Binomial distribution with changing success rate

[poker_n00b]

Searched the internets, couldn't find anything.

I am kind of trying to quantify the "running bad concept".

Let's say I lose one 70/30 flip, one 80/20 flip and one 50/50 flip. The success rate of winning, p(winning), changes in all three examples.

How can I calculate the probability of losing those 3 flips?

Note:

I want to exclude the amount of money or chips put it, because it has a bias. For example, I may lose 10 times AA vs 72o for 10$, but win 72o vs AA for 1 million. The amount of money you end up having is in no way representative of how you are running. I may be wrong on this one.

[btmagnetw]

multiply them.

[Silent A]

Chances of losing all three flips = (0.3)(0.2)(0.5) = 0.03 = 3%

[pzhon]

Suppose you have n flips, and win with proabilities p_1, p_2, ..., p_n.

The expected number that you win is p_1+p_2+...+p_n.
The variance is p_1(1-p_1) + p_2(1-p_2) + ... +p_n(1-p_n), which is less than or equal to n*p_average(1-p_average).
The standard deviation is the squareroot of the variance.

To approximate the probability of seeing at most w wins using a normal approximation, translate this into winning w+0.5 or less, then express this as a number of standard deviations below the average, and look up the value on a table (or web page) of the cumulative distribution function of a standard normal distribution. E.g., with an average of 5, standard deviation of 2, and w=3, translate 3.5 into 0.75 standard deviations below the mean. In the linked page, integrate from -10 to -0.75.

The absolute error will be small with enough trials. If you are estimating a small probability, the proportional error may be too large. Then you may want to use a spreadsheet or computer algebra package to compute the exact probabilities.