Re: A Gambling Math Question

In article <1169579987.759066.43860@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
JyBrdnj@xxxxxxxxx wrote:
Let p be the probability of success and q = 1-p = probability of failure.
If N is the number of trials, the expected number of successes is
obviously N*p. The standard deviation is approximately sqrt(N*p*q). This
comes from the well-known normal approximation to the binomial
distribution, which you can easily read about using a Google search.

I don't understand why the standard deviation formula would be
sqrt(N*p*q). Why would you need to multiply both the probability of a
win and the probability of a loss. Wouldn't you just need one side,
either the sqrt (N*p) or the sqrt (N*q).

The variance, which is the square of the standard deviation, can be
computed as the mean of the squares minus the square of the mean.
If something is 1 with probability p and 0 with probability q = 1-p,
then the mean and the mean of the squares are both p. Thus, the
variance is p - p^2 = p(1-p) = pq. The variance of n trials is n
times the variance of one trial. Thus, the variance is Npq, so the
standard deviation is sqrt(Npq).

Rob Johnson <rob@xxxxxxxxxxxxxx>
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