## Monday, October 01, 2018

### When is defense more valuable than offense?

Is it possible, as a general rule, for a run prevented to be worth more than a run scored?

I don't think so.

Suppose every team in the league scored one fewer run, and allowed one fewer run. If runs prevented were more valuable than runs scored, every team would improve. But, then, the league would no longer balance out to .500.

But the values of offensive and defensive runs *are* different for individual teams.

Suppose a team scores 700 runs and allows 600. That's an expected winning percentage of .57647 (Pythagoras, exponent 2).

Suppose it gains a run of offense, so it scores 701 instead of 700. At 701-600, its expectation becomes .57717, an improvement of .00070.

Now, instead, suppose its extra run comes on defense, and it goes 700-599. Now, its expectation is .57728, an improvement of .00081.

So, for that team, the run saved is more valuable than the run scored.

It turns out that if a team scores more than it allows, a run on defense is more valuable than a run on offense. If a team allows more than it scores, the opposite is true.

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Just recently, I figured out an intuitive way to show why that happens, without having to use Pythagoras at all. I'm going to switch from baseball to hockey, because if you assume that goals scored have a Poisson distribution, the explanation works out easier.

Suppose the Edmonton Oilers score 5 goals per game, and allow 4. If they improve their offense by a goal a game, the 5-4 advantage becomes 6-4. If they improve their defense by a goal, the 5-4 becomes 5-3.

Which is better?

Even though both scenarios have the Oilers scoring an average two more goals than the opposition, that doesn't happen every game, because there's random variation in how the goals are distributed among the games. With zero variation, the Oilers win every game 5-3 or 6-4. But, with the kind of variation that actually occurs, there's a good chance that the Oilers will lose some games.

For instance, Edmonton might win one game 7-1, but lose the next 5-3. Over those two games, the Oilers do indeed outscore their opponents by two goals a game, on average, but they lose one of the two games.

The average is "Oilers finish the game +2". The Oilers lose when the result is at least two goals against them. In other words, when the result varies from expectation by -2 goals or greater.

The more variation around the mean of +2, the greater the chance the Oilers lose. Which  means the team with the advantage wants less variation in scores, and the underdog wants more variation.

Now, let's go to the assumption that goals follow a Poisson distribution.*

(*Poisson is the distribution you get if you assume that in any given moment, each team has its own fixed probability of scoring, independent of what happened before. In hockey, that's a reasonable approximation -- not perfect, but close enough to be useful.)

For a Poisson distribution, the SD of the difference in goals is exactly the square root of the total goals scored.

In the 5-3 case, the SD of goal differential is the square root of 8. In the 6-4 case, the SD is the square root of 10. Since root-10 is higher than root-8, the underdog should prefer 6-4, but the favored Oilers should prefer 5-3.

Which means, for the favorite, a goal of defense is more valuable than a goal of offense.

This "proof" is only for Poisson, but, for the other sports, the same logic holds. In baseball, football, soccer, and basketball, the more goals/runs/points per game, the more variation around the expectation.

Think about what a two goal/point/run spread means in the various sports leagues. In the NBA, where 200 points are scored per game, a 2-point spread is almost nothing. In the NFL, it means more. In MLB, it means a lot more. In the NHL, more still. And, in soccer, where the average is fewer than three goals per game, a two-goal advantage is almost insurmountable.

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