THE BOOK--Playing The Percentages In Baseball

I don’t think that’s right ... you can’t go by the average, you have to go by the marginal.  I think it’s just that francophone players from Québec are better than the rest of the world, in general, and so they’re disproportionally represented at the top, as compared to the bottom.

More arguments at my blog here.

If both are the same kind of distributions (i.e., no Gretzky outlier that is not counteracted by a Lemieux outlier), then the marginal and the average will work out to the same thing.

The marginal and the average will not work out to the same thing if the means are different.  If Québec players score more, it might just be that they’re better players—that is, they have a higher mean.

Suppose Quebec’s mean is 1 point higher than the rest of Canada (RoC) mean.  At the margin, they’ll be the same, but the average Quebecker will be higher.  For instance, Quebec has:

10 players at 3 points
5 players at 4
2 players at 5
1 player at 6.

RoC is the same, but the points are 1 lower.

If the cutoff to make the team is 3 points, then the average Quebecker is 3.7 (the average of the above players), while the average RoCer is 3.5 (5 players at 3, 2 players at 4, 1 player at 5).

But the marginal player both ways is a 3.

Phil.... hmmm… interesting.  For that particular distribution, it worked out as you said.  Had we taken a normal distribution, would it still work out the same?

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I read Phil’s blog.  Interesting again.  The one thing that I disagree with is his use of the word “racism”.  First off, I don’t know that being unilingual french is a “race”.  Secondly, racial bias, or culture bias, is not the same thing as racism or culturism.

***

Finally, in order to prove my point, we should see a drop in per capita rates at each quality level.  For example, there are (by my gut) far more 1st and 2nd line Russian players than there are 4th line Russian players.  It simply make little sense for a team to invest in a Russian player (time and money) to make a marginal impact, when he can just go to Kingston, and get humself a grinder.

Indeed, we know this to be true, because half of the 1st round picks in the last several years have been European/Russian, while only one-third of all NHLers are so.

Anyway, Phil’s examples are worth exploring more, and applying it to where we don’t expect bias (say OHL v WHL, or perhaps even Canada v USA) to see if his theory has practical application.

It is also worth considering that hockey players are not only chosen for their ability to score. My general impression is that in junior hockey, the Quebec league places far more emphasis on scoring, while the OHL and WHL are more focused on defense. Since 3rd and 4th line players tend to have a more defensive role, one would expect that they would come from junior leagues that focus on teaching those skills.

>“Had we taken a normal distribution, would it still work out the same?”

Good question.  I’m pretty sure it would, but can’t say for sure.

Another good point on the word “racism”.  I thought I read it in the original article, but it’s not there.  I shouldn’t have used it.

>>“Had we taken a normal distribution, would it still work out the same?”

>Good question.  I’m pretty sure it would, but can’t say for sure.

Let me clarify what I mean: I’m pretty sure that if I think about it a bit, I will be able to come up with an informal “proof” that it would.

In any case, the illustration shows that it CAN work out that way, which is enough to refute the point.

Found it!  “Racism” is in the first paragraph of the National Post story.  It’s the reporter’s word, not Sirois’s word.

>>“Had we taken a normal distribution, would it still work out the same?”

I’ve thought about it, and I think, yes, it would.

Two part argument.  Consider two normal distributions with the same SD, but one has a mean higher than the other.  Say, X with mean 100, and Y with mean 105.

1.  The farther right in the curve you go, the bigger the proportional difference between the higher mean and the lower mean.

That is: At 105, Y might be 1% higher than X.  At 110, it might be 3% higher.  At 130, it might be 10% higher.  At 150, it might be 30% higher.  And so on.

I can’t prove this formally, but if you look at a normal distribution table and run some numbers, you’ll see it’s true.

2.  If you choose a cutoff of (say) 130, you’ll find that the curve Y is higher than X for everything to the right of the cutoff (this is obvious).  Now, if Y were the same percentage higher than X everywhere on that right tail, then the average would have to be the same. 

But, from (1), we know that Y is a higher percentage higher than X the farther right you go.

Therefore, the average Y must be higher than the average X, above an arbitrary point.