FanGraphs Baseball

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  1. Eric, could you explain how you got the big play clutch scores? I’m not getting the same numbers as you are. I’m seeing Howard’s big hit worth about .55 “clutch”, and for Mora I’m getting about .32 “clutch”.

    I’m just calculating his clutch score minus the one big play, or I think doing (WPA / avg(pLI for the season)) – (WPA/LI) to find the value of an individual play would work too.

    Comment by David Appelman — May 23, 2008 @ 11:57 am

  2. David, I see where the error is and I’m correcting it now. What I had done was WPA/(pLI – WPA/LI) instead of (WPA/pLI) – (WPA/LI). Everything should make more sense now.

    With pLI I am looking at the individual game in which the big play took place. So, Burrell had a 0.899 WPA on the big play, with a 3.56 LI, and in that May 2 game, he had a 1.38 pLI.

    When plugged in, (WPA/pLI) – (WPA/LI) = 0.6514-0.2525 = 0.40.

    So, that play accounted for a 0.40 clutch, and the net sum of all other plays, entering last night, would be 0.93, since he had a 1.33.

    Comment by Eric Seidman — May 23, 2008 @ 4:44 pm

  3. With Mora:

    .418 WPA/1.92 pLI in the game of his big play = .2177
    .418 WPA/5.14 LI of the big play = .0813

    WPA/pLI – WPA/LI = ..136, or .14

    Comment by Eric Seidman — May 23, 2008 @ 5:07 pm

  4. I would use the pLI for the season when looking at an individual clutch play instead of for just that game since clutch isn’t calculated on a per game basis and then added up. Ideally you’d want to pretend like that play never happened and then calculate what his clutch score would have been without it entirely.

    You’re probably going to be undervaluing the “clutch” aspect of the play a little to considerably by using just that games pLI because chances are it’s going to be a much higher pLI than his “normal” pLI would be.

    For instance, if a player had only one plate appearance in that game and his LI in that appearance was let’s say 5, and then he hit a home run worth .6 wins, you’d be calculating his clutch score for that plate appearance at 0.

    Comment by David Appelman — May 23, 2008 @ 5:27 pm

  5. Okay, gotcha’, we’re all clear on that. I’ll go in and re-calculate.

    Comment by Eric Seidman — May 23, 2008 @ 5:36 pm

  6. All numbers corrected, my apologies to anyone for the confusion.

    Comment by Eric Seidman — May 23, 2008 @ 6:33 pm

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