Mixing
probability to have 2 pair throwing 5 dices
$begingroup$
We throw 5 dices. What is the probability to have 2 pairs ?
My solution says it $frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$, where as for me it's $$frac{6cdot 5cdot 4cdot 5!}{6^5cdot 2!cdot 2!}.$$
I do as follow : Throwing 5 dices is the same thing as throwing one dice 5 times. What we want is $$AABBC$$
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $frac{5!}{2!2!}$. At the end, I get
$$frac{6cdot 5cdot 4 cdot 5!}{6^5cdot 2!cdot 2!}.$$
What's wrong in my argument ?
probability
asked Jan 20 at 13:54
user623855user623855
1457
$endgroup$
add a comment |
$begingroup$
We throw 5 dices. What is the probability to have 2 pairs ?
My solution says it $frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$, where as for me it's $$frac{6cdot 5cdot 4cdot 5!}{6^5cdot 2!cdot 2!}.$$
I do as follow : Throwing 5 dices is the same thing as throwing one dice 5 times. What we want is $$AABBC$$
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $frac{5!}{2!2!}$. At the end, I get
$$frac{6cdot 5cdot 4 cdot 5!}{6^5cdot 2!cdot 2!}.$$
What's wrong in my argument ?
probability
asked Jan 20 at 13:54
user623855user623855
1457
$endgroup$
add a comment |
$begingroup$
We throw 5 dices. What is the probability to have 2 pairs ?
My solution says it $frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$, where as for me it's $$frac{6cdot 5cdot 4cdot 5!}{6^5cdot 2!cdot 2!}.$$
I do as follow : Throwing 5 dices is the same thing as throwing one dice 5 times. What we want is $$AABBC$$
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $frac{5!}{2!2!}$. At the end, I get
$$frac{6cdot 5cdot 4 cdot 5!}{6^5cdot 2!cdot 2!}.$$
What's wrong in my argument ?
probability
asked Jan 20 at 13:54
user623855user623855
1457
$endgroup$
We throw 5 dices. What is the probability to have 2 pairs ?
My solution says it $frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$, where as for me it's $$frac{6cdot 5cdot 4cdot 5!}{6^5cdot 2!cdot 2!}.$$
I do as follow : Throwing 5 dices is the same thing as throwing one dice 5 times. What we want is $$AABBC$$
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $frac{5!}{2!2!}$. At the end, I get
$$frac{6cdot 5cdot 4 cdot 5!}{6^5cdot 2!cdot 2!}.$$
What's wrong in my argument ?
probability
probability
asked Jan 20 at 13:54
user623855user623855
1457
asked Jan 20 at 13:54
user623855user623855
1457
asked Jan 20 at 13:54
user623855user623855
1457
asked Jan 20 at 13:54
user623855user623855
1457
asked Jan 20 at 13:54
user623855user623855
1457
1457
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-dfrac16=dfrac56$
Taking your approach, you would say the probability should be $dfrac{6cdot 5 cdot 2!}{6^2cdot 1!cdot 1!} = dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
- You could say there are ${6 choose 2}=dfrac{6 times 5}{2}$ ways of choosing $A$ and $B$
- Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
answered Jan 20 at 14:25
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$frac{binom{6}{2}cdot 4 cdot frac{5!}{2! 2!}}{6^5}=frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$$
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
$endgroup$
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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active
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active
oldest
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$begingroup$
Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-dfrac16=dfrac56$
Taking your approach, you would say the probability should be $dfrac{6cdot 5 cdot 2!}{6^2cdot 1!cdot 1!} = dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
- You could say there are ${6 choose 2}=dfrac{6 times 5}{2}$ ways of choosing $A$ and $B$
- Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
answered Jan 20 at 14:25
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-dfrac16=dfrac56$
Taking your approach, you would say the probability should be $dfrac{6cdot 5 cdot 2!}{6^2cdot 1!cdot 1!} = dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
- You could say there are ${6 choose 2}=dfrac{6 times 5}{2}$ ways of choosing $A$ and $B$
- Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
answered Jan 20 at 14:25
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-dfrac16=dfrac56$
Taking your approach, you would say the probability should be $dfrac{6cdot 5 cdot 2!}{6^2cdot 1!cdot 1!} = dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
- You could say there are ${6 choose 2}=dfrac{6 times 5}{2}$ ways of choosing $A$ and $B$
- Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
answered Jan 20 at 14:25
HenryHenry
101k481168
$endgroup$
Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-dfrac16=dfrac56$
Taking your approach, you would say the probability should be $dfrac{6cdot 5 cdot 2!}{6^2cdot 1!cdot 1!} = dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
- You could say there are ${6 choose 2}=dfrac{6 times 5}{2}$ ways of choosing $A$ and $B$
- Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
answered Jan 20 at 14:25
HenryHenry
101k481168
answered Jan 20 at 14:25
HenryHenry
101k481168
answered Jan 20 at 14:25
HenryHenry
101k481168
answered Jan 20 at 14:25
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
$begingroup$
You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$frac{binom{6}{2}cdot 4 cdot frac{5!}{2! 2!}}{6^5}=frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$$
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
$endgroup$
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
add a comment |
$begingroup$
You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$frac{binom{6}{2}cdot 4 cdot frac{5!}{2! 2!}}{6^5}=frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$$
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
$endgroup$
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
add a comment |
$begingroup$
You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$frac{binom{6}{2}cdot 4 cdot frac{5!}{2! 2!}}{6^5}=frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$$
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
$endgroup$
You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$frac{binom{6}{2}cdot 4 cdot frac{5!}{2! 2!}}{6^5}=frac{6cdot 5cdot 4cdot 5! }{6^5cdot 2cdot 2!cdot 2!}$$
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
edited Jan 20 at 14:19
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
answered Jan 20 at 13:58
Robert ZRobert Z
99.2k1068139
99.2k1068139
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
add a comment |
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
Do you agree that I can do $frac{5!}{2!2!}$ differents word with $AABBC$ ? If yes, I already consider $AABBC$ and $BBAAC$... sorry, I don't get the point
$endgroup$
– user623855
Jan 20 at 14:01
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I edit my answer. Is it better now?
$endgroup$
– Robert Z
Jan 20 at 14:17
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
I'm still a bit confuse... but I have to think about it.
$endgroup$
– user623855
Jan 20 at 14:18
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
$begingroup$
@user623855 : Don't think about permutation. What you want to do is exactely AABBC in this order. If you do, $11223$ or $22113$ you exactly have the form $AABBC$ but both are the same. So you count them twice.
$endgroup$
– Surb
Jan 20 at 14:22
add a comment |
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