# Pirates, 60%

*By clayd On October 1, 2013 · 2 Comments*

A very quick look at the Reds/Pirates game tonight.

Reds starter Johnny Cueto, this season, had a .207 eqa against left-handed hitters and a .234 eqa against righties. He is himself a RH.

Russell Martin, the Pirates catcher, is right-handed, and hit for a .275 eqa against right-handed pitchers.

Now, if a hitter with a .275 eqa goes against a pitcher who allows a .234 eqa, the expected eqa of the result should be roughly (275*234 / 260), or .248. This value is consistent with converting each of the eqas into winning percentages, letting them go head to head with the log5 method, and converting back into an eqa.

Repeating for the whole Pirate roster

C Martin (R) .275v234 = .248

1b Morneau (L) .297v207 = .236

2B Walker (S) .296v207 = .236

SS Mercer (R) .247×234 = .222

3B Alvarez(L) .299v.207 = .238

LF Marte (R) .278v234 = .250

CF McCutcheon (R) .322v234 = .290

RF Byrd (R) .291v234 = .262

The average value, for the whole Pirate team, is .249; eqas don’t sum linearly, they sum by the 2.5 power. That’s not a big deal here (a straight mean is .248), but will be for the opposite calculation.

Doing the same thing for the Reds against Francisco Liriano is trickier, because his EQA against lefties is so low as to be below zero. The eqa is low enough that a lineup of average hitters would be better off without that hitter in it, even if they got fewer chances as a result. Using an alternate form of the eqr equation, one that can’t go sub-zero, yields an eqa of .132. Against righties , he’s got a decidedly pedestrian .265 eqa allowed.

C Hanigan (R) 225v265 = 229

1b Votto (L) 303v132 = 154

2b Phillips (R) 271v265 = 276

SS Cozart (R) 245v265 = 250

3B Frazier (R) 278v265 = 283

LF Ludwick (R) 278v265 = 283

CF Choo (L) 243v132 = 123

RF Bruce (L) 262v132 = 133

The average value here works out to a .230 eqa when using the 2.5 power rule, quite a bit higher than the .216 you’d get from a straight average.

The win chance of a .249 eqa team (the Pirates) against a .230 team (the Reds) is .598.

Extending this into bullpens would probably only help the Pirates more. They had a composite .240 eqa from their bullpen in 2013, compared to the Reds .245; their likely top 4 dominate the Reds top 4, in a game theory sense:

Melancon 190 v Chapman 207

Watson 207 v Hoover 231

Wilson 217 v LeCure 232

Grilli 225 v Simon 225

### 2 Responses to *Pirates, 60%*

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Can you show the calc. to get from .249 EQA vs. .230 EQA, to .598 win prob.? Also, what is the proper way to convert EQA into runs per 9 innings? Thanks.

You know the James pythagorean method, yes?

Wpct = R^2 / (R^2 + RA^2)

Well, R is proportional to eqa to the 2.5 power, so

Wpct = EQA^5 / (eqa^5 +eqaa^5)

To the second question,

EqR = 5 * eqa^2.5 * outs, so for nine innings it is 135 * eqa^2.5