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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

Tree Diagram

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.

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  • $\begingroup$ Your reasoning shows the probability $w$ that you win solves $$w=p^2\cdot1+2pq\cdot w+q^2\cdot0$$ hence $$w=\frac{p^2}{1-2pq}=\frac{p^2}{p^2+q^2}$$ $\endgroup$
    – Did
    Commented Aug 18, 2018 at 0:03
  • $\begingroup$ I am a bit confused by your comment. Because the 2pq would be if I drew which means that I am effectively starting from 0-0 as i mentioned or 1-1 which is why. Could you please expamded you comment into an answer? $\endgroup$
    – james2018
    Commented Aug 18, 2018 at 7:36
  • $\begingroup$ I am a bit confused by your confusion: $2pq$ is the probability that after two games, the players are again even. When this happens, your chances to win are exactly what they were at the start of the game, hence the term $2pq\cdot w$. What is unclear there? $\endgroup$
    – Did
    Commented Aug 18, 2018 at 12:10
  • $\begingroup$ Yes that I understand that. But my reasoning is that probability will not effect the overall chance of me winning as so to speak the game resets. Which is why I never included the probability and noticed that it formed a geometric sum for the overall. That how I see the question itself $\endgroup$
    – james2018
    Commented Aug 18, 2018 at 12:38

1 Answer 1

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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.

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