# Beginner probability question: Can you help me refine my understanding of the Linearity of Expectation?

In the beginning I was really impressed by the Linearity of Expectation working even with crazy amounts of dependency.

But then I heard this example from youtube:

Say n people sit around a spinner (a “lazy-Susan”) with n different dishes. Spin randomly. How many people do we expect will get same dish as initially? The answer is one! Because you can break it up into n indicator variables each with expected value 1/n. Note that in reality however, either everyone will get their same dish or no one will get their same dish. Either n or 0. But expected value is 1.

My unrefined train of thinking, can you point out illogical bits in the following?

• It seems to me the expected value of 1 in the lazy susan problem above is not very useful information... in fact, someone might be easily misled by this result. It seems to me that therefore expected value is not so powerful, so linearity of expectation is also not so powerful. Therefore I shouldn't be so impressed by linearity of expectation.

• LOE overcomes dependency by ignoring it in the first place. And the impressive feelings I had for the ability to overcome dependency was my failure to realize the expected value is for an initial condition with no updated information (conditioning). Also, in the lazy susan example above, the expected value was not so terribly useful.

• Then again, I am impressed how it works for calculating the expected value of the hyper geometric RV and binomial so much faster than doing it the old school way with the pmf

• $$\sum_{x_1,\cdots , x_N}\sum_{k=1}^N x_k P(x_1,\cdots , x_N) = \sum_{k=1}^N \left<x_k\right>$$ – Count Iblis Sep 9 '17 at 20:57
• Expected values are averages and as such are no more or less useful than other types of averages. When you use them, you always discard potentially useful information. Also, remember that the expected value of a random variable might not even be one of the values that it can take on. E.g., if $X=\pm1$, each with probability 1/2, $E[X]=0$. Similarly, the expected value of a single standard die roll is 3.5, which is of course impossible to roll. – amd Sep 9 '17 at 22:28