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 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