Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs 
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 (ie. k and (k+1)th game) wins the match. If the chance that the match is won at the rth game is $k$, then find relation between $k$ and $r$

I haven’t really understood the question or what it wants to do, so I don’t know how to begin. Can anyone explain the question and give a starter hint?
 A: Note:  what follows assumes that the match ends with the game between $P_m$ and $P_1$ even if no victor for the match is determined.  If, to the contrary, one is meant to replay the thing until a victor is set, then just use the infinite case below, setting $P_i=P_j$ if $i\equiv j\pmod m$. If you want the probability that a specified player wins, just sum the relevant probabilities.  If all you want is the probability that the match ends in a particular game, then the stated formula holds.
For simplicity, let us first consider the case where $m=\infty$ (Note that $P_1$ can never win this match).
A match can be considered a binary string, where a $0$ means that the player with the lower index won, and a $1$ means that the player with the higher index won.  Thus the match $0010$  means that the winners were $P_1,P_2,P_4, P_4$ so $P_4$ is the winner of the match.   in this way, we see that the match typically ends with the appearance of the first $10$ (player $P_1$ messes this up a little).
It is easy to count the strings with no $10$.  They must all be of the form $0^a1^b$  So there are $n+1$ such of length $n$.
The probability that a string of length $r$ has it's first $10$ ending in slot $r$ is then $$\frac {r-2+1}{2^{r-2}}\times \frac 14=\boxed {\frac {r-1}{2^r}}$$
Sanity Check:  Note that $$\sum_{r=2}^{\infty}\frac {r-1}{2^r}=1$$ so, in the case of infinite $m$, the match is eventually resolved with probability $1$.
The one exception to this analysis arises if $m$ is finite so $P_1$ might win.  In that case we need that there be no winner in the first $m-1$ games, but that $P_1$ must win the first and last.  Thus we have a word of the form $0W0$ where $W$ is a word of length $m-2$ with no $10$.  The probability of that is $$\frac 12\times \frac {m-2+1}{2^{m-2}}\times \frac 12=\frac {m-1}{2^m}$$
so the same formula holds in this case.
Sanity Check (for the exceptional case):  suppose $m=4$.  Then the paths in which $P_1$ wins are $$(P_1, P_2,P_3,P_1)\quad(P_1,P_2,P_4, P_1)\quad(P_1,P_3,P_4,P_1)$$ so the answer would be $\frac 3{16}$ in this case, matching the formula.
Remark:  there is no need for the sum of these probabilities to be $1$ since, in the finite case, the match could certainly be decided without a victor.
A: @Aditya: An attempt to simplify the presentation. We can forget about m, work out for $r=5$, and then generalize
The last two games have to be $\fbox{LW}\fbox{WL}$
and the first $(5-2)$ can either be $\fbox{WL}\fbox{WL}\fbox{WL}$ or
any  of the $3\;\fbox{WL}$ can be replaced by $\fbox{LW}$ with the caveat that once a $\fbox{LW}$ has been inserted, all subsequent ones also must change to $\fbox{LW}$
This is the rationale of $(r-2+1)$ valid strings
For this example, $Pr = \dfrac{5-2+1}{2^5}$
Putting in the general symbols, we get $Pr = \dfrac{r-1}{2^r}$
A: A chart of what happens in the first few games would perhaps help supplement lulu's answer above:

Note that the outcome is split by the probabilities of the last two games, since that's how we know that the match finishes: when the "new" player wins one round and then - as the "old" player - wins the next round, shown as NO. Here the options for the new player winning are gathered into a single item by "forgetting" past history - so for example the ...ON and ...NN outcomes from game $3$ have a combined probability of $3/8$, which is then split into two $3/16$ probability outcomes of game $4$.
A: The question is simply if the match ends at game $r$ who won?  If it ends on the second game, only players $1$ or $2$ could have won because they are the only players who could have played twice.  If it ends on the third game only the third player could have won.  If player $1$ wins the first game, player $3$ must win the second or the match ends at the second game.  Then the third game has to be won by player $3$ to end.  In general, if it ends at game $r$ in the first round, it is player $r$ who wins.  If the competition continues, the same is true.  Game $m$, the one between the winner of game $m-1$ and player $1$ can only be the last game if player $m$ wins it (as well as the previous).  The match is won by player $r \pmod {m}$ unless it is won in the second game, when it might be player $1$.
