Rolling a die

Aquafina · #1 $Old$ November 2nd, 2009, 06:52 AM

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?

Robb · #2 $Old$ November 2nd, 2009, 07:29 AM

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 $\frac{2}{6}$ of winning to $\frac{1}{6}$

So assuming that A goes first;

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

Aquafina · #3 $Old$ November 2nd, 2009, 08:41 AM

Quote:

Originally Posted by Robb $View Post$

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 $\frac{2}{6}$ of winning to $\frac{1}{6}$

So assuming that A goes first;

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

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?

Robb · #4 $Old$ November 2nd, 2009, 08:53 AM

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
$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}=\frac{25}{36}\cdot\frac{1}{6}$
$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}=\left(\frac{25}{36}\right)^2 \cdot\frac{1}{6}$
etc.

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