Probability of labelling people in a lineup
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I was watching this video, in which a "professional astrologer" is tasked with matching up people with their star signs.
(Spoiler alert!) At around the 5:30 mark, 4 of the 12 participants reveal that their star sign was correctly identified.
This leads to the question:
What is the probability of him getting 4 or more star signs correct, and can this be nicely generalised to getting $x$ or more correct answers out of a possible $y$ people?
Note that each of the twelve participants have a different star sign, and so do each of the 12 cards.
There is a similar question here - coincidentally inspired by the same series - however, this focuses on the expected value rather than the general probability.
Thanks,
Sam
probability combinatorics
asked Nov 19 at 20:06
Samuel Banks
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0
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favorite
I was watching this video, in which a "professional astrologer" is tasked with matching up people with their star signs.
(Spoiler alert!) At around the 5:30 mark, 4 of the 12 participants reveal that their star sign was correctly identified.
This leads to the question:
What is the probability of him getting 4 or more star signs correct, and can this be nicely generalised to getting $x$ or more correct answers out of a possible $y$ people?
Note that each of the twelve participants have a different star sign, and so do each of the 12 cards.
There is a similar question here - coincidentally inspired by the same series - however, this focuses on the expected value rather than the general probability.
Thanks,
Sam
probability combinatorics
asked Nov 19 at 20:06
Samuel Banks
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was watching this video, in which a "professional astrologer" is tasked with matching up people with their star signs.
(Spoiler alert!) At around the 5:30 mark, 4 of the 12 participants reveal that their star sign was correctly identified.
This leads to the question:
What is the probability of him getting 4 or more star signs correct, and can this be nicely generalised to getting $x$ or more correct answers out of a possible $y$ people?
Note that each of the twelve participants have a different star sign, and so do each of the 12 cards.
There is a similar question here - coincidentally inspired by the same series - however, this focuses on the expected value rather than the general probability.
Thanks,
Sam
probability combinatorics
asked Nov 19 at 20:06
Samuel Banks
1
I was watching this video, in which a "professional astrologer" is tasked with matching up people with their star signs.
(Spoiler alert!) At around the 5:30 mark, 4 of the 12 participants reveal that their star sign was correctly identified.
This leads to the question:
What is the probability of him getting 4 or more star signs correct, and can this be nicely generalised to getting $x$ or more correct answers out of a possible $y$ people?
Note that each of the twelve participants have a different star sign, and so do each of the 12 cards.
There is a similar question here - coincidentally inspired by the same series - however, this focuses on the expected value rather than the general probability.
Thanks,
Sam
probability combinatorics
probability combinatorics
asked Nov 19 at 20:06
Samuel Banks
1
asked Nov 19 at 20:06
Samuel Banks
1
asked Nov 19 at 20:06
Samuel Banks
1
asked Nov 19 at 20:06
Samuel Banks
1
asked Nov 19 at 20:06
Samuel Banks
1
1
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add a comment |
1 Answer
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Not an explicit answer, but some references.
What you're looking for is $X = $the number of fixed points in a random permutation. (I.e. if the astrologer guesses a random permutation, how many would he/she guess correctly?)
The distribution of $X$ can be obtained via the Inclusion-Exclusion Principle. Here is one reference that includes the formula for $P(X = k)$:
http://demonstrations.wolfram.com/TheNumberOfFixedPointsInARandomPermutation/
And https://en.wikipedia.org/wiki/Random_permutation_statistics contains a ton more details on related problems, but AFAICT not the explicit formula.
answered Nov 19 at 22:38
antkam
1,373112
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Not an explicit answer, but some references.
What you're looking for is $X = $the number of fixed points in a random permutation. (I.e. if the astrologer guesses a random permutation, how many would he/she guess correctly?)
The distribution of $X$ can be obtained via the Inclusion-Exclusion Principle. Here is one reference that includes the formula for $P(X = k)$:
http://demonstrations.wolfram.com/TheNumberOfFixedPointsInARandomPermutation/
And https://en.wikipedia.org/wiki/Random_permutation_statistics contains a ton more details on related problems, but AFAICT not the explicit formula.
answered Nov 19 at 22:38
antkam
1,373112
add a comment |
up vote
1
down vote
Not an explicit answer, but some references.
What you're looking for is $X = $the number of fixed points in a random permutation. (I.e. if the astrologer guesses a random permutation, how many would he/she guess correctly?)
The distribution of $X$ can be obtained via the Inclusion-Exclusion Principle. Here is one reference that includes the formula for $P(X = k)$:
http://demonstrations.wolfram.com/TheNumberOfFixedPointsInARandomPermutation/
And https://en.wikipedia.org/wiki/Random_permutation_statistics contains a ton more details on related problems, but AFAICT not the explicit formula.
answered Nov 19 at 22:38
antkam
1,373112
add a comment |
up vote
1
down vote
up vote
1
down vote
Not an explicit answer, but some references.
What you're looking for is $X = $the number of fixed points in a random permutation. (I.e. if the astrologer guesses a random permutation, how many would he/she guess correctly?)
The distribution of $X$ can be obtained via the Inclusion-Exclusion Principle. Here is one reference that includes the formula for $P(X = k)$:
http://demonstrations.wolfram.com/TheNumberOfFixedPointsInARandomPermutation/
And https://en.wikipedia.org/wiki/Random_permutation_statistics contains a ton more details on related problems, but AFAICT not the explicit formula.
answered Nov 19 at 22:38
antkam
1,373112
Not an explicit answer, but some references.
What you're looking for is $X = $the number of fixed points in a random permutation. (I.e. if the astrologer guesses a random permutation, how many would he/she guess correctly?)
The distribution of $X$ can be obtained via the Inclusion-Exclusion Principle. Here is one reference that includes the formula for $P(X = k)$:
http://demonstrations.wolfram.com/TheNumberOfFixedPointsInARandomPermutation/
And https://en.wikipedia.org/wiki/Random_permutation_statistics contains a ton more details on related problems, but AFAICT not the explicit formula.
answered Nov 19 at 22:38
antkam
1,373112
answered Nov 19 at 22:38
antkam
1,373112
answered Nov 19 at 22:38
antkam
1,373112
answered Nov 19 at 22:38
antkam
1,373112
1,373112
add a comment |
add a comment |
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