THE BOOK--Playing The Percentages In Baseball

Thursday, February 21, 2008

Home field advantage in the playoffs

By Tangotiger, 12:12 PM

I always thought that one of the big benefits of the HFA is the travelling. That if we looked at the playoffs, the HFA would be quite a bit less. While I don’t care too much about the individual game-by-game splits (managers have high selective sampling here), the overall results seem exactly in line with the regular season HFA. Anyone have the relevant stats in NFL, NBA, and NHL?

• Sabermetrics • Talent_Distribution

#1 MGL 2008/02/21 @ 06:43 PM

I don’t know what this means:

This tends to make sense as, in theory, the better team in each league will win a seven game series, against a lesser opponent in their league, regardless of whether or not they are at home.

As Pizza suggests in the comments, or at least questions, I don’t think any of the samples is large enough to generate any conclusions or inferences with any degree of confidence. In 250 games, ONE standard deviation for the difference between H/R wp between two samples is over 4%!

Even the difference between that 48 and 63% for 52 games is around 1.5 SD, not considered particularly significant.

#2 Eric J. Seidman 2008/02/21 @ 06:51 PM

If we look at Binomial Probabilities, these are the results we get -

Div. G1 = 0
Div. G2 = +1.80
Div. G3 = -0.14
Div. G4 = +0.36
Div. G5 = 0

Ch. G1 = +0.45
Ch. G2 = -0.75
Ch. G3 = +1.36
Ch. G4 = 0
Ch. G5 = +0.32
Ch. G6 = -0.38
Ch. G7 = +1.11

WS. G1 = +2.07
WS. G2 = +1.42
WS. G3 = +0.76
WS. G4 = 0
WS. G5 = 0
WS. G6 = +1.68
WS. G7 = 0

#3 Eric J. Seidman 2008/02/21 @ 06:55 PM

MGL,

Yeah I definitely agree we cannot make any sort of conclusion. It was just interesting finding the numbers in certain games and how the percentages of the home team winning have differed in certain instances.

I’ll have to explore this again in 200 years to find some concrete conclusions.

#4 Eric J. Seidman 2008/02/21 @ 07:12 PM

Forgot to clarify, those are the Binomial Z-Scores.

#5 MGL 2008/02/22 @ 12:19 AM

Just for your edification, Eric, when you want to test the “significance” of the difference between (or sum of) two sample values (like the difference between the 48% and the 63%), you take the sum of the variances, and then take the square root of that (to get the standard deviation by chance for the difference).

#6 Eric J. Seidman 2008/02/22 @ 12:21 AM

MGL,

Thanks for the info! I was unsure how to truly test the significance, as I am sure you could tell, but now it’s cleared up.

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