Dice Roll Probability Game Winner

Question

I've just started studying Probability and Statistics and I am stuck in the following problem:

There is a game, in which Players A and B roll a pair of dice one after 
another. The Game ends if Player A gets number 9 or if Player B gets number 6. 
Whoever gets his number first wins the game.
If player A starts always first, find the probability of Player A rolling 
the dices last.

So Basically, I have to find the probability of Player A winning the game.

We notice that:

• Player A needs to get { (3,6) , (4,5) , (5,4) , (6,3) } in order to win: P(A) = 4/36 .
• Player B needs to get { (1,5) , (2,4) , (3,3) , (4,2) , (5,1) } in order to win: P(B) = 5/36 .

We also notice that if Player A doesn't get 9, the probability of playing again is P(B') = 1 - P(B) = 31/36 , because he will only play if Player B doesn't get 6.

However, I don't see how we can find the propability of the player A winning the game If the number of the rounds is unlimited.

Any help?

Thanks in advance.

Answer 1

Let $p$ be the probability that $A$ wins the game. Then $$p = \frac{4}{36} + \left(\frac{32}{36}\right) \left(\frac{31}{36}\right) p $$ since after two rolls, the game resets.

So just solve for $p$.

Answer 2

$P(A$ wins$)=P(A$ wins on their first throw$)+P(A$ wins on their second throw$)+P(A$ wins on their third throw$)+...$

$P(A$ wins on their first throw$)=\frac {4}{36}$

$P(A$ wins on their second throw$)=P(A$ doesn't win on their first throw$)\times P(B$ doesn't win on their first throw$)\times P(A$ wins on their second throw$)=\frac {32}{36}\frac {31}{36}\frac {4}{36}$

$P(A$ wins on their third throw$)=\frac {32}{36}\frac {31}{36}\frac {32}{36}\frac {31}{36}\frac {4}{36}$

$P(A$ wins$)=\frac {4}{36}\left ( 1+\left(\frac {31}{36}\frac {32}{36}\right)+\left(\frac {31}{36}\frac {32}{36}\right)^2+...\right)$

Sum of geometric sequence for you to work out...

Answer 3

Let $R_i$ be the event that the game ends on roll $i$. If player A wins, we know the roll ends on an odd number.

$P(A) = \sum\limits_{k=0}^{\infty} P(R_{2k+1}) = \sum\limits_{k=0}^{\infty} (\frac{32}{36} \frac{31}{36})^k \frac{4}{36}$

$\sum\limits_{k=0}^{\infty} (\frac{32}{36} \frac{31}{36})^k$ is a geometric series, so everything can be rewritten as

$= \cfrac{1}{1-(\frac{32}{36} \frac{31}{36})}\frac{4}{36}$

Answer 4

Let $p=4/36$, $q=5/36$, and $r=(1-p)(1-q)$. The probability that player $A$ wins the game is $$ \sum_{n=1}^{\infty}r^{n-1}p=p\sum_{n=0}^{\infty}r^n=p\frac{1}{1-r}=\frac{9}{19}. $$ Note that $r^{n-1}p$ is the probability that player $A$ wins the game on the $n$-th trial, and we have taken the sum over all possible trials, applying the formula for a geometric series.