Mixing
Calculating combinations of three of a kind in Texas Hold 'Em
$begingroup$
The question sounds quite simple - in a game of Texas Hold'Em (or simply out of ${52 choose 7}$), how many combinations of cards are there, that would result in three of a kind being the highest hand?
The Wikipedia article about poker probabilities says, that the frequency of three of a kind in a 7-card poker game is 6,461,620.
I have tried to arrive at that solution by first calculating all possible three card combinations: ${4 choose 3} 13 = 52$, and then multiplying that number by the amount of possible combinations of the other 4 cards, however quickly realised that 6,461,620 isn't divisible by 52 which throws my plan out of the window.
How do I arrive at the correct amount of possible combinations?
combinations card-games poker
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
$endgroup$
add a comment |
$begingroup$
The question sounds quite simple - in a game of Texas Hold'Em (or simply out of ${52 choose 7}$), how many combinations of cards are there, that would result in three of a kind being the highest hand?
The Wikipedia article about poker probabilities says, that the frequency of three of a kind in a 7-card poker game is 6,461,620.
I have tried to arrive at that solution by first calculating all possible three card combinations: ${4 choose 3} 13 = 52$, and then multiplying that number by the amount of possible combinations of the other 4 cards, however quickly realised that 6,461,620 isn't divisible by 52 which throws my plan out of the window.
How do I arrive at the correct amount of possible combinations?
combinations card-games poker
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
$endgroup$
add a comment |
$begingroup$
The question sounds quite simple - in a game of Texas Hold'Em (or simply out of ${52 choose 7}$), how many combinations of cards are there, that would result in three of a kind being the highest hand?
The Wikipedia article about poker probabilities says, that the frequency of three of a kind in a 7-card poker game is 6,461,620.
I have tried to arrive at that solution by first calculating all possible three card combinations: ${4 choose 3} 13 = 52$, and then multiplying that number by the amount of possible combinations of the other 4 cards, however quickly realised that 6,461,620 isn't divisible by 52 which throws my plan out of the window.
How do I arrive at the correct amount of possible combinations?
combinations card-games poker
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
$endgroup$
The question sounds quite simple - in a game of Texas Hold'Em (or simply out of ${52 choose 7}$), how many combinations of cards are there, that would result in three of a kind being the highest hand?
The Wikipedia article about poker probabilities says, that the frequency of three of a kind in a 7-card poker game is 6,461,620.
I have tried to arrive at that solution by first calculating all possible three card combinations: ${4 choose 3} 13 = 52$, and then multiplying that number by the amount of possible combinations of the other 4 cards, however quickly realised that 6,461,620 isn't divisible by 52 which throws my plan out of the window.
How do I arrive at the correct amount of possible combinations?
combinations card-games poker
combinations card-games poker
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
asked Jan 23 at 2:24
Mantas KandrataviciusMantas Kandratavicius
155
155
add a comment |
add a comment |
1 Answer
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$begingroup$
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives
$$13cdot 4 cdot 48 cdot 44 cdot 40 cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240cdot 5 cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
add a comment |
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1 Answer
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$begingroup$
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives
$$13cdot 4 cdot 48 cdot 44 cdot 40 cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240cdot 5 cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
add a comment |
$begingroup$
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives
$$13cdot 4 cdot 48 cdot 44 cdot 40 cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240cdot 5 cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
add a comment |
$begingroup$
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives
$$13cdot 4 cdot 48 cdot 44 cdot 40 cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240cdot 5 cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
$endgroup$
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives
$$13cdot 4 cdot 48 cdot 44 cdot 40 cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240cdot 5 cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
edited Jan 23 at 4:16
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
answered Jan 23 at 3:00
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
add a comment |
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I'm not interested in calculating the chance tho, I just want to know how to arrive at the number of possible combinations that result in three of a kind being the highest hand (namely 6,461,620)
$endgroup$
– Mantas Kandratavicius
Jan 23 at 3:02
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
$begingroup$
I read "highest hand" as the highest hand at the table. This is more tractable. I will update.
$endgroup$
– Ross Millikan
Jan 23 at 3:30
add a comment |
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