Rolling fair die game The question is: Peter and Mary take turns rolling a fair die. If Peter rolls 1 or 2 he wins and the game stops. If Mary rolls 3,4,5, or 6, she wins and the game stops. They keep rolling until one of them wins. Suppose Peter rolls first.
a) What is the probability that Peter wins and rolls at most 4 times?
Here is the solution: We want to find $A=\\\{ \text{Peter wins and rolls at most 4 times} \\\}$. We decompose the event into $A_k$= {Peter wins on his kth roll}. Then $A=\cup^4_{k=1}$ and since the events $A_k$ are mutually exclusive, $P(A)=\sum^4_{k=1} P(A_k)$. My book says that the ratio of favorable alternatives over the total number of alternatives yields $$P(A_k)=\frac{(4 \cdot 2)^{k-1}\cdot 2}{(6 \cdot 6)^{k-1} \cdot 6}=\left(\frac{8}{36} \right)^{k-1}\frac{2}{6}=\left( \frac{2}{9}\right)^{k-1}\frac{1}{3}$$ Then the solution if sound using the geometric sum.
I am having problems understanding this answer. Why is the 2/6 not to the k power? and why are we multiplying the probability of failure of Peter and Mary inside the k-1 power? I'd really appreciate it if someone could break down how we obtain $A_k$.
 A: The probability that Peter wins in any given round is $\frac{2}{6}$. The probability that Peter and Mary both don't win on any given round is
$$\mathbb{P}(\text{Peter loses and Mary loses}) = \mathbb{P}(\text{Peter loses}) \cdot \mathbb{P}(\text{Mary loses}) = \frac{4}{6} \cdot \frac{2}{6}.$$
If we want Peter to win, then we need Mary to lose every time, and Peter to win at the end. We use independence to turn intersection ("and") into multiplication.
$$\mathbb{P}(\text{Peter wins on first roll}) = \frac{2}{6} $$
$$
\begin{align*}
\mathbb{P}(\text{Peter wins on the second roll}) &= \mathbb{P}(\text{Peter and Mary both lose on round 1 and Peter wins on round 2}) \\ &= \mathbb{P}(\text{Peter and Mary both lose}) \cdot \mathbb{P}(\text{Peter wins}) \\ &= \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{2}{6}
\end{align*}
$$
$$
\begin{align*}
\mathbb{P}(\text{Peter wins on the third roll}) &= \mathbb{P}(\text{Peter and Mary both lose on round 1 and 2 and Peter wins on round 3}) \\ &= \mathbb{P}(\text{Peter and Mary lose})^2 \cdot \mathbb{P}(\text{Peter wins}) \\ &= \left(\frac{4}{6} \cdot \frac{2}{6}\right)^2 \cdot \frac{2}{6}
\end{align*}$$
$$
\begin{align*}
\mathbb{P}(\text{Peter wins on the fourth roll}) &= \mathbb{P}(\text{Peter and Mary both lose on round 1 and 2 and 3 and Peter wins on round 4}) \\ &= \mathbb{P}(\text{Peter and Mary lose})^3 + \mathbb{P}(\text{Peter wins}) \\ &= \left(\frac{4}{6} \cdot \frac{2}{6}\right)^3 \cdot \frac{2}{6}
\end{align*}$$
Add up these. Hopefully the pattern makes sense--you multiply probability that Peter and Mary lose for the first $k-1$ rounds, then finally probability that Peter wins on round $k$.
A: For a roll, the probability that Peter fail is 4/6 and Mary fail 2/6.
If Peter wins on his kth roll, Peter should fail on his first k-1 rolls and Mary fail on her first k-1 rolls. The probability is $( \frac{2}{6} )^{k-1}(\frac{4}{6})^{k-1}$.
And, on k-th roll, the probability that Peter wins is 2/6.
Thus the probability is, $P(A_k)=(\frac{2}{6})^{k-1}(\frac{4}{6})^{k-1}\frac{2}{6}$
