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By Tangotiger 10:43 AM

?It just seems so arbitrary no? I mean, why 7-4-3-2-1? Why not 5-4-3-2-1? Why not 10-7-4-2-1? Why not anything else? Well, we can figure it out, and it’s going to be ridiculously easy to do so. For this, I will use WAR at Baseball Reference, but someone out there can replicate this using Fangraphs WAR, or any other system you would like.

What you need is the top 6 pitchers in whatever system you are looking at. I chose 1998-2016, AL, NL. That gives me 38 sets of 6 pitchers. We figure out the average WAR (or whatever point system you are using) for each ranked player.

The #1 player has an average of 8.01 WAR. The #2 player has 6.94 WAR. All the way down to the #6 WAR player at 5.14 WAR. It looks like this:

8.01 6.94 6.36 5.92 5.49 5.14

We know that with only 5 spots on the ballot, the 6th (and lower) placed player can only be worth 0 points. The 6th place player is essentially a “replacement” level player, the baseline to which everyone is compared to. So, from each of them, we remove 5.14 wins (or whatever point system you are using). We have this:

2.88 1.81 1.22 0.79 0.36 0.00

We can simply stop here, but we’d like to have integers, and we’d like the 5th place guy to be worth 1 point. So, we just divide all the numbers by 0.36. It keeps the proportion the same. We get:

8.0 5.0 3.4 2.2 1.0 0.0

So, this approach would suggest a point system of 8-5-3-2-1, rather than 7-4-3-2-1. However, you will note that 3.4 and 2.2 in there. If we were to take the above scale and multiply it by 84%, we get:

6.8 4.2 2.9 1.8 0.8 0.0

As you can see, everything now rounds off by 0.1 or 0.2 in one direction or the other, rather than two of the spots doing all the rounding. And rounded to the nearest integer, we get…. 7… 4… 3… 2… 1.

So, yeah, 7-4-3-2-1 may SEEM arbitrary, but it’s really a choice between that and 8-5-3-2-1. In either case, Porcello would take it over Verlander.