# Number of biased coin tosses needed to match headscount for a number of tosses of a different biased coin

I have two different biased coins with probabilities $p_1$ and $p_2$. Coin 1 I toss $n$ times. I would like to know how often I should toss coin 2 to be $p_3$% sure I'll have more heads from coin 2 than from coin 1.

I have read up on binomial distributions and I could figure out the answer by summing those from a guessed starting point and going up or down by trial and error on the computer, but I'm hoping there is an easier way.

Context: Actually the coin flips are archers in a computer game facing off against opposing archers, each having a chance to incapacitate an opponent, I expect there will be anywhere from 1 to 1000 archers on each side. A cautious artificial intelligence wants to know how many archers it has to take so it's reasonably save to go near and have a positive outcome where fewer of his archers fall than those of the enemy.

• The problem is complicated. One could get answers by simulation. If the $p_i$ are not very far away from $1/2$ (and for example $0.2$ is not far away), and the numbers of tosses involved are large, one can use the normal approximation to get good explicit estimates. – André Nicolas Mar 7 '14 at 18:17

An approximation could be developed using normal distributions, where $$P[B<A] =P[B-A<0]= \Phi \left( {\mu_A - \mu_B } \over {\sqrt{\sigma^2_A + \sigma^2_B}} \right)$$