Probability of successful Secret Santa selection
$begingroup$
In the following Secret Santa Scenario:
There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.
What is the probability that no one picks out their own name?
At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.
probability permutations combinations
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
$endgroup$
add a comment |
$begingroup$
In the following Secret Santa Scenario:
There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.
What is the probability that no one picks out their own name?
At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.
probability permutations combinations
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16
add a comment |
$begingroup$
In the following Secret Santa Scenario:
There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.
What is the probability that no one picks out their own name?
At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.
probability permutations combinations
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
$endgroup$
In the following Secret Santa Scenario:
There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.
What is the probability that no one picks out their own name?
At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.
probability permutations combinations
probability permutations combinations
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
asked Dec 15 '18 at 7:11
M. WeateM. Weate
445
445
$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16
add a comment |
$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16
$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16
$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)
Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by
$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$
among other potential relations. This is commonly denoted $!n$.
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
$begingroup$
Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.
The answer is
$$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
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
$begingroup$
There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)
Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by
$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$
among other potential relations. This is commonly denoted $!n$.
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
$begingroup$
There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)
Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by
$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$
among other potential relations. This is commonly denoted $!n$.
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
$begingroup$
There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)
Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by
$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$
among other potential relations. This is commonly denoted $!n$.
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)
Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by
$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$
among other potential relations. This is commonly denoted $!n$.
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
answered Dec 15 '18 at 7:15
Eevee TrainerEevee Trainer
10.4k31742
10.4k31742
add a comment |
add a comment |
$begingroup$
Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.
The answer is
$$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.
The answer is
$$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.
The answer is
$$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.
The answer is
$$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 15 '18 at 7:17
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16