so, the question is as follows:
players $P_1,P_2...P_m$ of equal skill , play a game consecutively in pairs as $$ P_1P_2 , P_2P_3,...P_(m-1)P_m,P_mP_1,...$$
and any player who wins two consecutive games (i.e $k$ and $k+1$ game ) wins the match. if the chance that the match is won at the $r^{th}$ game is p then we need to find $p(r)$
my approach; for small value of r I was easily able to find that $$p(1)=0/2 , p(2)=1/4, p(3)=2/8 $$ which intuitively seems that $p(r)=(r-1)/2^r$ but i can't prove it . kindly help me out.
In the last two games, the second and first player must win, respectively, if the match is to end in $r \gt 1$ games. Let $g < r - 1$ be the first game that has the second player winning; then the remainder of the games $g + 1, \ldots, r - 2$ must have the second player winning as well (otherwise, the match would be finished sooner). This means that the number of valid sequences is defined by the possible values of $g$, with $g$ ranging from $0$ (no game prior to the last one has the first player win) to $r - 2$ (only game $r - 1$ has the second player win). Since each sequence is equally likely to occur, the probablity $P(r)$ of the match ending in $r$ games equals:
$$P(r) = \frac{r - 1}{2^n}$$