Mixing
Probability Question - A large number, $N$ , people go to a convention at a hotel. Each person is assigned...
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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 Jan 3 at 17:11
dmtri
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 17:11
dmtri
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
$endgroup$
7
$begingroup$
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.
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– lulu
Jan 3 at 16:11
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@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34
add a comment |
$begingroup$
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 Jan 3 at 17:11
dmtri
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
$endgroup$
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 Jan 3 at 17:11
dmtri
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
edited Jan 3 at 17:11
dmtri
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
edited Jan 3 at 17:11
dmtri
1,4522521
edited Jan 3 at 17:11
dmtri
1,4522521
edited Jan 3 at 17:11
dmtri
1,4522521
1,4522521
asked Jan 3 at 16:08
BlankQQBlankQQ
211
asked Jan 3 at 16:08
BlankQQBlankQQ
211
asked Jan 3 at 16:08
BlankQQBlankQQ
211
211
7
$begingroup$
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.
$endgroup$
– lulu
Jan 3 at 16:11
$begingroup$
@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34
add a comment |
7
$begingroup$
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.
$endgroup$
– lulu
Jan 3 at 16:11
$begingroup$
@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34
7
7
$begingroup$
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.
$endgroup$
– lulu
Jan 3 at 16:11
$begingroup$
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.
$endgroup$
– lulu
Jan 3 at 16:11
$begingroup$
@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34
$begingroup$
@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34
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 Jan 4 at 10:51
lulululu
39.7k24677
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 Jan 4 at 10:51
lulululu
39.7k24677
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 10:51
lulululu
39.7k24677
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 10:51
lulululu
39.7k24677
$endgroup$
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 Jan 4 at 10:51
lulululu
39.7k24677
answered Jan 4 at 10:51
lulululu
39.7k24677
answered Jan 4 at 10:51
lulululu
39.7k24677
answered Jan 4 at 10:51
lulululu
39.7k24677
39.7k24677
add a comment |
add a comment |
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7
$begingroup$
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.
$endgroup$
– lulu
Jan 3 at 16:11
$begingroup$
@lulu Please turn your comment into an answer.
$endgroup$
– leonbloy
Jan 4 at 3:34