# $15$ people were assigned seats in a room, however they sit randomly, what's the expected number of people seating in their original assigned seats?

15 people have assigned seats in a room, however they sit randomly, whats the expected number of people seating in their original assigned seats?
It can't be $\frac{1}{15} 15 = 1$ because that would be too easy...

$x_1$ siting on seat #1 will have probability of $\frac{1}{15}$, conditional on this information, $x_2$ on seat #2 will only have probably of $\frac{1}{14}$ these probabilities are not independent.

Can someone explain to me how is "linearity of expectation"used in this example? I think "Linearity of expectation" means that the expectation is linear even if the underlyings are dependent?

• Maybe too easy, but the expectation is indeed $1$. One needs to prove it. Dec 9, 2015 at 3:37
• Check this out: math.stackexchange.com/questions/627913/…
– user940
Dec 9, 2015 at 3:41

The first person has $\frac{1}{15}$ chance of sitting where they sat originally.
The second person has a $\frac{14}{15} \cdot \frac{1}{14}$ chance of sitting where they sat originally ($\frac{14}{15}$ because there is a $\frac{1}{15}$ chance that the first person sat in the second person's seat!)