# I need to know how to get this %

• Dec 2nd 2012, 05:01 AM
I need to know how to get this %
Hello,

i have a problem calculating this:

I have a bankroll of : 100 $i have 65% chance to win 10$
i have 35% chance to lose 10$and i have unlimited tries. what is the % of my bankroll to become 0 example: first hit 65% for my bankroll to become 110$ and 35% chance for my bankroll to become 90$thanks. • Dec 2nd 2012, 06:07 AM HallsofIvy Re: I need to know how to get this % If you win i times then you must lose i+ 10 times, out of a total of n= 2i+ 10 times or i=(n-10)/2. It looks like it is a binomial distribution with p= .65, q= .35. That is, the probability of losing all your money in n turns is$\displaystyle \begin{pmatrix}n \\ (n-10)/2\end{pmatrix}(.65)^{(n-10)/2}(.35)^{(n+10)/2}$. For that to be possible n must be even, say n= 2m, so the probability of eventually losing all your money is$\displaystyle \sum_{m=0}^\infty \begin{pmatrix} 2m \\ m- 5\end{pmatrix}(.65)^{m- 5}(.35)^{m+5}\$.
• Dec 2nd 2012, 06:26 AM