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

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The discussion revolves around calculating the mean and variance of the number of successes, Vn, in a scenario where n people randomly select their identities from a box. The probability of k individuals picking their own identity is derived from permutations, leading to specific calculations for n=2 and n=3, both yielding an expected value (mean) of 1 and a variance of 1. The user notes a consistent pattern emerging, suggesting that for larger n, the mean and variance may also equal 1. This insight indicates a potential generalization for the problem across different values of n. The conversation highlights the mathematical approach to solving probability problems in combinatorial contexts.
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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!
 
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Mike2323 said:
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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