Probability Question - A large number, $N$ , people go to a convention at a hotel. Each person is assigned...
A large number, $N$, people go to a convention at a hotel. Each person is assigned one of $N$ hotel rooms. The go to a party, but before they had their key to a doorman. On the way out of the party, the doorman hands the keys back at random, what is the probability that at least one person is given his/her original key.
The general equation I have worked out should be
$$P(X>=1) = 1 - P(X=0)$$
$$P(X=0) = (1-(1/N))^N$$
Adding up all the probabilities that one person gets the right key. I have checked with the answer key and this is a solution. However, it does not work for the case where $N = 2$.
The probability that at least one person gets the right key should be $50% $ ,but this equation returns $75%$ . Can anyone explain what logical error I am making?
probability discrete-mathematics contest-math
edited 2 days ago
dmtri
1,4321521
asked 2 days ago
BlankQQ
161
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A large number, $N$, people go to a convention at a hotel. Each person is assigned one of $N$ hotel rooms. The go to a party, but before they had their key to a doorman. On the way out of the party, the doorman hands the keys back at random, what is the probability that at least one person is given his/her original key.
The general equation I have worked out should be
$$P(X>=1) = 1 - P(X=0)$$
$$P(X=0) = (1-(1/N))^N$$
Adding up all the probabilities that one person gets the right key. I have checked with the answer key and this is a solution. However, it does not work for the case where $N = 2$.
The probability that at least one person gets the right key should be $50% $ ,but this equation returns $75%$ . Can anyone explain what logical error I am making?
probability discrete-mathematics contest-math
edited 2 days ago
dmtri
1,4321521
asked 2 days ago
BlankQQ
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
7
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
@lulu Please turn your comment into an answer.
– leonbloy
yesterday
add a comment |
A large number, $N$, people go to a convention at a hotel. Each person is assigned one of $N$ hotel rooms. The go to a party, but before they had their key to a doorman. On the way out of the party, the doorman hands the keys back at random, what is the probability that at least one person is given his/her original key.
The general equation I have worked out should be
$$P(X>=1) = 1 - P(X=0)$$
$$P(X=0) = (1-(1/N))^N$$
Adding up all the probabilities that one person gets the right key. I have checked with the answer key and this is a solution. However, it does not work for the case where $N = 2$.
The probability that at least one person gets the right key should be $50% $ ,but this equation returns $75%$ . Can anyone explain what logical error I am making?
probability discrete-mathematics contest-math
edited 2 days ago
dmtri
1,4321521
asked 2 days ago
BlankQQ
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A large number, $N$, people go to a convention at a hotel. Each person is assigned one of $N$ hotel rooms. The go to a party, but before they had their key to a doorman. On the way out of the party, the doorman hands the keys back at random, what is the probability that at least one person is given his/her original key.
The general equation I have worked out should be
$$P(X>=1) = 1 - P(X=0)$$
$$P(X=0) = (1-(1/N))^N$$
Adding up all the probabilities that one person gets the right key. I have checked with the answer key and this is a solution. However, it does not work for the case where $N = 2$.
The probability that at least one person gets the right key should be $50% $ ,but this equation returns $75%$ . Can anyone explain what logical error I am making?
probability discrete-mathematics contest-math
probability discrete-mathematics contest-math
edited 2 days ago
dmtri
1,4321521
asked 2 days ago
BlankQQ
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
dmtri
1,4321521
asked 2 days ago
BlankQQ
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
dmtri
1,4321521
edited 2 days ago
dmtri
1,4321521
edited 2 days ago
dmtri
1,4321521
1,4321521
asked 2 days ago
BlankQQ
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
BlankQQ
161
asked 2 days ago
BlankQQ
161
161
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
BlankQQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
7
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
@lulu Please turn your comment into an answer.
– leonbloy
yesterday
add a comment |
7
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
@lulu Please turn your comment into an answer.
– leonbloy
yesterday
7
7
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
@lulu Please turn your comment into an answer.
– leonbloy
yesterday
@lulu Please turn your comment into an answer.
– leonbloy
yesterday
add a comment |
1 Answer
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The error in your method lies in the fact that you treat "$A$ getting the wrong key " and "$B$ getting the wrong key" as independent events. They aren't. For instance, if you only have two people then knowing that the first one gets the wrong key makes it certain that the second also gets the wrong key.
To solve the problem, use Derangements. The keys people are given can be thought of as a permutation of the names, so your question becomes "what is the probability that a (uniformly) randomly chosen permutation is a derangement?" for which the answer is clearly $$frac {!n}{n!}$$
Note that for large $n$ this approaches $frac 1e$ so, again for large $n$, the numerical probability is about $36.79%$
answered yesterday
lulu
39.2k24677
add a comment |
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1 Answer
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The error in your method lies in the fact that you treat "$A$ getting the wrong key " and "$B$ getting the wrong key" as independent events. They aren't. For instance, if you only have two people then knowing that the first one gets the wrong key makes it certain that the second also gets the wrong key.
To solve the problem, use Derangements. The keys people are given can be thought of as a permutation of the names, so your question becomes "what is the probability that a (uniformly) randomly chosen permutation is a derangement?" for which the answer is clearly $$frac {!n}{n!}$$
Note that for large $n$ this approaches $frac 1e$ so, again for large $n$, the numerical probability is about $36.79%$
answered yesterday
lulu
39.2k24677
add a comment |
The error in your method lies in the fact that you treat "$A$ getting the wrong key " and "$B$ getting the wrong key" as independent events. They aren't. For instance, if you only have two people then knowing that the first one gets the wrong key makes it certain that the second also gets the wrong key.
To solve the problem, use Derangements. The keys people are given can be thought of as a permutation of the names, so your question becomes "what is the probability that a (uniformly) randomly chosen permutation is a derangement?" for which the answer is clearly $$frac {!n}{n!}$$
Note that for large $n$ this approaches $frac 1e$ so, again for large $n$, the numerical probability is about $36.79%$
answered yesterday
lulu
39.2k24677
add a comment |
The error in your method lies in the fact that you treat "$A$ getting the wrong key " and "$B$ getting the wrong key" as independent events. They aren't. For instance, if you only have two people then knowing that the first one gets the wrong key makes it certain that the second also gets the wrong key.
To solve the problem, use Derangements. The keys people are given can be thought of as a permutation of the names, so your question becomes "what is the probability that a (uniformly) randomly chosen permutation is a derangement?" for which the answer is clearly $$frac {!n}{n!}$$
Note that for large $n$ this approaches $frac 1e$ so, again for large $n$, the numerical probability is about $36.79%$
answered yesterday
lulu
39.2k24677
The error in your method lies in the fact that you treat "$A$ getting the wrong key " and "$B$ getting the wrong key" as independent events. They aren't. For instance, if you only have two people then knowing that the first one gets the wrong key makes it certain that the second also gets the wrong key.
To solve the problem, use Derangements. The keys people are given can be thought of as a permutation of the names, so your question becomes "what is the probability that a (uniformly) randomly chosen permutation is a derangement?" for which the answer is clearly $$frac {!n}{n!}$$
Note that for large $n$ this approaches $frac 1e$ so, again for large $n$, the numerical probability is about $36.79%$
answered yesterday
lulu
39.2k24677
answered yesterday
lulu
39.2k24677
answered yesterday
lulu
39.2k24677
answered yesterday
lulu
39.2k24677
39.2k24677
add a comment |
add a comment |
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7
Your method is not correct as the events you consider are dependent. With two people, say, knowing that the first one got the wrong key implies the second one did as well. The relevant concept is Derangements.
– lulu
2 days ago
@lulu Please turn your comment into an answer.
– leonbloy
yesterday