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Probability of at least two people choosing the same menu
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A group of $n>4$ people gathers for a feast. Each
person brings $n$ dishes of his/her own unique stew. Then, each person
forms his/her own menu by choosing 4 distinct dishes (of different
stews), randomly and independently of any of the other people. What is the probability that at least two people chose the
exact same menu?
What I got so far is since the probability for a person to choose a specific menu is $p=frac1n choose 4$, if $X$ is the random variable of the number of people choosing "that" menu, then $Pr(Xgeq 2)=sum_k=2^n n choose kp^k(1-p)^n-k\$, but even after simplyfing this expression it doesn't resemble any of the answers (it's a multiple choice).
So I could really use some guidance here.
probability combinatorics
asked Jul 25 at 17:11
gbi1977
1247
add a comment |Â
up vote
1
down vote
favorite
A group of $n>4$ people gathers for a feast. Each
person brings $n$ dishes of his/her own unique stew. Then, each person
forms his/her own menu by choosing 4 distinct dishes (of different
stews), randomly and independently of any of the other people. What is the probability that at least two people chose the
exact same menu?
What I got so far is since the probability for a person to choose a specific menu is $p=frac1n choose 4$, if $X$ is the random variable of the number of people choosing "that" menu, then $Pr(Xgeq 2)=sum_k=2^n n choose kp^k(1-p)^n-k\$, but even after simplyfing this expression it doesn't resemble any of the answers (it's a multiple choice).
So I could really use some guidance here.
probability combinatorics
asked Jul 25 at 17:11
gbi1977
1247
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A group of $n>4$ people gathers for a feast. Each
person brings $n$ dishes of his/her own unique stew. Then, each person
forms his/her own menu by choosing 4 distinct dishes (of different
stews), randomly and independently of any of the other people. What is the probability that at least two people chose the
exact same menu?
What I got so far is since the probability for a person to choose a specific menu is $p=frac1n choose 4$, if $X$ is the random variable of the number of people choosing "that" menu, then $Pr(Xgeq 2)=sum_k=2^n n choose kp^k(1-p)^n-k\$, but even after simplyfing this expression it doesn't resemble any of the answers (it's a multiple choice).
So I could really use some guidance here.
probability combinatorics
asked Jul 25 at 17:11
gbi1977
1247
A group of $n>4$ people gathers for a feast. Each
person brings $n$ dishes of his/her own unique stew. Then, each person
forms his/her own menu by choosing 4 distinct dishes (of different
stews), randomly and independently of any of the other people. What is the probability that at least two people chose the
exact same menu?
What I got so far is since the probability for a person to choose a specific menu is $p=frac1n choose 4$, if $X$ is the random variable of the number of people choosing "that" menu, then $Pr(Xgeq 2)=sum_k=2^n n choose kp^k(1-p)^n-k\$, but even after simplyfing this expression it doesn't resemble any of the answers (it's a multiple choice).
So I could really use some guidance here.
probability combinatorics
asked Jul 25 at 17:11
gbi1977
1247
asked Jul 25 at 17:11
gbi1977
1247
asked Jul 25 at 17:11
gbi1977
1247
asked Jul 25 at 17:11
gbi1977
1247
1247
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
There are $N=binom n4$ possible possibilities for a menu, and there are enough dishes of each type so that
the persons are free to choose. So the problem can be restated, and let us better ask for the complementary probability:
$n$ persons choose in the same time, independently one of $N$ menus, that exist in overflow. Which is the probability for $n$ different choices?
The answer is
$$
frac1N^ncdot (N-0)(N-1)dots(N-(n-1))
=
left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright)
.
$$
The complement is the answer to the OP
$$
1-left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright) ,qquad
N = binom n4=frac 124n(n-1)(n-2)(n-3)
.
$$
answered Jul 25 at 17:26
dan_fulea
4,1421211
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
There are $N=binom n4$ possible possibilities for a menu, and there are enough dishes of each type so that
the persons are free to choose. So the problem can be restated, and let us better ask for the complementary probability:
$n$ persons choose in the same time, independently one of $N$ menus, that exist in overflow. Which is the probability for $n$ different choices?
The answer is
$$
frac1N^ncdot (N-0)(N-1)dots(N-(n-1))
=
left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright)
.
$$
The complement is the answer to the OP
$$
1-left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright) ,qquad
N = binom n4=frac 124n(n-1)(n-2)(n-3)
.
$$
answered Jul 25 at 17:26
dan_fulea
4,1421211
add a comment |Â
up vote
3
down vote
accepted
There are $N=binom n4$ possible possibilities for a menu, and there are enough dishes of each type so that
the persons are free to choose. So the problem can be restated, and let us better ask for the complementary probability:
$n$ persons choose in the same time, independently one of $N$ menus, that exist in overflow. Which is the probability for $n$ different choices?
The answer is
$$
frac1N^ncdot (N-0)(N-1)dots(N-(n-1))
=
left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright)
.
$$
The complement is the answer to the OP
$$
1-left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright) ,qquad
N = binom n4=frac 124n(n-1)(n-2)(n-3)
.
$$
answered Jul 25 at 17:26
dan_fulea
4,1421211
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
There are $N=binom n4$ possible possibilities for a menu, and there are enough dishes of each type so that
the persons are free to choose. So the problem can be restated, and let us better ask for the complementary probability:
$n$ persons choose in the same time, independently one of $N$ menus, that exist in overflow. Which is the probability for $n$ different choices?
The answer is
$$
frac1N^ncdot (N-0)(N-1)dots(N-(n-1))
=
left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright)
.
$$
The complement is the answer to the OP
$$
1-left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright) ,qquad
N = binom n4=frac 124n(n-1)(n-2)(n-3)
.
$$
answered Jul 25 at 17:26
dan_fulea
4,1421211
There are $N=binom n4$ possible possibilities for a menu, and there are enough dishes of each type so that
the persons are free to choose. So the problem can be restated, and let us better ask for the complementary probability:
$n$ persons choose in the same time, independently one of $N$ menus, that exist in overflow. Which is the probability for $n$ different choices?
The answer is
$$
frac1N^ncdot (N-0)(N-1)dots(N-(n-1))
=
left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright)
.
$$
The complement is the answer to the OP
$$
1-left(1-frac 1Nright)
left(1-frac 2Nright)
dots
left(1-frac n-1Nright) ,qquad
N = binom n4=frac 124n(n-1)(n-2)(n-3)
.
$$
answered Jul 25 at 17:26
dan_fulea
4,1421211
answered Jul 25 at 17:26
dan_fulea
4,1421211
answered Jul 25 at 17:26
dan_fulea
4,1421211
answered Jul 25 at 17:26
dan_fulea
4,1421211
4,1421211
add a comment |Â
add a comment |Â
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