Y.
So my answer is: (52*3*48*3*44)/(5!)
This doesn't even end up as an integer. So where did my logic go wrong?
probability combinatorics combinations
asked Dec 17 '18 at 16:25
smasma
162
$endgroup$
I've read the answer on counting the number of possible two pair hands in poker and I understand it. However, I don't know why my initial approach was wrong. Can someone point to the flaw in my approach.
Here's my logic. I'm going to assume a hand in the particular order XXYYZ, and then divide by 5! to account for the fact that the order of cards can be in any order.
The first card X is any one of the 52 cards.
The second card X is any one of the remaining 3 cards that would make the initial pair XX.
The third card Y is any one of the 48 cards remaining in the deck that's not X.
The fourth card Y is any one of the remaining 3 cards that would make the second pair YY.
The fifth card is any one of the 44 cards remaining in the deck that's not X nor Y.
So my answer is: (52*3*48*3*44)/(5!)
This doesn't even end up as an integer. So where did my logic go wrong?
probability combinatorics combinations
probability combinatorics combinations
asked Dec 17 '18 at 16:25
smasma
162
asked Dec 17 '18 at 16:25
smasma
162
asked Dec 17 '18 at 16:25
smasma
162
asked Dec 17 '18 at 16:25
smasma
162
asked Dec 17 '18 at 16:25
smasma
162
162
$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
1
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55
add a comment |
$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
1
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55
$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
1
1
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If order matters, you should multiply (not divide) by 5! You generate all possible permutations from a set of different elements.
However, David G. Stork's comment is still legit:
5♢5♠6♣6♡J♣ will generate all permutations including 5♢5♠6♣6♡J♣
6♡6♣5♠5♢J♣ will generate all permutations including 5♢5♠6♣6♡J♣
So, each permutation will be counted 8 times (2!×2!×2!). One 2! comes from permutation inside the first pair, one 2! comes from the second pair and one 2! from permutations between pairs (we could have selected 6 first and 5 second).
The number of tuples containing two pairs is then:
$$n=frac12frac{52times3}2timesfrac{48times3}2times44times5!=14,826,240$$
If we divide by total number of tuples (52!/47!), we will get the probability of two-pair hand:
$$p=frac{n}{52!/47!} = frac{198}{4165}approx 4.75,%$$
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
$begingroup$
If order matters, you should multiply (not divide) by 5! You generate all possible permutations from a set of different elements.
However, David G. Stork's comment is still legit:
5♢5♠6♣6♡J♣ will generate all permutations including 5♢5♠6♣6♡J♣
6♡6♣5♠5♢J♣ will generate all permutations including 5♢5♠6♣6♡J♣
So, each permutation will be counted 8 times (2!×2!×2!). One 2! comes from permutation inside the first pair, one 2! comes from the second pair and one 2! from permutations between pairs (we could have selected 6 first and 5 second).
The number of tuples containing two pairs is then:
$$n=frac12frac{52times3}2timesfrac{48times3}2times44times5!=14,826,240$$
If we divide by total number of tuples (52!/47!), we will get the probability of two-pair hand:
$$p=frac{n}{52!/47!} = frac{198}{4165}approx 4.75,%$$
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
$begingroup$
If order matters, you should multiply (not divide) by 5! You generate all possible permutations from a set of different elements.
However, David G. Stork's comment is still legit:
5♢5♠6♣6♡J♣ will generate all permutations including 5♢5♠6♣6♡J♣
6♡6♣5♠5♢J♣ will generate all permutations including 5♢5♠6♣6♡J♣
So, each permutation will be counted 8 times (2!×2!×2!). One 2! comes from permutation inside the first pair, one 2! comes from the second pair and one 2! from permutations between pairs (we could have selected 6 first and 5 second).
The number of tuples containing two pairs is then:
$$n=frac12frac{52times3}2timesfrac{48times3}2times44times5!=14,826,240$$
If we divide by total number of tuples (52!/47!), we will get the probability of two-pair hand:
$$p=frac{n}{52!/47!} = frac{198}{4165}approx 4.75,%$$
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
$begingroup$
If order matters, you should multiply (not divide) by 5! You generate all possible permutations from a set of different elements.
However, David G. Stork's comment is still legit:
5♢5♠6♣6♡J♣ will generate all permutations including 5♢5♠6♣6♡J♣
6♡6♣5♠5♢J♣ will generate all permutations including 5♢5♠6♣6♡J♣
So, each permutation will be counted 8 times (2!×2!×2!). One 2! comes from permutation inside the first pair, one 2! comes from the second pair and one 2! from permutations between pairs (we could have selected 6 first and 5 second).
The number of tuples containing two pairs is then:
$$n=frac12frac{52times3}2timesfrac{48times3}2times44times5!=14,826,240$$
If we divide by total number of tuples (52!/47!), we will get the probability of two-pair hand:
$$p=frac{n}{52!/47!} = frac{198}{4165}approx 4.75,%$$
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
$endgroup$
If order matters, you should multiply (not divide) by 5! You generate all possible permutations from a set of different elements.
However, David G. Stork's comment is still legit:
5♢5♠6♣6♡J♣ will generate all permutations including 5♢5♠6♣6♡J♣
6♡6♣5♠5♢J♣ will generate all permutations including 5♢5♠6♣6♡J♣
So, each permutation will be counted 8 times (2!×2!×2!). One 2! comes from permutation inside the first pair, one 2! comes from the second pair and one 2! from permutations between pairs (we could have selected 6 first and 5 second).
The number of tuples containing two pairs is then:
$$n=frac12frac{52times3}2timesfrac{48times3}2times44times5!=14,826,240$$
If we divide by total number of tuples (52!/47!), we will get the probability of two-pair hand:
$$p=frac{n}{52!/47!} = frac{198}{4165}approx 4.75,%$$
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,6791312
2,6791312
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
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$begingroup$
What if your first card were $5 diamondsuit$ and second card $5 clubsuit$... or they were in the reverse order? How would you count those?
$endgroup$
– David G. Stork
Dec 17 '18 at 16:29
$begingroup$
Although I assume an order of the cards initially, I thought dividing by 5! would make the count be of any order.
$endgroup$
– sma
Dec 17 '18 at 16:32
$begingroup$
Nope. For the $XX$ you must divide by $2!$... and likewise for the $YY$. And what if your cards appeared $YYZXX$? You must figure out all these orders to get your answer.
$endgroup$
– David G. Stork
Dec 17 '18 at 16:45
1
$begingroup$
Much easier to pick the two paired ranks, $binom {13}2$, pick the two suits within each rank, $binom 42^2$, and then pick the odd man out, $44$.
$endgroup$
– lulu
Dec 17 '18 at 16:55