looks right
[ 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 & 3. But Frames 1 & 4 are independent. So, I count only 17 pairs of correlated frames. Just find the Variance of a frame, the covariance of frames 1&2 and of 1&3, and you should be on your way [img]/images/graemlins/smile.gif[/img]
[Answer is then 10Var(frame) + 2X9Cov(f1,f2) + 2X8Cov(f1,f3)]
Have fun with that!
alThor
|