In a game, A and B play against each other. They throw a die alternatively until they get the face of 6. Once one gets 6, he wins. i.e., there is only one winner and they keep playing until one of them wins. Consider that B plays first. What is the probability of A winning?
I don't even have a clue how to start, but I'm quite sure that it is concerned with the geometric progression. Pls do help me, any answers will be greatly appreciated. ( but if possible, pls do explain in detail )
On
He either wins in the first round ...or second round...or third round...so on.
It's probability will be
$$P=\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^2\times \left(\frac{5}{6}\cdot \frac16 \right)+\left(\frac{5}{6}\right)^3\times \left( \left(\frac{5}{6}\right)^2\cdot \frac 16\right) \ldots $$
Now this sum can be calculated by solving the infinite GP sum
On
For $A$ to win on their first turn we need $B$ to not get a 6 and $A$ to get a 6, for probability $\frac{5}{6}\cdot\frac{1}{6}$.
For $A$ to win on their second turn, we need both to miss on their first try, $B$ to miss on their second try, and $A$ to get 6 on their second try, for probability $\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}$.
For $A$ to win on their third turn, we need $B$, $A$, $B$, $A$, $B$ to miss in that order and then $A$ to get a 6, for probability $\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}$.
This pattern continues infinitely so the probability $A$ wins the whole thing is $$\frac{5}{6}\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}+\cdots = \dfrac{\frac{5}{6}\cdot\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2} = \dfrac{5}{11}.$$
On
B plays first, so if you want the probability of A winning, then B must not roll a 6. This has probability $\tfrac56$. Then A can win with probability $\tfrac16$, but alternatively he may lose, with probability $\tfrac56$, in which case it is back to the situation we started with. So we can represent the total probability with an infinite sum:
$$\begin{align}P(\text{A wins})&=\frac56\frac16+\left(\frac56\frac56\right)\frac56\frac16+\left(\frac56\frac56\right)\left(\frac56\frac56\right)\frac56\frac16+...\\&=\frac5{36}\frac1{1-\frac{25}{36}}\\&=\frac5{36-25}\\&=\frac5{11} \end{align}$$
Hint:
Let $B$ play first, then the probability he/she wins in the first move is $\frac{1}{6}$. The probability that he/she does not win in the first move is $\frac{5}{6}$.
Suppose $B$ does not win in the first move (that has $\frac{5}{6}$ probability), in order for $B$ to win in the second, $A$ has to not win in their first move (that has $\frac{5}{6}$ probability) and then $B$ has to roll a six (that has $\frac{1}{6}$ probability).
Hence the probability of $B$ winning in their second move is $(\frac{5}{6})^2(\frac{1}{6})$.
Suppose $B$ does not win in their first move or second move...(that has ...probability)...