question

[VanVeen]

what is the probability of being 780bbs off expectation after 120 all-ins when you're risking 20bb/all-in? (so 39 units off in 120 runs)

how would i quickly calculate this in future?

thanks

[ig06]

I think for your example, the probability is going to be very very close to 0.
You are essentially just tossing weighted coins
The simplest case is when the weighting is constant. So suppose probability (heads) or probability (you win) if you prefer, is equal to p. Number of runs is n. This is a Bernouilli process and is described by a binomial distribution, i.e.

Prob(k heads in n) = C(n,k)p^k(1-p)^(n-k)

So you can solve the problem exactly since

Prob( number of heads in n runs is <= j and >=l) = sum(i=j to l) C(n,i)p^i(1-p)^(n-i)

There are functions in excel to support this kind of calculation.

There is also an approximate method (which I think is more useful for the general case, see below). It is based on the central limit theorem (You might well know this, otherwise see wikipedia or something), which indicates that as n increases, our distribution derived from independent random (here binomially distributed) events can be more and more closely approximated by a normal distribution. So specifically, the number of heads observed is close to normally distributed, with mean = np and standard deviation = sqrt(np(1-p))
The easiest way to compute confidence intervals is probably with excel (they used to use tables) since you need to integrate the normal distribution over the range of interest and it isn’t analytically solvable. In excel, use the following function:

NORMDIST(X,MEAN,STDEVIATION,TRUE)

This finds the integral for your distribution from –infinity to X, where the mean is np and standard deviation is sqrt(np(1-p)). TRUE is written in text (it’s a Boolean value meaning you want the cumulative distribution, which is the integral I described above). I note that in your example if you win 79 times you end up 38 units up and if you win 80 times you end up 40 units up if I’ve understood correctly. So I will use 80 times.
If I put p=0.5 for example, I want

NORMDIST( 80, 60, 5.48,TRUE) which is equal to 0.999869. So the probability of being up by 40 units or more is ~1-0.999869 =0.000131.
In that case the probability of finishing not more than 40 units up or down is ~1-2*0.000131=1-0.000262 =0.999738.

If I solve it exactly with the binomial distribution I get probability not finishing more than 40 units up or down is 0.999835.

The value of the approximate method using the normal approximation is when the value of p changes with each run. I think the normal approximation is valid here. In this case the mean number of successes is equal to sum(i=1 to n) (P_i) where P_i is the probability of success for each of the n runs. The variance is sum(i=1 to n) (P_i*(1-P_i)).

Hope this helps, sorry if my explanations aren’t great.

[ig06]

I said:
[ QUOTE ]
I note that in your example if you win 79 times you end up 38 units up and if you win 80 times you end up 40 units up if I’ve understood correctly. So I will use 80 times.

[/ QUOTE ]

Sorry, I had already assumed (in my head) for the sake of example that p=0.5. Obviously since you are comparing with the expectation, this has to be changed accordingly if p is different!