1. Dice throwing problem

The problem is following:
A and B takes turns throwing a dice. A begins. Who throws six first wins. What's the probability that A wins?

I tried to examine probabilities that A wins on different rounds. Something like this:
1. $\displaystyle \frac{1}{6}$

2. $\displaystyle \frac{5}{6}*\frac{5}{6}*\frac{1}{6}$

3. $\displaystyle \frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\f rac{1}{6}$

But I can't solve the general form (or something). I think it has something to do with binomial probability and geometric series... Any help is appreciated, thanks!

2. You're correct. Add up all the values you've gotten. This is a geometric series:
(1/6) * [ 1 + x + x^2 +x ^3 + ...]
where x = (5/6)(5/6).

3. Originally Posted by Greg98
The problem is following:
A and B takes turns throwing a dice. A begins. Who throws six first wins. What's the probability that A wins?

I tried to examine probabilities that A wins on different rounds. Something like this:
1. $\displaystyle \frac{1}{6}$

2. $\displaystyle \frac{5}{6}*\frac{5}{6}*\frac{1}{6}$

3. $\displaystyle \frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\f rac{1}{6}$

But I can't solve the general form (or something). I think it has something to do with binomial probability and geometric series... Any help is appreciated, thanks!
As the game can go on indefinately, you need to evaluate the infinite geometric series.

The probability that A wins is

$\displaystyle \frac{1}{6}+\frac{5}{6}*\frac{5}{6}*\frac{1}{6}+\f rac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\fra c{1}{6}+........$

The first term is $\displaystyle \frac{1}{6}$ and the common ratio is $\displaystyle \left(\frac{5}{6}\right)^2$

For a geometric series,

$\displaystyle S_n=\frac{a\left(1-r^n\right)}{1-r}$

$\displaystyle S_{\infty}=\frac{a}{1-r}$