# what is the probability of a double chance?

I came with this question solving a programming problem, but my question is math related.

The problem is a bowling system which the wording of the question says:

The application will randomly choose the number of pins knocked down, with a double chance to the pins $7,8,9$ and $10$.

So the pins $1$ to $6$ have a chance of $1/2$ to be knocked down.

So I came with something like this:

if random(0 ~ 100) < 50 => pin dropped


And my first thought of what would be a "double chance" was $2/3$.

if random(0 ~ 100) < 66 => pin dropped


But after writing it down I realized that the difference of the pins $1$-$6$ to pins $7$-$10$ was less than I though, just $15\%$ greater. So I started wondering if I may be misinterpreting the question.

I was thinking if a probability of two consecutive $1/2$ could be $(1/2)^2 = 25\%$ and $75\%$

I may be totally wrong, I need a second though on this.

-

The way I understand the problem, you want the probability of $n=1,2,...,6$ pins dropping $P_1$ to equal half of the probability of $n=7,8,...,10$ pins dropping $P_2$. And you want the total probability $$\sum_{n=1}^6 P_1 + \sum_{n=7}^{10} P_2=1.$$ Substituting $2P_1=P_2$ into the above expression yields $$6P_1 + 8P1 = 1.$$ So $$P_1 = \frac{1}{14},\;P_2=\frac{1}{7}.$$ I leave it to you to implement this programatically so you can model the pins as needed.