Probability of 5 people choosing a unique person randomly
There are 5 people in a circle and on the count of 3 they will all point at someone randomly and simultaneously.They cannot point to themselves.
What is the probability of them all pointing at someone different?
I appreciate this is probably quite a straightforward question for those who know probability.
I can see the total variations is $4^5$.
What I don't know how to work out is how many of these variations give an answer to the problem.
probability
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
add a comment |
There are 5 people in a circle and on the count of 3 they will all point at someone randomly and simultaneously.They cannot point to themselves.
What is the probability of them all pointing at someone different?
I appreciate this is probably quite a straightforward question for those who know probability.
I can see the total variations is $4^5$.
What I don't know how to work out is how many of these variations give an answer to the problem.
probability
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
add a comment |
There are 5 people in a circle and on the count of 3 they will all point at someone randomly and simultaneously.They cannot point to themselves.
What is the probability of them all pointing at someone different?
I appreciate this is probably quite a straightforward question for those who know probability.
I can see the total variations is $4^5$.
What I don't know how to work out is how many of these variations give an answer to the problem.
probability
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
There are 5 people in a circle and on the count of 3 they will all point at someone randomly and simultaneously.They cannot point to themselves.
What is the probability of them all pointing at someone different?
I appreciate this is probably quite a straightforward question for those who know probability.
I can see the total variations is $4^5$.
What I don't know how to work out is how many of these variations give an answer to the problem.
probability
probability
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
asked Dec 4 '18 at 9:51
Ben FranksBen Franks
261110
261110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$D_n$, Derangment is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place.
What we are looking for would be
$$frac{D_5}{4^5}=frac{44}{4^5}=frac{11}{4^4}.$$
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
|
show 1 more comment
Let us label the people A through E, and let $(A,B)$ mean that A pointed to B.
In order for everyone to point and be pointed to, we require that the 5 people be divided up into closed loops, for instance $(A,B)(B,C)(C,D)(DA)$ would be a closed loop of 4 members. In a closed loop of $n$ members, there are $(n-1)!$ possible orderings of the loop.
One possibility is for everyone to be in a 5 member loop. In a 5 member loop, there are $4!=24$ possible ways to divide the people into ordered pairs.
It is impossible to have a 4 member loop, as that would force the fifth person to be in a loop by himself.
It is possible to have a 3 member loop and a 2 member loop. There are $frac{5!}{3!2!}=10$ ways of dividing the people into those loops, and for each of those 10 ways, there are 2 ways of ordering the 3 person loop, for a total of 20 possibilities of a 3 member loop and a 2 member loop.
If we have a 2 member loop, we could either divide the 3 remaining people into a 3 member loop or into at least 2 loops. We already considered the possibility of a 3 and 2 member loop, and dividing 3 people into 2 or more loops would result in having a 1 person loop. Therefore this adds no new possibilities.
It is impossible to have a 1 member loop.
Therefore, we found a total of 44 scenarios where each person points to someone else. The probability is $frac{44}{4^5}=frac{11}{256}$
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$D_n$, Derangment is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place.
What we are looking for would be
$$frac{D_5}{4^5}=frac{44}{4^5}=frac{11}{4^4}.$$
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
|
show 1 more comment
$D_n$, Derangment is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place.
What we are looking for would be
$$frac{D_5}{4^5}=frac{44}{4^5}=frac{11}{4^4}.$$
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
|
show 1 more comment
$D_n$, Derangment is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place.
What we are looking for would be
$$frac{D_5}{4^5}=frac{44}{4^5}=frac{11}{4^4}.$$
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
$D_n$, Derangment is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place.
What we are looking for would be
$$frac{D_5}{4^5}=frac{44}{4^5}=frac{11}{4^4}.$$
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
edited Dec 4 '18 at 10:07
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 4 '18 at 9:56
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
|
show 1 more comment
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
1
1
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
It should be $frac{11}{4^4}$, not $frac{11}{4^6}$
– MoKo19
Dec 4 '18 at 10:06
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
oops, thanks. i made a mistake.
– Siong Thye Goh
Dec 4 '18 at 10:07
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
I have been trying to wrap my head around this derangement and how it applies to this problem. My issue is that the order is allowed to be natural though. ABCDE is a possible answer.
– Ben Franks
Dec 4 '18 at 10:10
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
they cannot point to themselves right?
– Siong Thye Goh
Dec 4 '18 at 10:11
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
no, I meant A to B to C etc.
– Ben Franks
Dec 4 '18 at 10:20
|
show 1 more comment
Let us label the people A through E, and let $(A,B)$ mean that A pointed to B.
In order for everyone to point and be pointed to, we require that the 5 people be divided up into closed loops, for instance $(A,B)(B,C)(C,D)(DA)$ would be a closed loop of 4 members. In a closed loop of $n$ members, there are $(n-1)!$ possible orderings of the loop.
One possibility is for everyone to be in a 5 member loop. In a 5 member loop, there are $4!=24$ possible ways to divide the people into ordered pairs.
It is impossible to have a 4 member loop, as that would force the fifth person to be in a loop by himself.
It is possible to have a 3 member loop and a 2 member loop. There are $frac{5!}{3!2!}=10$ ways of dividing the people into those loops, and for each of those 10 ways, there are 2 ways of ordering the 3 person loop, for a total of 20 possibilities of a 3 member loop and a 2 member loop.
If we have a 2 member loop, we could either divide the 3 remaining people into a 3 member loop or into at least 2 loops. We already considered the possibility of a 3 and 2 member loop, and dividing 3 people into 2 or more loops would result in having a 1 person loop. Therefore this adds no new possibilities.
It is impossible to have a 1 member loop.
Therefore, we found a total of 44 scenarios where each person points to someone else. The probability is $frac{44}{4^5}=frac{11}{256}$
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
add a comment |
Let us label the people A through E, and let $(A,B)$ mean that A pointed to B.
In order for everyone to point and be pointed to, we require that the 5 people be divided up into closed loops, for instance $(A,B)(B,C)(C,D)(DA)$ would be a closed loop of 4 members. In a closed loop of $n$ members, there are $(n-1)!$ possible orderings of the loop.
One possibility is for everyone to be in a 5 member loop. In a 5 member loop, there are $4!=24$ possible ways to divide the people into ordered pairs.
It is impossible to have a 4 member loop, as that would force the fifth person to be in a loop by himself.
It is possible to have a 3 member loop and a 2 member loop. There are $frac{5!}{3!2!}=10$ ways of dividing the people into those loops, and for each of those 10 ways, there are 2 ways of ordering the 3 person loop, for a total of 20 possibilities of a 3 member loop and a 2 member loop.
If we have a 2 member loop, we could either divide the 3 remaining people into a 3 member loop or into at least 2 loops. We already considered the possibility of a 3 and 2 member loop, and dividing 3 people into 2 or more loops would result in having a 1 person loop. Therefore this adds no new possibilities.
It is impossible to have a 1 member loop.
Therefore, we found a total of 44 scenarios where each person points to someone else. The probability is $frac{44}{4^5}=frac{11}{256}$
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
add a comment |
Let us label the people A through E, and let $(A,B)$ mean that A pointed to B.
In order for everyone to point and be pointed to, we require that the 5 people be divided up into closed loops, for instance $(A,B)(B,C)(C,D)(DA)$ would be a closed loop of 4 members. In a closed loop of $n$ members, there are $(n-1)!$ possible orderings of the loop.
One possibility is for everyone to be in a 5 member loop. In a 5 member loop, there are $4!=24$ possible ways to divide the people into ordered pairs.
It is impossible to have a 4 member loop, as that would force the fifth person to be in a loop by himself.
It is possible to have a 3 member loop and a 2 member loop. There are $frac{5!}{3!2!}=10$ ways of dividing the people into those loops, and for each of those 10 ways, there are 2 ways of ordering the 3 person loop, for a total of 20 possibilities of a 3 member loop and a 2 member loop.
If we have a 2 member loop, we could either divide the 3 remaining people into a 3 member loop or into at least 2 loops. We already considered the possibility of a 3 and 2 member loop, and dividing 3 people into 2 or more loops would result in having a 1 person loop. Therefore this adds no new possibilities.
It is impossible to have a 1 member loop.
Therefore, we found a total of 44 scenarios where each person points to someone else. The probability is $frac{44}{4^5}=frac{11}{256}$
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
Let us label the people A through E, and let $(A,B)$ mean that A pointed to B.
In order for everyone to point and be pointed to, we require that the 5 people be divided up into closed loops, for instance $(A,B)(B,C)(C,D)(DA)$ would be a closed loop of 4 members. In a closed loop of $n$ members, there are $(n-1)!$ possible orderings of the loop.
One possibility is for everyone to be in a 5 member loop. In a 5 member loop, there are $4!=24$ possible ways to divide the people into ordered pairs.
It is impossible to have a 4 member loop, as that would force the fifth person to be in a loop by himself.
It is possible to have a 3 member loop and a 2 member loop. There are $frac{5!}{3!2!}=10$ ways of dividing the people into those loops, and for each of those 10 ways, there are 2 ways of ordering the 3 person loop, for a total of 20 possibilities of a 3 member loop and a 2 member loop.
If we have a 2 member loop, we could either divide the 3 remaining people into a 3 member loop or into at least 2 loops. We already considered the possibility of a 3 and 2 member loop, and dividing 3 people into 2 or more loops would result in having a 1 person loop. Therefore this adds no new possibilities.
It is impossible to have a 1 member loop.
Therefore, we found a total of 44 scenarios where each person points to someone else. The probability is $frac{44}{4^5}=frac{11}{256}$
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
edited Dec 4 '18 at 10:20
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
answered Dec 4 '18 at 10:04
MoKo19MoKo19
1914
1914
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
add a comment |
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
+1 thanks for the answer. Given answer to Siong Thye Goh as he answered first.
– Ben Franks
Dec 4 '18 at 10:22
add a comment |
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