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Find the probability that if we roll a die 6 times, we find exactly two numbers repeated twice(e.g. 121234,335422)
Clash Royale CLAN TAG#URR8PPP
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1
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The probability that we have 1 number twice is : 2/6
The probability that we have two numbers twice is 2/6*2/6
And the other 2 must be random: 2/6*2/6*1/6*1/6
However the answer is 0.347
How should I proceed to get such answer?
probability
asked Jul 14 at 22:10
Rayri
485
add a comment |Â
up vote
1
down vote
favorite
The probability that we have 1 number twice is : 2/6
The probability that we have two numbers twice is 2/6*2/6
And the other 2 must be random: 2/6*2/6*1/6*1/6
However the answer is 0.347
How should I proceed to get such answer?
probability
asked Jul 14 at 22:10
Rayri
485
1
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The probability that we have 1 number twice is : 2/6
The probability that we have two numbers twice is 2/6*2/6
And the other 2 must be random: 2/6*2/6*1/6*1/6
However the answer is 0.347
How should I proceed to get such answer?
probability
asked Jul 14 at 22:10
Rayri
485
The probability that we have 1 number twice is : 2/6
The probability that we have two numbers twice is 2/6*2/6
And the other 2 must be random: 2/6*2/6*1/6*1/6
However the answer is 0.347
How should I proceed to get such answer?
probability
asked Jul 14 at 22:10
Rayri
485
edited Jul 14 at 22:23
asked Jul 14 at 22:10
Rayri
485
asked Jul 14 at 22:10
Rayri
485
asked Jul 14 at 22:10
Rayri
485
485
1
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37
add a comment |Â
1
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37
1
1
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37
add a comment |Â
2 Answers
2
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oldest
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up vote
0
down vote
accepted
The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled,
and then $binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $frac6!2!$ ways to arrange those numbers
So the probability one exactly one repeat is
$$P(1)= frac6binom 54times frac 6!2!6^6 approx 0.23$$
For two repeats ...
there are $binom 62 = 15$ ways to choose the numbers to be doubled,
and then $binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $frac6!2!2! $ ways to arrange those numbers
SO ...
$$P(2)= fracbinom 62 binom 42times frac 6!2!2!6^6 approx 0.347$$
answered Jul 14 at 22:36
WW1
6,4821712
add a comment |Â
up vote
1
down vote
Taking $R$ for repeated numbers and $S$ for single numbers, there are $binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45cdot 360 = 16200$ options.
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $dfrac1620046656=dfrac2572 approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
answered Jul 14 at 22:35
Joffan
31.9k43169
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled,
and then $binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $frac6!2!$ ways to arrange those numbers
So the probability one exactly one repeat is
$$P(1)= frac6binom 54times frac 6!2!6^6 approx 0.23$$
For two repeats ...
there are $binom 62 = 15$ ways to choose the numbers to be doubled,
and then $binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $frac6!2!2! $ ways to arrange those numbers
SO ...
$$P(2)= fracbinom 62 binom 42times frac 6!2!2!6^6 approx 0.347$$
answered Jul 14 at 22:36
WW1
6,4821712
add a comment |Â
up vote
0
down vote
accepted
The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled,
and then $binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $frac6!2!$ ways to arrange those numbers
So the probability one exactly one repeat is
$$P(1)= frac6binom 54times frac 6!2!6^6 approx 0.23$$
For two repeats ...
there are $binom 62 = 15$ ways to choose the numbers to be doubled,
and then $binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $frac6!2!2! $ ways to arrange those numbers
SO ...
$$P(2)= fracbinom 62 binom 42times frac 6!2!2!6^6 approx 0.347$$
answered Jul 14 at 22:36
WW1
6,4821712
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled,
and then $binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $frac6!2!$ ways to arrange those numbers
So the probability one exactly one repeat is
$$P(1)= frac6binom 54times frac 6!2!6^6 approx 0.23$$
For two repeats ...
there are $binom 62 = 15$ ways to choose the numbers to be doubled,
and then $binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $frac6!2!2! $ ways to arrange those numbers
SO ...
$$P(2)= fracbinom 62 binom 42times frac 6!2!2!6^6 approx 0.347$$
answered Jul 14 at 22:36
WW1
6,4821712
The total number of equally probable arrangements of 6 dice rolls is $6^6$
To get exactly on double:
there are 6 ways to choose the number to be doubled,
and then $binom 54 =5$ ways to choose the remaining 4 non repeated numbers
finally there are $frac6!2!$ ways to arrange those numbers
So the probability one exactly one repeat is
$$P(1)= frac6binom 54times frac 6!2!6^6 approx 0.23$$
For two repeats ...
there are $binom 62 = 15$ ways to choose the numbers to be doubled,
and then $binom 42 =6$ ways to choose the remaining 2 non repeated numbers
finally there are $frac6!2!2! $ ways to arrange those numbers
SO ...
$$P(2)= fracbinom 62 binom 42times frac 6!2!2!6^6 approx 0.347$$
answered Jul 14 at 22:36
WW1
6,4821712
answered Jul 14 at 22:36
WW1
6,4821712
answered Jul 14 at 22:36
WW1
6,4821712
answered Jul 14 at 22:36
WW1
6,4821712
6,4821712
add a comment |Â
add a comment |Â
up vote
1
down vote
Taking $R$ for repeated numbers and $S$ for single numbers, there are $binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45cdot 360 = 16200$ options.
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $dfrac1620046656=dfrac2572 approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
answered Jul 14 at 22:35
Joffan
31.9k43169
add a comment |Â
up vote
1
down vote
Taking $R$ for repeated numbers and $S$ for single numbers, there are $binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45cdot 360 = 16200$ options.
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $dfrac1620046656=dfrac2572 approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
answered Jul 14 at 22:35
Joffan
31.9k43169
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Taking $R$ for repeated numbers and $S$ for single numbers, there are $binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45cdot 360 = 16200$ options.
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $dfrac1620046656=dfrac2572 approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
answered Jul 14 at 22:35
Joffan
31.9k43169
Taking $R$ for repeated numbers and $S$ for single numbers, there are $binom 62 = 15$ patterns to choose from (e.g. $RRSRRS, SRRRSR$). Then within the repeated numbers $a,b$ there are $binom 31=3$ patterns when we have an $a$ first: $aabb, abab, abba$.
So we have $15cdot 3=45$ templates to fill from the available numbers. So we can in each case choose the numbers $6!/2! = 360$ ways, giving a total of $45cdot 360 = 16200$ options.
By contrast there are $6^6=46656$ unrestricted options for the outcome of rolling a die six times.
Thus we have a probability of $dfrac1620046656=dfrac2572 approx 0.3472$
Note this is for exactly two numbers repeated exactly twice each. The question could also be interpreted as exactly two numbers repeated at least twice each.
answered Jul 14 at 22:35
Joffan
31.9k43169
answered Jul 14 at 22:35
Joffan
31.9k43169
answered Jul 14 at 22:35
Joffan
31.9k43169
answered Jul 14 at 22:35
Joffan
31.9k43169
31.9k43169
add a comment |Â
add a comment |Â
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1
How does 121134 qualify as exactly two numbers repeated twice ?
– WW1
Jul 14 at 22:15
Sorry I meant 121234
– Rayri
Jul 14 at 22:21
Do the numbers have to be repeated exactly twice, or would say $111223$ qualify?
– Badam Baplan
Jul 14 at 22:31
Yes they should be repeated exactly twice
– Rayri
Jul 14 at 22:32
Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:37