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?
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}
$
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