Suppose in a single round, I win with probability p, and you win with probability 1 − p. We play repeatedly and keep score until one of us is two ahead (for example a score of 2 − 0, 3 − 1, 2 − 4, etc.). What is the probability that I win overall?
I have been working through this probability problem. But I am not sure if my method is the most efficient or even may not be even correct and was wondering if someone could advise if I am carrying this out correctly, and if there more effective way of approaching this type of question?
My method for calculation
If I have understood the above question correctly, the only way to win is if you I am 2 ahead?
So one way I could win two time in a row would if I have a probability of $p^2$.
But if I get a $p$ then a $q$ or the other persons throws and $q$ then $p$ we have drawn, so to me this is like starting again i.e $0-0$ again so for me two win now I would need the following $pqpp=p^3 q$ or if I draw again then I would need to win the following $pqpqpp=p^4 q^2$ and so fourth.
As probabilities are independent (each future try is not effected by the previous) then I can add the following probabilities to find my overall chance of winning.
$P(Over \:all \:win)=p^2+p^3q+p^4q^2+.....=p^2+p^3(1-p)+p^3(1-p)^2+....$
so now using the infite geometric sum $S=\frac{p^2}{1-p(1-p)}=\frac{p^2}{1-p+p^2}$.
So my probability for winning over all is:
$$P(Over \:all \:win)=\frac{p^2}{1-p+p^2}$$
I am bit rusty with probability as it something I don’t do too often. But as mentioned is there more efficient way of approaching this type of question, as I was thinking of extending this question and exploring if I had to be 3 ahead what would the over all win be.
The case is random walk start at 0, reach 2 before -2. Recall Gambler's ruin . The proof if you learned martingale is trivial, otherwises you can use $P_k=pP_{k+1}+(1-p)P_{k-1}$ and bound condition prob 0,1 to derive.