Understanding Expectation and Variance

Understanding Expectation and Variance

When we project Keenan Allen to score 14.5 points in DraftKings’ scoring system, what does that mean?

It doesn't mean that we expect him to score precisely 14.5 points. That’s possible, but it’s very unlikely. Even if 14.5 is more likely than any other specific number, that exact outcome occupies an exceedingly small slice of probability space.

What it means in theory is that if you take each fantasy point total Allen could conceivably get, multiplied it by the respective probability of getting that score, and add all of those products up, you'd get a sum of 14.5. (Using the same procedure, we'd project the roll of a six-sided die to produce a value of 3.5, because 1*1/6 + 2*1/6 + … + 6*1/6 = 3.5. Even though the die lacks a side with 3.5 on it, 3.5 is a good projection in the sense that it would be the fair over/under at even odds.)

I say “in theory” because nobody actually does projections that way. If you consult this book’s chapter on projections, you won't see anyone estimating the probability that Keenan Allen will score 0.0 points, and then doing the same for 0.1 points, 0.2 points, and so on all the way up to 60+ points before doing some multiplication and addition to get a projection of 14.5 points.


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