5-Card Poker Two-Pair Probability Calculation
$begingroup$
Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$
Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?
probability discrete-mathematics poker
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
$endgroup$
add a comment |
$begingroup$
Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$
Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?
probability discrete-mathematics poker
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
$endgroup$
1
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32
add a comment |
$begingroup$
Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$
Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?
probability discrete-mathematics poker
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
$endgroup$
Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$
Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?
probability discrete-mathematics poker
probability discrete-mathematics poker
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
edited Aug 30 '15 at 4:53
Graham Kemp
85.8k43378
85.8k43378
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
asked Aug 29 '15 at 23:05
Adam TachéAdam Taché
255
255
1
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32
add a comment |
1
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32
1
1
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
$endgroup$
add a comment |
$begingroup$
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
binom{13}{2}
cdot binom{4}{2}
cdot binom{4}{2}
cdot binom{13 - 2}{1}
cdot binom{4}{1}
= 123,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
$endgroup$
add a comment |
$begingroup$
First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:
- Choose two face values out of 13 possible face values for the pairs (eights and aces):
13 choose 2 = 78 - Choose two cards from the smaller face value (8(clubs), 8(spades)):
4 choose 2 = 6 - Choose two cards from the larger face value (A(spades),A(clubs));
4 choose 2 = 6 - Choose one card from those remaining (9(diamond)):
44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)
By the product rule, H=78*6*6*44=123552
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
$endgroup$
add a comment |
$begingroup$
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
$endgroup$
add a comment |
$begingroup$
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
$endgroup$
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
answered Aug 29 '15 at 23:17
jorikijoriki
171k10188348
171k10188348
add a comment |
add a comment |
$begingroup$
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
binom{13}{2}
cdot binom{4}{2}
cdot binom{4}{2}
cdot binom{13 - 2}{1}
cdot binom{4}{1}
= 123,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
$endgroup$
add a comment |
$begingroup$
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
binom{13}{2}
cdot binom{4}{2}
cdot binom{4}{2}
cdot binom{13 - 2}{1}
cdot binom{4}{1}
= 123,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
$endgroup$
add a comment |
$begingroup$
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
binom{13}{2}
cdot binom{4}{2}
cdot binom{4}{2}
cdot binom{13 - 2}{1}
cdot binom{4}{1}
= 123,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
$endgroup$
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
binom{13}{2}
cdot binom{4}{2}
cdot binom{4}{2}
cdot binom{13 - 2}{1}
cdot binom{4}{1}
= 123,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
answered Sep 1 '15 at 12:59
vonbrandvonbrand
20k63159
20k63159
add a comment |
add a comment |
$begingroup$
First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:
- Choose two face values out of 13 possible face values for the pairs (eights and aces):
13 choose 2 = 78 - Choose two cards from the smaller face value (8(clubs), 8(spades)):
4 choose 2 = 6 - Choose two cards from the larger face value (A(spades),A(clubs));
4 choose 2 = 6 - Choose one card from those remaining (9(diamond)):
44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)
By the product rule, H=78*6*6*44=123552
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
add a comment |
$begingroup$
First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:
- Choose two face values out of 13 possible face values for the pairs (eights and aces):
13 choose 2 = 78 - Choose two cards from the smaller face value (8(clubs), 8(spades)):
4 choose 2 = 6 - Choose two cards from the larger face value (A(spades),A(clubs));
4 choose 2 = 6 - Choose one card from those remaining (9(diamond)):
44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)
By the product rule, H=78*6*6*44=123552
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
add a comment |
$begingroup$
First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:
- Choose two face values out of 13 possible face values for the pairs (eights and aces):
13 choose 2 = 78 - Choose two cards from the smaller face value (8(clubs), 8(spades)):
4 choose 2 = 6 - Choose two cards from the larger face value (A(spades),A(clubs));
4 choose 2 = 6 - Choose one card from those remaining (9(diamond)):
44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)
By the product rule, H=78*6*6*44=123552
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
$endgroup$
First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:
- Choose two face values out of 13 possible face values for the pairs (eights and aces):
13 choose 2 = 78 - Choose two cards from the smaller face value (8(clubs), 8(spades)):
4 choose 2 = 6 - Choose two cards from the larger face value (A(spades),A(clubs));
4 choose 2 = 6 - Choose one card from those remaining (9(diamond)):
44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)
By the product rule, H=78*6*6*44=123552
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
answered Dec 13 '18 at 14:08
Shamsul HudaShamsul Huda
12
12
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
add a comment |
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:26
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
$begingroup$
Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
$endgroup$
– hardmath
Dec 13 '18 at 14:28
add a comment |
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1
$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32