Re: A Gambling Math Question


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).

The variance of the sum of independent random variables is the sum of
their individual variances. So, let r be a random variable that can
take the value 0 with probability q, or the value 1 with probability p.
The variance of r is easily seen to be p*q, so from the additive
property the sum of N such random variables has variance N*p*q. But the
sum of N such random variables has a binomial(N, p) distribution, so
the variance of that distribution is also 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). Interestingly enough if you
did that for 100 trials the standard deviation would be 7.07, and if
you multiplied that by 3, you would get 21.21. I once heard the most
ever recorded losses in a row on a roulette table, betting black or
red, was like 22 or 23 losses in a row. Interesting.

.

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