eli5: Do expected values of various kinds of discrete probability distribution always have the highest probability?

In: 3

As the other response demonstrated already: the expected value is potentially not even possible to get at all, and even if it is not the most likely.

Instead, the EV minimizes the expected “distance” to the outcome; in more formal words, the _variance_. To see why and what this even means, note the variance sums the squared differences (X-E)² for all outcomes X, each weighted with their respective chance. In yet other interpretation, this is akin to the Pythagorean theorem, where distances are found by summing squares as well.

Or in formulas: variance is the expected value of (X-E)².

So one might wonder which potential other options for E minimizes this value. But it turns out with some basic calculus that this happens exactly if E is already the expected value itself.

No.

We have a random variable that can take on three values 1,2, and 3.

1 and 3 have a 40% probability and 2 has a 10% probability.

The expected value is 1×0.4 + 2×0.1 + 3×0.4=1.8.

Not only is the EV closest to the least probable value, it’s not even a value the distribution can take on at all.

Hell, consider even a uniform distribution. The EV of a standard die roll is 3.5.