#11
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Re: sci.math, reality, Bruce\'s Example, Independence
[ QUOTE ]
In Bruce's example, 50% strikes and 90% spares, the correct way to use the formula is to set x = 0.90 s = 0.45 average = 90+100*0.45+100*0.5+20*0.45*0.5+100*0.5^2+10*0.5^3 = 215.75 [/ QUOTE ] Right, sorry about that. That average fits much better with those stats. The thing is that we are only considering single pin spares for scoring purposes, but a bowler's average for all spares will be significantly lower than for single pin spares. For example, a PBA record for single pin spare percentage is about 99.5%, while the record for all spares is 86%, for multi-pin spares it is 67%, and for splits it is only 25%. See pba records. [ QUOTE ] It seems the more strikes you get, the more likely you are to get another strike. That lack of independence would increase your average. [/ QUOTE ] Right, this happens due to changing oil patterns. Once you've found a spot on the lane that gets the ball to hook into the pocket, it will stay good for awhile assuming you can physically deliver the ball the same way to that spot. Then when the oil starts to break down, you may lose your ball reaction for a few shots until you figure out how to adjust again, so misses will also be more likely to follow misses. |
#12
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Re: Bowling Average
I don't have anything to add directly, but the Journal of Recreational Mathematics has included articles on random bowling games. I think the problems I saw analyzed there were the expected scores if each frame or each ball is chosen uniformly from among the possibilities. The articles were longer than would be needed to compute these, so they might have included other things.
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