When $A$ and $B$ flip coins, the one coming closest to a given line wins $1$ penny from the other. If $A$ starts with $3$ and $B$ with $7$ pennies, what is the probability that $A$ winds up with all of the money if both players are equally skilled? What if A were a better player who won $60\%$ percent of the time?
This question I found in the end of the Sheldon Ross introduction (3rd chapter) to probability book. The given answer is $0.3$ but I am not sure for which question exactly, however neither of my answers is equal to that and I am not sure how to exactly apply the basic terms of probility "events" to this type of questions.
My attempt:
first question - $$ P(A)=P(B)=\frac{1}{2}$$ $$ P(A \,wins \,all) := 0 \,losses : AAAAAAA$$ $$ 1\, loss :BAAAAAAAA, ABAAA..... $$ $$ 2\, loss :....$$
and so on , which can be formulated by:
$$ P(A)= \frac{1}{2^7}*\sum_{0}^{\infty} \frac{1}{2^{2n}}= \frac{1}{2^7}*\frac{4}{3}=0.01$$ Now, I am not sure if I should consider that if A losses 3 matches in a row in the beginning it basically loses I guess?
Second question, I used a similar method: $$ 0.6^7*\sum_{0}^{\infty} 0.6^n*0.4^n= 0.6^7*\frac{1}{1-\frac{6}{25}} = 0.037 $$
Any hints, or advice greatly appreciated!
For $1\le n\le 9$, let $p_n$ be the probability that $A$ wins if $A$ has $n$ pennies before starting the next round.
Then we have the following system of $5$ equations in $5$ unknowns . . .
\begin{cases} p_1=\frac{1}{2}p_2\\[4pt] p_2=\frac{1}{2}p_3+\frac{1}{2}p_1\\[4pt] p_3=\frac{1}{2}p_4+\frac{1}{2}p_2\\[4pt] p_4=\frac{1}{2}p_5+\frac{1}{2}p_3\\[4pt] p_5=\frac{1}{2}\;\;\;\text{[by symmetry]}\\ \end{cases} Solving the system yields $p_3=\frac{3}{10}$
If instead of equal skills, we assume that $A$ has probability $\frac{3}{5}$ of winning each round, we get the following system of $9$ equations in $9$ unknowns . . . \begin{cases} p_1=\frac{3}{5}p_2 \qquad\qquad\qquad\;\; \\[4pt] p_2=\frac{3}{5}p_3+\frac{2}{5}p_1\\[4pt] p_3=\frac{3}{5}p_4+\frac{2}{5}p_2\\[4pt] p_4=\frac{3}{5}p_5+\frac{2}{5}p_3\\[4pt] p_5=\frac{3}{5}p_6+\frac{2}{5}p_4\\[4pt] p_6=\frac{3}{5}p_7+\frac{2}{5}p_5\\[4pt] p_7=\frac{3}{5}p_8+\frac{2}{5}p_6\\[4pt] p_8=\frac{3}{5}p_9+\frac{2}{5}p_7\\[4pt] p_9=\frac{3}{5}+\frac{2}{5}p_8 \end{cases} Solving the system yields $p_3={\large{\frac{41553}{58025}}}\approx 0.7161223611$