Probability of having six side first

Question

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!

Answer 1

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 $6/11$.

Answer 2

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:

$$ \frac{11 P}{6} = 1 \implies P = \frac{6}{11} $$

Answer 3

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 $a$.

Answer 4

Hint: $$ P(A\text{ wins}) = P(A \text{ wins in round 1}) + P(A\text{ wins in round 2}) + P(A\text{ wins in round 3}) + \cdots $$