How to compute these probabilities? 
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*A pair of dice is cast until either the sum of seven or eight appears. How to compute the probability of a seven before an eight? 

*Now, if this pair of dice is cast until a seven appears twice or until each of a six and eight have appeared at least once. How to compute the probability of the six and eight occurring before two sevens? 
 A: Problem 1: The chance for a seven is $6/36 = 1/6$ and the chance for an eight is $5/36$. The chance for neither is $25/36$.
We want to compute the chance of getting neither $k$ times and then getting a seven, where $k$ goes from $0$ up. The chance is thus:
$$\frac{1}{6} + \frac{25}{36}·\frac{1}{6} + \frac{25^2}{36^2}·\frac{1}{6} + \ldots = \\ \frac{1}{6}\sum_{k=0}^\infty \left(\frac{25}{36}\right)^k = \frac{1}{6}·\frac{1}{1-\frac{25}{36}} = \frac{6}{11}$$
Problem 2
This is a bit more tricky. We are going to compute the chance for the alternative: to het two sevens but not both a six and an eight. Then, if $p$ is the chance for that to happen, what we want is $1-p$.
We have two cases:
We have a sequence of $k$ numbers, none of wich are seven or six (but can be eight), then a seven, then another sequence of $k'$ numbers that are not six nor seven, and then another seven.
OR
We have a sequence of $k$ numbers, none of wich are seven or eight (but can be six), then a seven, then another sequence of $k'$ numbers that are not eight nor seven, and then another seven.
(note that both cases are mutually exclusive)
The chance for a six is $5/36$ and the chance for nor six nor seven is $25/36$.
So we basically have to add all the numbers of the form:
$$\underbrace{\frac{25}{36}·\frac{25}{36}\ldots \frac{25}{36}}_{k \text{times}}·\frac{1}{6}·\underbrace{\frac{25}{36}·\frac{25}{36}\ldots \frac{25}{36}}_{k' \text{times}}·\frac{1}{6}$$
$$\sum_{k=0}^\infty \sum_{k'=0}^\infty \left(\frac{25}{36}\right)^k\frac{1}{6}\left(\frac{25}{36}\right)^{k'}\frac{1}{6}=\frac{1}{6}\sum_{k=0}^\infty \left(\frac{25}{36}\right)^k·\frac{1}{6}\sum_{k'=0}^\infty \left(\frac{25}{36}\right)^{k'} = \left(\frac{6}{11}\right)^2 = \frac{36}{121}$$
That's the chance for one of the cases, but since both cases have the same chance, the total chance of the alternative is $\displaystyle\frac{72}{121}$, therefore, the chance we want is $1-\displaystyle\frac{72}{121} = \frac{49}{121}$
A: For problem $1$ suppose the game ends on the $n^{th}$ throw. There are six chances of getting a $7$ on the $n^{th}$ throw to five of getting an $8$, so the probability of ending on a $7$ is $\cfrac 6{11}$.
Since this is independent of $n$ it is also the probability for all throws taken together.
