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?


1 Answer 1



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!)

And so forth...

  • $\begingroup$ This is not true because if the first person sits on the place that belongs to the second person then the prob for the second person is zero. I recommend this link:math.stackexchange.com/questions/627913/… $\endgroup$
    – James LT
    Nov 12, 2017 at 20:42
  • $\begingroup$ @James This is in fact what I said in my answer. One can also use linearity of expectation. $\endgroup$
    – MT_
    Nov 13, 2017 at 3:23

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