Probability of having six side first I have an exercise as follows: A and B alternately throw a dice (which has six sides numbered from 1 to 6). A starts firstly. What is the probability that A will be the first person who has side 6?
Thanks for any help!
 A: Assuming $A$ throws first:
Prob on 1st throw is $1/6$.  Prob on second throw is $(5/6)^2 1/6$ because both $A$ and $B$ have to fail.  You end up getting a series:
$$\begin{align}\frac{1}{6} + \left(\frac{5}{6}\right)^2 \frac{1}{6} + \left(\frac{5}{6}\right)^4 \frac{1}{6} + \ldots=\frac{1}{6} \frac{1}{1-\left(\frac{5}{6}\right)^2}\end{align}$$
which is $6/11$.
A: Let $P$ be the probability.  The probability of $A$ winning is given by $$P = \frac{1}{6} + \frac{5}{6}(1 - P)$$
This is because either $A$ wins on the first roll, or it's as if $B$ started first and we want the probability he doesn't win, which is $1-P$.
Solving this gives:
$$ \frac{11 P}{6} = 1 \implies P = \frac{6}{11} $$
A: Hint:  What is the chance that A wins on the first throw?  If A doesn't, his chance of winning is the same as B's chance at the start.  Let $a$ be A's chance to win.  Then B's chance to win is $1-a$.  This gives you a linear equation for $a$.
A: Hint:
$$
P(A\text{ wins}) = P(A \text{ wins in round 1}) + P(A\text{ wins in round 2}) + P(A\text{ wins in round 3}) + \cdots
$$
