[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.