Higher BA than OBP ??

[Detroit-Tigers]

[quote:f78e0aa984="old__toriihunter"][quote:f78e0aa984="mjf309"]How does this happen arithmetically?[/quote:f78e0aa984]

OBP = (Hits + BB + HBP) / (AB + BB + HBP + SAC)

...so SAC is included in the calc and can therefore lower the OBP. Since BB is in the numerator and the denominator, it cancels itself out. Therefore if your SAC > BB, your OBP can theoretically be lower than your BA.

BA = (Hits/AB), excludes BB, HPB, and SAC

Phil Niekro did it one year, I believe.[/quote:f78e0aa984]

Maybe I am a little slow, but mathematically I don't think BB and BB cancels each other out. (e.g. (2 hits + 2 BB) / (8AB + 2BB) = .400 OBP, but if you didn't have the 2BB you would have 2 / 8 = .250 OBP)

Looking for OBP > Average
OBP = (Hits + BB + HBP) / (AB + BB + HBP + SAC)

Therefore what we are looking for is
Average > OBP
Average > (Hits + BB + HBP) / (AB + BB + HBP + SAC)

For an assumption of .250 average
.250 > (25 + BB + HBP) / (100 + BB + HBP + SAC)
25 + .25BB + .25 HBP +.25 SAC > 25 + BB + HBP
.25BB + .25 HBP +.25 SAC > BB + HBP
.25 SAC > .75 (BB + HBP)
SAC > 3 (BB + HBP)

[b:f78e0aa984][i:f78e0aa984]So, as a general rule, anytime the # of sacrifices is greater than 1 / Average - 1 (or 2 - 3 times) your BB + HBP, you can have a situation where average > OBP.[/i:f78e0aa984][/b:f78e0aa984]

(1 / Av - 1) > BB + HBP