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$A$ and $B$ play a series of games. Each game is independently won by $A$ with probability $p$ and by $B$ with probability $1−p$. They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series.

Q:Find the probability that A is the winner of the series.

Could anyone please help me out with this question and let me know how to approach it? BTW, I've already seen the answer from the book, but I don't know how to get it. Thanks.

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  • $\begingroup$ ma.utexas.edu/users/geir/teaching/m362k/weeklyhw4solns.pdf $\endgroup$ – lab bhattacharjee Sep 28 '13 at 19:34
  • $\begingroup$ Hint: Find the probability that A is the winner of the series when A is one game ahead, when A is one game behind, and when A and B won the same number of games. $\endgroup$ – Did Sep 28 '13 at 19:41
  • $\begingroup$ @labbhattacharjee, Can you explain how you get $P({F}_{3})$=$2p(1-p)$? I have seen the solution and I don't comprehend how they got the answer for it. $\endgroup$ – Kururugi Suzaku Sep 28 '13 at 21:00
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    $\begingroup$ @KururugiSuzaku: In the solution given in labbhattacharjee's link they state that $P(F_3)=2p(1-p)$ because $F_3$ covers two scenarios: $A$ wins the first, $B$ wins the second or the other way around. Denote the first by $AB$ and the other by $BA$. Then $F_3=\{AB,BA\}$ and $P(AB)=p(1-p)=P(BA)$ showing that $P(F_3)=P(AB)+P(BA)=2p(1-p)$. $\endgroup$ – String Sep 28 '13 at 22:08
  • $\begingroup$ @String, Thanks for the explanation. Now, it makes more sense. But I thought that there must be four rounds meaning 4 spaces, which makes it impossible not for one of them to find a round. How come the sample space was reduced to only 2 spaces. Would you mind to explain that part? Thanks. :) $\endgroup$ – Kururugi Suzaku Sep 28 '13 at 23:50
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Here is how I would have solved it. Let $f_n$ denote the probability that $A$ wins given that $A$ at the current state has won $n$ times more than $B$. Then $$ \begin{align} f_{-2}&=0\\ f_{-1}&=(1-p)\cdot f_{-2} +p\cdot f_0\\ f_0&=(1-p)\cdot f_{-1}+p\cdot f_1\\ f_1&=(1-p)\cdot f_0+p\cdot f_2\\ f_2&=1 \end{align} $$ This system of linear equations can be solved for $f_0$ by first eliminating $f_{-2}$ and $f_2$: $$ \begin{align} f_{-1}&=p\cdot f_0\\ f_0&=(1-p)\cdot f_{-1}+p\cdot f_1\\ f_1&=(1-p)\cdot f_0+p \end{align} $$ and then plugging the expressions for $f_{-1}$ and $f_1$ into the expression for $f_0$ and solve for it: $$ f_0=(1-p)\cdot p\cdot f_0+p\cdot (1-p)\cdot f_0+p^2\\ \Updownarrow\\ f_0=\frac{p^2}{1-2p(1-p)} $$

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