probability question on casting a pair of dice

Question

A pair of dice is cast until a seven appears twice or until each of a six and eight has appeared at least once. Show that the probability of the six and eight occurring before two sevens is 0.546.

An attempt made was to consider 2 cases , one in which a 7 can happen and the other in which no 7 appears. In each cases the end outcome should be 6 or 8. There has to be atleast one 8 or 6 in remaining positions.I have considered it as a random variable and sum over 1 to infinity. But the answer obtained is not the one required. Please help

The equation I arrive at is: x be the position at which we end our experiment . i.e. I have a 6,8 as sum

(x-2)(6/36)(x-1)(5/36)(25/36)^(x-3)(5/36) + (x-1)(5/36)(5/36)(25/36)(x-2) which needs to be sum over from 1 to infinity.

Answer 1

You have a Markov chain with six states: none of $6,7,8$ has appeared, $7$ has appeared but no $6$ or $8$, one of $6$ or $8$ has appeared but no $7$, both $7$ and (one of $6$ or $8$) has appeared, two $7$s have appeared (final), a $6$ and $8$ have both appeared (final). You can ignore all other rolls and compute the transition probabilities. For example, from start you move to one $7$ with chance $\frac 6{16}$ because there are $6$ ways to get $7$ and $10$ ways to get $6$ or $8$ and to one of $6$ or $8$ with chance $\frac {10}{16}$.

Answer 2

The probability of rolling a $6$, $7$, and $8$ in a single roll of a pair of dice is $5/36$, $6/36$, and $5/36$ respectively. Since we ignore all other outcomes, you can think of the problem as instead rolling a single $3$-sided die that yields $6$, $7$, and $8$ with probability $5/16$, $6/16$, and $5/16$ respectively.

Suppose the first roll is a $6$; this happens with probability $5/16$.

• If the next non-$6$ roll is a $7$ (which happens with probability $6/11$), then the next non-$6$ roll after that must be an $8$ (which happens with probability $5/11$).
• Otherwise the next non-$6$ roll is an $8$ (probability $5/11$).

So if the first roll is $6$, you win with probability $\frac{5}{16} (\frac{6}{11} + 1) \frac{5}{11}$.

If the first roll is $8$, you obtain the same probability as well.

If the first roll is $7$ (probability $6/16$), the next roll must be $6$ or $8$ (probability $10/16$). If this roll was $6$, then we need the next non-$6$ roll to be $8$ (probability $5/11$); similarly if this roll was $8$, then the next non-$8$ roll must be $6$ (probability $5/11$). So if the first roll is $7$ you win with probability $\frac{6}{16} \cdot \frac{10}{16} \cdot \frac{5}{11}$.

Adding up these three cases yields $4225/7744 \approx 0.546$.