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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?

Silvia Ghinassi
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MadMath
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1 Answers1

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Hint

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...

MT_
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  • 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:https://math.stackexchange.com/questions/627913/question-on-the-hat-check-problem?noredirect=1&lq=1 – James LT Nov 12 '17 at 20:42
  • @James This is in fact what I said in my answer. One can also use linearity of expectation. – MT_ Nov 13 '17 at 03:23