THE BOOK--Playing The Percentages In Baseball

Player A

Played A=.92*.900=(.828+.750)/2=.789 OPS

Player B=.92*.810=(.745+.810)/2=.778 OPS

Could be wrong, and you could apply the factors in another way.

I’m not sure what you mean by “true park factor”—was (1) OPS inflated by 8% in home games, or (2) by 16% in home games, thus we adjust the overall stats by 8%?

If it’s two players from the same park, it doesn’t really matter for the comparison.

In scenario (1), (900/1.08+750)/2=792 for Player A and (810/1.08+810)/2=780 for Player B.

In scenario (2), (900+750)/2/1.082=764 for Player A and (810+810)/2/1.08=750 for Player B.

Either way, Player A comes out with the higher adjusted OPS—just as he has the higher absolute OPS.  Any park adjustment given to one will be given to the other, based on your parameters.

I see what you’re going for (I think)—but the thing is that while Player B is 8% on the road than Player A is, Player A is 11% better at home.  So Player A comes out ahead, given your assumptions, no matter what you do. 

The trickier way of looking at it is if:

Player A:  870 at home, 750 road
Player B:  810 at home, 810 road

Again, both in the same park, same amount of games, etc.

Both players have an 810 absolute OPS overall.

(1) Adjusting Player A is (870/1.08+750)/2=778, and Player B is (810/1.08+810)/2=780.  Player B comes out ahead, slightly—because his road OPS is +8% better than Player A’s, but Player A’s home OPS is only 7% better than Player B’s.  So B goes ahead.

(2) Adjusting Player A is (870+750)/2/1.08=750 and Player B is (810+810)/2/1.08=750, so they remain equal.

So the question depends on whether or not your park factor is meant to apply to home stats only, or overall stats.

Depends what you’re trying to measure. 

If you’re trying to figure which is the better hitter, you want to take 1/16 of home park stats and add it to the road park stats.  That tells you how each would do playing equally in all parks.  Obviously, player B is more likely to be the better hitter overall.

That’s probably not what you’re asking, though.

Cut the award in half. They’re dead even.

If A’s overall OPS is 825 (mean of 900 and 750) and B’s is 810, then A is superior.
Their park adjustments are equal so they cancel each other out, don’t they?

I figured it a bit differently but selected player A also.

I chose to first separate their home/road performances and compare them to a standard. I chose a road game standard of .750 OPS, which happens to translate to an .810 OPS in the home park (.750 * 1.08).

Player A is a 1.00 on the road (.750/.750) and 1.1111… at home (.900/.810).

Player B is a 1.00 at home (.810/.810) and 1.08 on the road (.810/.750).

Player A on the road is equal to player B at home. Player A is a slightly better offensive player because his edge at home is larger than player B’s edge on the road.

Of course a simpler method would be to just use a season total. Both players are seeing the same overall park effects. How extreme would a park need to play in order to create a serious discrepancy in the apparent value of two players on the same team?

I think Bowie/5 has it.  Both players play under equal conditions, so there’s no need to adjust. 

You wouldn’t adjust for park any more than you’d adjust for weather.  Even if it’s easier to hit in July, both guys played in July, so it’s not a factor.

What about the refrain you constantly hear from writers and analysts?

“The home park didn’t affect player X’s stats, as you can see from his home/road splits.”

Applying that to player B…

I wasn’t going to go there, but you asked for it: Player B was more consistent.

The 1.08 park factor is the “true” (exactly what you would expect going forward) ratio of home OPS to the entire league (including the home park) OPS, as it usually is.

"I think Bowie/5 has it.  Both players play under equal conditions, so there’s no need to adjust.”

Well, if you were doing a traditional park adjustment, of course you would have to adjust each player.

What if one player had an OPS of zero at home and 1.500 on the road and the other had .750 and .750?

Or .100 at home and 1.400 on the road and .750/.750?

After adjusting each player’s home (and road) stats, they would NOT have the same park neutral OPS.

I’ll throw something else out there:  We know that parks do not affect all players the same.  Is there a suggestion in my example that player A is affected much more than player B, such that we should adjust player A’s stats differently (perhaps by 1.10) than player B (perhaps by 1.06)?

#11/MGL: If we’re determining past value, I do not believe there needs to be a different adjustment. It’s the “Juan Pierre in Coors Field” situation. Even though Pierre’s game isn’t predicated on driving the ball into gaps or over the fence (things that Coors accentuates), his performance is still worth something compared to the Coors Field average.

Player A’s .000 OPS in a hitter’s environment is still worth some amount relative the OPS environment the park provides, regardless of how he’s affected by the park. I found it more difficult to explain in words here, but this article I penned should explain my thoughts a bit better (pardon the self-promotion):

http://www.calltothepen.com/2010/08/06/saber-slant-17-the-juan-pierre-question/

I have a sneaking suspicion that this is a Carlos Gonzalez-inspired post…
Just give the award to the guy with more GWRBI.

I put in a response to MGl’s question, but due to a link, it’s been spam-filtered. In short, I think no additional adjustment needs to be made to the “true” OPS factor, because that player’s past value on offense still needs to be compared to the average of that park, regardless of how he attains it.

"because that player’s past value on offense still needs to be compared to the average of that park, regardless of how he attains it.”

Right, for an MVP type award, if a player can somehow take advantage of his home park, then it should not be held against him.  At least that is one point of view, and probably a more reasonable one that the opposite one…

Player B is more valuable because more of his offensive contribution came on the road where the run environment is lower.  Thus Player B contributed more wins even though he contributed fewer runs.

The simple answer is player A.  They played in the same environment and had the same stats.

You should consider the run environment, it takes more runs to turn into a win at home, this will bring the two closer, but A is still in the lead.

Turning OPS crudely into runs, A is +30 over B at home and -20 on the road.  Maybe the PF indicates it takes 11 runs to turn into a win in the hitter’s park instead of 10.  That means he’s 0.7 wins ahead overall instead of 1.0, but you’d have to construct a much more extreme example to make player B more valuable.

If a run gained in a high scoring run environment is worth less than one in a neutral, does the same follow for defense?  Does UZR or TotalZone take this into account by park adjusting the runs saved?

UZR does not.  It is not designed to determine retrospective win value. It is designed to infer true talent defensive value in an unknown or context neutral environment.

TZ does not.  Plays saved in the 60’s are worth fewer runs than plays in 2000.  But those runs are worth more wins.  Those considerations somewhat cancel out.  Probably not exactly but I did not think it was worth the effort to try and improve that sort of precision on an estimate that has a lot of uncertainty anyway.

The assumption of my WAR rankings is that X extra plays made = Y Wins, and that the X/Y value is generally constant in the different run environments through baseball history.

#16

Interesting, and that was a central point Dave Cameron made in a post about Carlos Gonzalez today.

#16 ignores the fact that player A has a much higher OPS at home.  Does that not count?

Perhaps some hitters have the ability to adjust to their home park conditions.  Example, a RHB who has mainly has a LD CF, CF-RF power stroke, and adjusts his swing to hit more LF FB at Fenway, or a LHB pull hitter who goes the other way more at Fenway.  These adjustments may hurt them on the road, but does not mean that if traded they could not adjust and hit just as well in their new home park.

On the other side of the equation, some hitters benefit from being in a park that suppresses HR and runs due to its’ size, or at least are unaffected.  Tony Gwynn got a sizeable park adjustment for playing in SDG which is reflected in his park adjusted stats, and he probably does not deserve that boost. 

The devil is in the adjustments, especially park adjustments for some individuals.

In this particular case, you really can’t say much due to SSS over a single season but if true I would vote for Player A who put up a higher combined OPS, and not penalize him because he hit better in the park he plays 1/2 of his games (either he adjusted or his hitting style was better suited to the park). 

Even if every park was designed the same, a hitter who hits the same at home as on the road is clearly under performing at home where the team/player has a HFA and is expected to win more than on the road, and hit better.  The average player has a 50 OPS advantage at home that has nothing to do with park.  Player B needs to make some adjustments, maybe have the wife leave him alone more, get more sleep, something.

So what’s the answer? 

There is no one definitive answer…