# Probability and counting dice roll

Two fair six-sided dice are rolled repeatedly until a score of 7, 9, or 11 is obtained.

You win if a score of 9 is the first of these three scores to occur.

Determine the probability that you win.

This is what I've done so far: Since probability is equally likely then $\frac {\text{# outcome in$E$}}{\text{outcomes in$S$}}$ since order is important and the die is being replaced after each rolled then I used the formula $n^k$ so the number of outcome in $S$ is equal to $6^2=36$ where $n=6$ and $k=2$. My problem is finding out the number of outcome in $E$.

• What is $E$? Is it rolling $9$, is it rolling any number in the set $\{7,9,11\}$, or is it something else? – David K Feb 18 '15 at 20:59
• $E$, the favoured space, contains the [redacted] outcomes where the dice add to $9$. $S$, the conditioned space, contains the [redacted] outcomes where the dice add to any of $7, 9,$ or $11$. – Graham Kemp Feb 18 '15 at 21:05

## 2 Answers

This is what I've done so far: Since probability is equally likely then $\frac {\text{# outcome in$E$}}{\text{outcomes in$S$}}$ since order is important and the die is being replaced after each rolled then I used the formula $n^k$ so the number of outcome in $S$ is equal to $6^2=36$ where $n=6$ and $k=2$. My problem is finding out the number of outcome in $E$.

No, $36$ is the count of all possible outcomes, but we are only interested in a restricted set of outcomes.

Let $n(7)$ count the (equally likely) outcomes of rolling two dice that result in a sum of $7$. These are $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$, so $n(7)=6$. Similiarly there are $n(9)$ outcomes that result in a sum of $9$, and $n(11)$ that result in a sum of $11$. We ignore all other outcomes as the experiment is repeated until the required condition occurs (That is: one of these three events) .

The favoured event space, $E$, contains all the outcomes that result in a sum of $9$.

Likewise the conditioned space, $S$, contains all the outcomes that result in a sum of $7, 9,$ or $11$.

So you have $$\frac{|E|}{|S|} = \frac {n(9)}{n(7)+n(9)+n(11)}$$

• thanks i got 1/3 a different way but wasn't sure – Zeus Feb 18 '15 at 20:59
• For this question you can easily get the right answer for the wrong reasons. For example, "There are three different scores that end the game, one of those scores wins, so probability $\frac13$," but that argument is invalid. So you should still check your work. – David K Feb 18 '15 at 21:05
• Yeah i did get it by adding p(7)+p(9)+p(11),which was wrong way of doing it. – Zeus Feb 18 '15 at 21:12
• Is there a way to approach this problem in the context of counting techniques? – Zeus Feb 18 '15 at 21:16

For variety, here is a different approach involving a recurrence.

Denote the probability of winning by "$P(w)$". Since there are $4$ out of $36$ ways of rolling a $9$, we win $\frac{4}{36}$ of the time on the first roll. The chance we lose on the first roll is $\frac{6}{36} + \frac{2}{36}$. So, the chance we don't win and don't lose on the first roll is $\frac{24}{36}$.

So, the chance of winning after not-winning and not-losing on the first roll is $\frac{24}{36} \cdot P(w)$. Hence, our total chance of winning is

$P(w) = \frac{4}{36} + \frac{24}{36} \cdot P(w)$

Solving for $P(w)$, we get $\frac{1}{3}$.