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A Rare Riley Riddle

N people drop their own key into a basket, after shuffling, every one randomly pick one, X is number of people get their own key. what is E(X),Var(X)? [closed]

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$begingroup$

$N$ people drop their own key into a basket, after shuffling, every one randomly pick one, $X$ is number of people get their own key. what is $E(X)$,$mathrm{Var}(X)$?

probability

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edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

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closed as off-topic by Leucippus, dantopa, StubbornAtom, jgon, Eevee Trainer Mar 16 at 0:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, dantopa, StubbornAtom, jgon, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

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  $begingroup$
  Where is your attempt? What is the probability that all persons get their own keys?
  $endgroup$
  – callculus
  Mar 15 at 18:42
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should find $E[X]$ easy to answer. What is the probability a particular individual gets their own key?
  $endgroup$
  – Henry
  Mar 15 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 15 at 19:07
  

  
  
  
  

add a comment | 

-1

$begingroup$

$N$ people drop their own key into a basket, after shuffling, every one randomly pick one, $X$ is number of people get their own key. what is $E(X)$,$mathrm{Var}(X)$?

probability

share|cite|improve this question

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

$endgroup$

closed as off-topic by Leucippus, dantopa, StubbornAtom, jgon, Eevee Trainer Mar 16 at 0:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, dantopa, StubbornAtom, jgon, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where is your attempt? What is the probability that all persons get their own keys?
  $endgroup$
  – callculus
  Mar 15 at 18:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should find $E[X]$ easy to answer. What is the probability a particular individual gets their own key?
  $endgroup$
  – Henry
  Mar 15 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 15 at 19:07
  

  
  
  
  

add a comment | 

$begingroup$

$N$ people drop their own key into a basket, after shuffling, every one randomly pick one, $X$ is number of people get their own key. what is $E(X)$,$mathrm{Var}(X)$?

probability

share|cite|improve this question

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

$endgroup$

share|cite|improve this question

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

share|cite|improve this question

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

edited Mar 15 at 20:34

Daniele Tampieri

2,58721022

asked Mar 15 at 18:36

walkerwalker

1

asked Mar 15 at 18:36

walkerwalker

1

2 Answers
2

active

oldest

votes

3

$begingroup$

Use linearity of expectation, indicator random variables, and remember that $Var(X) = E[X^2] - E[X]^2$

Let $X = X_1+X_2+dots+X_n$ where $X_i = begin{cases} 1&text{if person}~i~text{gets their key}\0&text{otherwise}end{cases}$

You correctly found that $Pr(X_i=1)=frac{1}{n}$ and as such $E[X] = ncdot frac{1}{n}=1$

Now, consider $X^2 = (X_1+X_2+dots+X_n)^2 = X_1^2+X_2^2+dots+X_n^2 + X_1X_2 + X_1X_3+dots+X_nX_{n-1}$

$X_i^2 = X_i$ so those terms are easy to deal with. $X_iX_j$ with $ineq j$ on the other hand will equal $1$ if and only if both persons $i$ and $j$ receive their keys.

We have for $ineq j$ that $Pr(X_iX_j=1) = Pr(X_i=1)Pr(X_j=1mid X_i=1) = frac{1}{n}cdot frac{1}{n-1}$

Now, taking note of the number of occurrences of each and letting $ineq j$ we have:

$E[X^2] = ncdot E[X_i] + n(n-1)cdot E[X_iX_j] = ncdot frac{1}{n} + n(n-1)cdot frac{1}{n}cdotfrac{1}{n-1} = 1 + 1=2$

So, $Var(X) = E[X^2]-E[X]^2 = 2 - 1^2 = 1$

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answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

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add a comment | 

1

$begingroup$

Expected value is easy to calculate; every person has an equal chance of pulling his own key.

The first is uniformly distributed. The second can only pull his key if the first guy did not pull it, and so on.

$E(X) = $
$$frac{1}{n} + frac{1}{n-1}*frac{n-1}{n}+ frac{1}{n-2}*frac{n-2}{n-1}*frac{n-1}{n} +...$$

$$=1$$

$VAR(X) = E(X^2) - E(X)^2 $

let $Y_i$ denote whether or not person $i$ managed to find his own key.

then $E(X^2) = E(Y_iY_j) $ for all $i,j$

note that distinct $i, j$ result in $E(Y_iY_j)=$ $$frac{1}{n(n-1)}$$
The number of distinct pairs is $n(n-1)/2$

By linearity of expectation you get that $E(X^2) = 2$
hence:

$VAR(X) = 2 - 1 = 1$

share|cite|improve this answer

answered Mar 15 at 19:33

loxlox

212

$endgroup$

add a comment | 

3

$begingroup$

Use linearity of expectation, indicator random variables, and remember that $Var(X) = E[X^2] - E[X]^2$

Let $X = X_1+X_2+dots+X_n$ where $X_i = begin{cases} 1&text{if person}~i~text{gets their key}\0&text{otherwise}end{cases}$

You correctly found that $Pr(X_i=1)=frac{1}{n}$ and as such $E[X] = ncdot frac{1}{n}=1$

Now, consider $X^2 = (X_1+X_2+dots+X_n)^2 = X_1^2+X_2^2+dots+X_n^2 + X_1X_2 + X_1X_3+dots+X_nX_{n-1}$

$X_i^2 = X_i$ so those terms are easy to deal with. $X_iX_j$ with $ineq j$ on the other hand will equal $1$ if and only if both persons $i$ and $j$ receive their keys.

We have for $ineq j$ that $Pr(X_iX_j=1) = Pr(X_i=1)Pr(X_j=1mid X_i=1) = frac{1}{n}cdot frac{1}{n-1}$

Now, taking note of the number of occurrences of each and letting $ineq j$ we have:

$E[X^2] = ncdot E[X_i] + n(n-1)cdot E[X_iX_j] = ncdot frac{1}{n} + n(n-1)cdot frac{1}{n}cdotfrac{1}{n-1} = 1 + 1=2$

So, $Var(X) = E[X^2]-E[X]^2 = 2 - 1^2 = 1$

share|cite|improve this answer

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

$endgroup$

add a comment | 

3

$begingroup$

Use linearity of expectation, indicator random variables, and remember that $Var(X) = E[X^2] - E[X]^2$

Let $X = X_1+X_2+dots+X_n$ where $X_i = begin{cases} 1&text{if person}~i~text{gets their key}\0&text{otherwise}end{cases}$

You correctly found that $Pr(X_i=1)=frac{1}{n}$ and as such $E[X] = ncdot frac{1}{n}=1$

Now, consider $X^2 = (X_1+X_2+dots+X_n)^2 = X_1^2+X_2^2+dots+X_n^2 + X_1X_2 + X_1X_3+dots+X_nX_{n-1}$

$X_i^2 = X_i$ so those terms are easy to deal with. $X_iX_j$ with $ineq j$ on the other hand will equal $1$ if and only if both persons $i$ and $j$ receive their keys.

We have for $ineq j$ that $Pr(X_iX_j=1) = Pr(X_i=1)Pr(X_j=1mid X_i=1) = frac{1}{n}cdot frac{1}{n-1}$

Now, taking note of the number of occurrences of each and letting $ineq j$ we have:

$E[X^2] = ncdot E[X_i] + n(n-1)cdot E[X_iX_j] = ncdot frac{1}{n} + n(n-1)cdot frac{1}{n}cdotfrac{1}{n-1} = 1 + 1=2$

So, $Var(X) = E[X^2]-E[X]^2 = 2 - 1^2 = 1$

share|cite|improve this answer

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

$endgroup$

add a comment | 

$begingroup$

Use linearity of expectation, indicator random variables, and remember that $Var(X) = E[X^2] - E[X]^2$

Let $X = X_1+X_2+dots+X_n$ where $X_i = begin{cases} 1&text{if person}~i~text{gets their key}\0&text{otherwise}end{cases}$

You correctly found that $Pr(X_i=1)=frac{1}{n}$ and as such $E[X] = ncdot frac{1}{n}=1$

Now, consider $X^2 = (X_1+X_2+dots+X_n)^2 = X_1^2+X_2^2+dots+X_n^2 + X_1X_2 + X_1X_3+dots+X_nX_{n-1}$

$X_i^2 = X_i$ so those terms are easy to deal with. $X_iX_j$ with $ineq j$ on the other hand will equal $1$ if and only if both persons $i$ and $j$ receive their keys.

We have for $ineq j$ that $Pr(X_iX_j=1) = Pr(X_i=1)Pr(X_j=1mid X_i=1) = frac{1}{n}cdot frac{1}{n-1}$

Now, taking note of the number of occurrences of each and letting $ineq j$ we have:

$E[X^2] = ncdot E[X_i] + n(n-1)cdot E[X_iX_j] = ncdot frac{1}{n} + n(n-1)cdot frac{1}{n}cdotfrac{1}{n-1} = 1 + 1=2$

So, $Var(X) = E[X^2]-E[X]^2 = 2 - 1^2 = 1$

share|cite|improve this answer

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

$endgroup$

share|cite|improve this answer

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

answered Mar 15 at 19:20

JMoravitzJMoravitz

48.7k43988

1

$begingroup$

Expected value is easy to calculate; every person has an equal chance of pulling his own key.

The first is uniformly distributed. The second can only pull his key if the first guy did not pull it, and so on.

$E(X) = $
$$frac{1}{n} + frac{1}{n-1}*frac{n-1}{n}+ frac{1}{n-2}*frac{n-2}{n-1}*frac{n-1}{n} +...$$

$$=1$$

$VAR(X) = E(X^2) - E(X)^2 $

let $Y_i$ denote whether or not person $i$ managed to find his own key.

then $E(X^2) = E(Y_iY_j) $ for all $i,j$

note that distinct $i, j$ result in $E(Y_iY_j)=$ $$frac{1}{n(n-1)}$$
The number of distinct pairs is $n(n-1)/2$

By linearity of expectation you get that $E(X^2) = 2$
hence:

$VAR(X) = 2 - 1 = 1$

share|cite|improve this answer

answered Mar 15 at 19:33

loxlox

212

$endgroup$

add a comment | 

1

$begingroup$

Expected value is easy to calculate; every person has an equal chance of pulling his own key.

The first is uniformly distributed. The second can only pull his key if the first guy did not pull it, and so on.

$E(X) = $
$$frac{1}{n} + frac{1}{n-1}*frac{n-1}{n}+ frac{1}{n-2}*frac{n-2}{n-1}*frac{n-1}{n} +...$$

$$=1$$

$VAR(X) = E(X^2) - E(X)^2 $

let $Y_i$ denote whether or not person $i$ managed to find his own key.

then $E(X^2) = E(Y_iY_j) $ for all $i,j$

note that distinct $i, j$ result in $E(Y_iY_j)=$ $$frac{1}{n(n-1)}$$
The number of distinct pairs is $n(n-1)/2$

By linearity of expectation you get that $E(X^2) = 2$
hence:

$VAR(X) = 2 - 1 = 1$

share|cite|improve this answer

answered Mar 15 at 19:33

loxlox

212

$endgroup$

add a comment | 

$begingroup$

Expected value is easy to calculate; every person has an equal chance of pulling his own key.

The first is uniformly distributed. The second can only pull his key if the first guy did not pull it, and so on.

$E(X) = $
$$frac{1}{n} + frac{1}{n-1}*frac{n-1}{n}+ frac{1}{n-2}*frac{n-2}{n-1}*frac{n-1}{n} +...$$

$$=1$$

$VAR(X) = E(X^2) - E(X)^2 $

let $Y_i$ denote whether or not person $i$ managed to find his own key.

then $E(X^2) = E(Y_iY_j) $ for all $i,j$

note that distinct $i, j$ result in $E(Y_iY_j)=$ $$frac{1}{n(n-1)}$$
The number of distinct pairs is $n(n-1)/2$

By linearity of expectation you get that $E(X^2) = 2$
hence:

$VAR(X) = 2 - 1 = 1$

share|cite|improve this answer

answered Mar 15 at 19:33

loxlox

212

$endgroup$

share|cite|improve this answer

answered Mar 15 at 19:33

loxlox

212

answered Mar 15 at 19:33

loxlox

212

answered Mar 15 at 19:33

loxlox

212