Consider a modified gambler's ruin problem where in each coin toss the player either wins one more coin with probability $p$ or lose all his coins with probability $1-p$.
a) Give an equation $P_m$ of winning starting with $m$ coins where $0<m<N$ with the law of total probability. Note boundary conditions.
b) Write an expression of probability $P_m$ in terms of $p$.
Consider another version of the gambler's ruin where the player has probability p of winning 1 coin, probability
c)Show that $P_m$ has the same probability as the usual unbalanced gambler ruin with a probability of winning a coin being $\tilde{p}=\frac{p}{p+q}$
So for a), I conditioned on winning the first toss. Suppose $P_m$ is the event of winning starting with $m$ coins. With the law of total probability, $P(P_m)=P(P_m|W)P(W)+P(P_m|W^C)P(W^C)$
Then in b) I substituted in $P(W)=p$, since it is not possible to win without winning the first toss, then $$P(P_m)=P(P_m|W)P(W)$$ $$=pP(P_m|W)$$ $$=p\frac{P(P_m \cap W)}{P(W)}$$ $$=p\frac{P(P_m \cap W)}{P(W)}$$ $$=p\frac{P(W|P_m)P(P_m)}{P(W)}$$ $$=p\frac{p^m}{p}$$ $$=p^m$$
I am not sure if the above logic is correct or how to go about solving the second modified (win one, lose one, staying put) version of the problem? Should I use conditioning as I did or trying to account for the next/previous toss like in the original gambler's ruin problem? i.e. $pP_i + qP_i = pP_{i+1} + qP_{i−1}$