You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?

Question

You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?

What I tried was \begin{align*} P(X=6) & = \frac{5}{36}\\ P(X=8) & = \frac{5}{36}\\ P(X=7) & = \frac{6}{36} = \frac{1}{6} \end{align*}

I try finding the probability of getting $6$ and $8$ and not $7$

$$P(X=6)+P(X=8)+P(X \neq 7) = \frac{5}{26} + \frac{5}{36} + \frac{5}{6} = 1.11$$

I stick since I know that the probability cannot be greater than one. The solution key says the answer is $0.5456$.

Answer 1

It is easier to compute the probability of first getting two $7's\;\;$ and take the complement

Totals other than $6,7,8$ don't matter, so we can simply take their odds in favor as $5:6:5$

Consider the three mutually exclusive starts $(a)\;\;7-7,\;\;(b)\;\; 7-(6\;\;or\;\; 8),\;\;(c)\;\; (6\;\; or\;\; 8)-7$

In $(a),\;\; 7$ wins with $Pr = \frac{6}{16}\frac6{16}$

For $(b)\;\;and\;\;(c)$, note that for subsequent throws, it becomes a direct race between $7$ and whichever of $6$ and $8$ has not yet appeared, thus the odds in favor of $7$ becomes $6:5$

In $(b),\;\; 7\;\;$ wins with $Pr = \frac6{16}\frac{10}{16}\frac6{11}$

In $(c),\;\; 7\;\;$ wins with $Pr = \frac{10}{16}\frac6{11}\frac6{11}$

Adding up, $7$ wins with $Pr = 0.4544$

Finally, answer = $1 - 0.4544 = 0.5456$

Answer 2

Everything other than $6$, $7$, or $8$ is irrelevant. So we can imagine that we are repeating an experiment where the probability of $6$ is $5/16$, as is the probability of $8$, while $7$ has probability $6/16$.

Condition on the result of the first toss. Suppose it is a $6$. Then $6$'s become irrelevant, and the race is between $7$ and $8$. The first now has probability $\frac{6}{11}$ and the second has probability $\frac{5}{11}$. If we get an $8$, we win. If we get a $7$, then the probability we win is $\frac{5}{11}$. So the contribution to the probability is $$\frac{5}{16}\left(\frac{5}{11}+\frac{6}{11}\frac{5}{11} \right).$$

There is the same contribution to the probability of winning from first toss is $8$.

Now we look at what happens if the first toss is $7$. If the next toss is $7$, we have lost. If it is $6$ or $8$, then our probability of winning is $\frac{5}{11}$. The contribution to the probability is $$\frac{6}{16}\left(\frac{10}{16}\frac{5}{11}\right).$$

We conclude that the contribution from $6$ or $8$ first is $\frac{850}{(16)(121)}$, and the contribution from $7$ first is $\frac{300}{(256)(11)}$. Add. The official answer seems to be right.

Answer 3

Based on the OP attempt, I try to find a more "elementary" way to tackle the problem, but it seems that the most natural and systematic way is to use the markov chain regardless.

We can model the game by Markov chain with the following transition matrix: $$ \begin{matrix} 1 & 6^c \cap 8^c \cap 7^c \\ 2 & 6 \cup 8 \\ 3 & 7 \\ 4 & (6 \cup 8) \cap 7 \\ 5 & 6 \cap 8 \\ 6 & 7 \cap 7 \end{matrix} \begin{bmatrix} \frac {5} {9} & \frac {5} {18} & \frac {1} {6} & 0 & 0 & 0 \\ 0 & \frac {25} {36} & 0 & \frac {1} {6} & \frac {5} {36} & 0 \\ 0 & 0 & \frac {5} {9} & \frac {5} {18} & 0 & \frac {1} {6} \\ 0 & 0 & 0 & \frac {25} {36} & \frac {5} {36} & \frac {1} {6} \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

where state $1$ is not having any of the $6, 8, 7$ yet, state $2$ is having the first $6$ or $8$ but not $7$ yet, state $3$ is having the first $7$ but no $6$ or $8$ yet, state $5$ is having both the first $6$ or $8$ and $7$. State $5$ and $6$ are the absorbing states, which corresponding to winning and losing respectively.

The transition probabilities are easy to understand here. The only crucial fact need to remind here is that after the getting the first $6$ or $8$ (transition from state $1$ to $2$ or $3$ to $4$ with probability $\frac {5} {18}$), the probability of getting another in the next cast (i.e. transition from state $2$ or $4$ to $5$) will be only $\frac {5} {36}$.

Let $p_i$ be the probability of winning given now we are at state $i$. By definition $p_5 = 1, p_6 = 0$ and we want to determine $p_1$. By using the first step analysis technique, we have the system:

$$ \begin{cases} \displaystyle p_1 = \frac {5} {9} p_1 + \frac {5} {18} p_2 + \frac {1} {6} p_3 \\ \displaystyle p_2 = \frac {25} {36} p_2 + \frac {1} {6} p_4 + \frac {5} {36} \\ \displaystyle p_3 = \frac {5} {9} p_3 + \frac {5} {18} p_4 \\ \displaystyle p_4 = \frac {25} {36} p_4 + \frac {5} {36} \\ \end{cases}$$

https://www.wolframalpha.com/input/?i=p1+%3D+5p1%2F9+%2B+5p2%2F18+%2B+p3%2F6,+p2+%3D+25p2%2F36+%2B+p4%2F6+%2B+5%2F36,+p3+%3D+5p3%2F9+%2B+5p4%2F18,+p4+%3D+25p4%2F36+%2B+5%2F36

On solving this (e.g. by backward substitution), we have $$p_1 = \frac {4225} {7744} \approx 0.54558367768, p_2 = \frac {85} {121}, p_3 = \frac {25} {88}, p_4 = \frac {5} {11} $$

which is the same as the answer key suggested.