MHB Help With Probability Problem: Find Mean & Variance of Vn

[Mike2323]

Hello and i need your help with the following statistic ploblem!

in a party there are n people (n >=2). each man cast identity in a common box.They mix and each person chooses identification at random from the box, without resetting. We say that we have success if one chooses his identity. If with Vn denote the number of successes, find the mean and variance of Vn

A tried to use Gambler's Ruin and i don't know if i get something the answer

I made this think

Pn=0 n=0
Pn=1 N=1

PPn+1+(1-P)Pn-1 (for 0<n<N)Please, if someone could help me i would appreciate it since i didn't answer this problem in an old test!

THank you so much with your time!

 

[I like Serena]

Hello and i need your help with the following statistic ploblem!

in a party there are n people (n >=2). each man cast identity in a common box.They mix and each person chooses identification at random from the box, without resetting. We say that we have success if one chooses his identity. If with Vn denote the number of successes, find the mean and variance of Vn

A tried to use Gambler's Ruin and i don't know if i get something the answer

I made this think

Pn=0 n=0
Pn=1 N=1

PPn+1+(1-P)Pn-1 (for 0<n<N)Please, if someone could help me i would appreciate it since i didn't answer this problem in an old test!

THank you so much with your time!

Hi Mike2323! Welcome to MHB! ;)The probability that $k$ persons in a group of $n$ pick their own identity is given by:
$$P(V_n=k)=\frac{\text{#Permutations that leaves $k$ of $n$ identities the same}}{\text{#Permutations of $n$ identities}}$$
where \(\text{#Permutations of $n$ identities}=n!\).For $n=2$ either the identities are swapped, or they are not.
So $P(V_2=0)=\frac{1}{2!},\ P(V_2=1)=\frac{0}{2!},\ P(V_2=2)=\frac{1}{2!}$.
That gives us the expectation:
$$E_2 = \frac 12 \cdot 0 + \frac 12 \cdot 2 = 1$$
and the variance:
$$\sigma_2^2 = \frac 12 \cdot (-1)^2 + \frac 12 \cdot (1)^2 = 1$$

For $n=3$ either the identities are cyclically shifted (2 possibilities), or 2 identities have been swapped (3 possibilities), or everyone gets their own identity (1 possibility).
So $P(V_3=0)=\frac{2}{3!},\ P(V_3=1)=\frac{3}{3!},\ P(V_3=2)=\frac{0}{3!},\ P(V_3=3)=\frac{1}{3!}$.
That gives us the expectation:
$$E_3 = \frac 26 \cdot 0 + \frac 36 \cdot 1 + \frac 16 \cdot 3 = 1$$
and the variance:
$$\sigma_3^2 = \frac 26 \cdot (-1)^2 + \frac 16 \cdot (2)^2 = 1$$

I think I'm seeing a pattern here...
If I do the same thing for $n=4$ and $n=5$, I'm getting again that $E=1$ and $\sigma^2=1$... (Thinking)