Combinatorics of two pairs in poker
$begingroup$
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
$endgroup$
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
2,3141311
$endgroup$
$begingroup$
Thanks, that makes sense.
$endgroup$
– sma
Dec 18 '18 at 19:45
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044139%2fcombinatorics-of-two-pairs-in-poker%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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
2,3141311
$endgroup$
$begingroup$
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
2,3141311
$endgroup$
$begingroup$
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
2,3141311
$endgroup$
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
2,3141311
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,3141311
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,3141311
answered Dec 17 '18 at 16:57
Vasily MitchVasily Mitch
2,3141311
2,3141311
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044139%2fcombinatorics-of-two-pairs-in-poker%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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