Bowling Average

[irchans]

Puzzle:

If, each frame, you get a strike with probability x, and get a spare with probability s, and get an open frame with probability (1-x-s), what is your average score and standard deviation? (Let's assume that you get 9 pins if you don't get a strike on the first ball.)

(I have figured out the correct average, but I think my standard deviation is wrong. I will post the average formulas in a few days if no one else does.)

Cheers,
Irchans

[irchans]

Well, I guess no one was interested, but here is the answer anyway. If the probability of a strike is x, the probability of a spare is s, and the first ball in a frame gets nine pins if it is not a strike, then the score in a given frame is:

30 for three strikes
29 for two strikes followed by a nonstrike
20 for a strike followed by a spare
19 for a strike followed by an open frame
20 for a spare followed by a strike
19 for a spare followed by a nonstrike
9 for an open frame

Converting each possibility to an expected value, we get

30*x*x*x +
29*x*x*(1-x) +
20*x*s +
19*x*(1-x-s) +
20*s*x
19*s*(1-x) +
9*(1-x-s).

That simplifies to

9 + 10*s + 10*x + 2*s*x + 10*x^2 + x^3.

So for a full game the average score would be

90 + 100*s + 100*x + 20*s*x + 100*x^2 + 10*x^3.

I have not been able to figure out the standard deviation without simulation. For a typical 140 average bowler, I get a standard deviation of about 23.

Any comments? Does the formula look reasonable?

Cheers,
Irchans

[Paul2432]

I think there may be a problem with your analysis. The score in each frame is not independent. For example, if you score 30 in the first frame, your score in the second frame must be at least 29.

I'll post some additional thoughts later.

Paul

[evanski]

This problem is more complicated than you make it out to be. Your scores on frames are inter-related, as you cannot follow a 30 frame with a 9, and what not. Im not quite sure hwo to calculate this, ill think about it.

evan

[cpk]

You don't have enough information. You also need...

f: Average first-ball score. This is so you can calculate the value of spares and doubles.
o: Average open-frame score. This is so you can calculate the value of strikes followed by open frames.

Let's say p = 1 - x - s to make the representation easier

The math is now easy:

9[(10 + (10+f(1-x)+10x)x + 10s + op)x + <font color="red">strikes</font>
(10 + 10x + f(1-x))s + <font color="red">spares</font>
+ op] <font color="red">open</font>
+
(10 + (10+f(1-x)+10x)x + 10s + op)x + <font color="red">X in frame 10</font>
(10 + 10x + f(1-x))s + <font color="red">/ in frame 10</font>
op <font color="red">open</font>

[cpk]

BTW, I have to admit that it took me a number of tries to get it right. I still may not have it right.

[alThor]

[ QUOTE ]

30 for three strikes
29 for two strikes followed by a nonstrike
20 for a strike followed by a spare
19 for a strike followed by an open frame
20 for a spare followed by a strike
19 for a spare followed by a nonstrike
9 for an open frame

Converting each possibility to an expected value, we get

30*x*x*x +
29*x*x*(1-x) +
20*x*s +
19*x*(1-x-s) +
20*s*x
19*s*(1-x) +
9*(1-x-s).

That simplifies to

9 + 10*s + 10*x + 2*s*x + 10*x^2 + x^3.

So for a full game the average score would be [ten times as much.]

[/ QUOTE ]

Looks right. (Others may not have realized that EV is additive, so you can calculate the EV of each frame separately, and just add them together, as you did.) It's been so long since I bowled, I got worried that you had to treat frame 10 differently, but I think you are ok.

Now to Std.Dev.: There is a way to solve it. Recall the formula Var(X+Y) + Var(X) + Var(Y) + 2Cov(X,Y)? If we have ten frames, the total variance will be the sum of the 10 variances, plus 2 times the coviance of each pair of frames. There are 45 pairs of frames! Fortunately, most of them are independent. Frames 1 and 2 are correlated, as are 1 &amp; 3. But Frames 1 &amp; 4 are independent. So, I count only 17 pairs of correlated frames. Just find the Variance of a frame, the covariance of frames 1&amp;2 and of 1&amp;3, and you should be on your way /images/graemlins/smile.gif

[Answer is then 10Var(frame) + 2X9Cov(f1,f2) + 2X8Cov(f1,f3)]

Have fun with that!

alThor

[BruceZ]

[ QUOTE ]
Well, I guess no one was interested, but here is the answer anyway. If the probability of a strike is x, the probability of a spare is s, and the first ball in a frame gets nine pins if it is not a strike, then the score in a given frame is:

30 for three strikes
29 for two strikes followed by a nonstrike
20 for a strike followed by a spare
19 for a strike followed by an open frame
20 for a spare followed by a strike
19 for a spare followed by a nonstrike
9 for an open frame

Converting each possibility to an expected value, we get

30*x*x*x +
29*x*x*(1-x) +
20*x*s +
19*x*(1-x-s) +
20*s*x
19*s*(1-x) +
9*(1-x-s).

That simplifies to

9 + 10*s + 10*x + 2*s*x + 10*x^2 + x^3.

So for a full game the average score would be

90 + 100*s + 100*x + 20*s*x + 100*x^2 + 10*x^3.

I have not been able to figure out the standard deviation without simulation. For a typical 140 average bowler, I get a standard deviation of about 23.

Any comments? Does the formula look reasonable?

Cheers,
Irchans

[/ QUOTE ]

This is correct, and I got the same per frame EV. What's throwing some people is that the frames are not independent, and in fact the scores in different frames can depend on the same roll, but this does not matter. The game score is the sum of the frame scores, and since the EV is linear, the EV of the sum is the sum of EVs for each frame. Since the EV of all frames are the same, the game average is just 10 times the EV for a single frame.

Let's check some cases for correctness:

Note that if we set x = 100% and s = 0% for a player who always strikes, then your formula correctly computes his average as 300.

For a player who always spares and never strikes or opens, s = 100%, x = 0%, and the formula correctly computes his average as 190.

For a player that always opens, then x = s = 0, and the formula correctly computes his average as 90.

Now for a realistic example. If his strike percentage is 50%, and his spare percentage is 90%, his average comes out to 265.25. From experience, this is an unrealistically high average for this player with realistic stats, and this is due to the assumption that he always makes 9 or 10 on his first ball, no splits, no buckets, no washouts, etc. Still this is a great start to a useful formula. Good job. See alThor's post for the variance. Make sure the scores are normal though before interpreting the standard deviation.

[irchans]

Thanks for the reply and feedback everyone.

As for realism, the assumption that the bowler always gets 9 pins or a strike on the first ball obviously is too optimistic. Maybe the formula should just be considered an upper bound. 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

because of the way I used s. I guess I used s as the "unconditional probability of getting a spare" in a given frame as opposed to the more natural definition "probability of getting a spare given that the first ball was not a strike." For me, x=.3 and s= .25 (got to work on those spares), and the formula predicts an average of 156 which seems a little high.

Over at sci.math, Gerry Myerson found a great paper on the lack of independence (http://www.economics.pomona.edu/GarySmith/bowling/bowling.html) of strikes. It seems the more strikes you get, the more likely you are to get another strike. That lack of independence would increase your average.

I am going to work on the standard deviation using alThor's covariance remarks. Before posting I tried to do standard deviation, but, as Paul2432 and Evanski pointed out, I forgot about covariance. I had just multiplied the single frame variance by 10. Oops.


Cheers,
Irchans

[irchans]

cpk,

You are right! Your improved formula,

10((10 + (10+f(1-x)+10x)x + 10s + o p)x +
(10 + 10x + f(1-x))s +
+ o p)

is exactly the same as the answer I gave in the special case f=9 and o=9, but of course your formula is much better because now it is easy to estimate the average of bowlers (like most of us) that sometimes get less than 9 pins on their first ball.

Very nice!

Irchans

[BruceZ]

[ 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 (http://www.pba.com/news/default.asp?ID=2572).


[ 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.

[pzhon]

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.