Hey jab1023.

For this, recall that the coin toss outcomes have a distribution that is Binomial(10,0.5) if we assume a fair coin (it may not be, but that's usually the assumption we make).

Now the average number of heads we get is 5 since 10*0.5 = 5.

Now lets say you want to figure out the number of outcomes given some probability. Suppose we want to look at a 90% confidence interval of how many different numbers of heads we would expect given this 90%.

If you use a normal approximation with sample mean x_bar and standard error se = SQRT(x_bar*(1-x_bar))/SQRT(n) and your interval will [x_bar - Z(a/2)*se, x_bar + z(a/2)*se] where z(a/2) refer to the z-values for the interval (two tailed). For 95% z(a/2) is roughly 1.95.

Now you take that and look at the interval it generates and you will get a lower and an upper end which will contain certain whole numbers corresponding to the number of successes: those numbers correspond to the events inside that interval.

So if you get say (3.776,6.224) for some a then the outcomes will include 3,4,5, and 6 successes for that confidence level.

You may also want to check out correction methods for using the normal approximation.