1. ## Rolling a die

There is a game with 2 players (A&B) who take turns to roll a die and have to roll a six to win. What is the probability of person A winning?

I got (5/6)^(2n) x (1/6)

Is this correct?

2. This question is similiar to a thread I created before... see the second post, change homer and marge to A and B, and the probability from $\displaystyle \frac{2}{6}$ of winning to$\displaystyle \frac{1}{6}$

So assuming that A goes first;

$\displaystyle P(A)=\frac{1}{6} \cdot \frac{1}{1-\frac{25}{36} } =\frac{1}{6}\cdot \frac{36}{11}=\frac{6}{11}$

3. Originally Posted by Robb
This question is similiar to a thread I created before... see the second post, change homer and marge to A and B, and the probability from $\displaystyle \frac{2}{6}$ of winning to$\displaystyle \frac{1}{6}$

So assuming that A goes first;

$\displaystyle P(A)=\frac{1}{6} \cdot \frac{1}{1-\frac{25}{36} } =\frac{1}{6}\cdot \frac{36}{11}=\frac{6}{11}$
How do you get a geometric sum? The variable n is not defined, so Homer/A could win the game in the first game or the 301st game.. etc?

4. Yup, exactly.. he can win on his first roll, his second roll, his third roll etc..

So, if we let W be the event that a 6 is rolled, and N be the event any other number, and A rolls first, then A will win if the following happens;
W
NNW
NNNNW
NNNNNNW
NNNNNNNNW
etc.
So the probability of these happening;
1/6
$\displaystyle \frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}=\frac{ 25}{36}\cdot\frac{1}{6}$
$\displaystyle \frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\f rac{5}{6}\cdot\frac{1}{6}=\left(\frac{25}{36}\righ t)^2 \cdot\frac{1}{6}$
etc.

So the Probability of A winning is the sum of the combinations of those events happening
$\displaystyle P(A winning)=\frac{1}{6}+\frac{25}{36}\cdot\frac{1}{6} +\left(\frac{25}{36}\right)^2 \cdot\frac{1}{6}+....$
Which is a geometric series