Evaluating probability that a player wins the game 
Players $P_1,P_2,P_3,...,P_m$ of equal skill,play a game consecutively in pairs as $P_1P_2,P_2P_3,P_3P_4,...,P_{m-1}P_m,P_mP_1,P_1P_2 \text{ and so on}...$, and any player who wins two consecutive games(i.e. $k$ and $(k+1)th$ game) wins the match.The question is to find $P_r$ where $P_r$ is the probability that player $P_r$ wins the match.

I tried to figure find out a recursion.$$P_r=(1-P_{r-1})\frac14$$ The probability that $P_r$ wins the match is same as the probability that $P_{r-1}$ loses the match and $P_r$ wins the next two successive matches.However I am not sure of the recursion and could not find the general formula.
 A: Let $\omega$ be a random variable equal to the number of winnder.
Your solution has two misses. At first you are trying to compute $p_k$ that is probability that game ends after $k$ matches. Then the probability of $r$th player win is $P\{\omega = r\} = p_r + p_{r + m} + p_{r + 2m} + \cdots$, not just $p_r$.
Secondly if game continues after $k - 1$ matches with probability $1 - p_{k - 1}$ then it can end after $k$ matches with unknown probability, because this probability doesn't give us information about the result of the last $(k - 1)$th match, and the conditional probability that it was won by player who continues playing is not $\frac12$ in general. (EDIT: Events "the game doesn't end after $k - 1$ matches" and "the game ends after $k$ matches" are not independent at all, since they share a subevent.)
For simplicity we will say that $P_i$ and $P_j$ is the same player if $i \equiv j \pmod m$.
Now it is easy to see that general game process is the following: $P_1$ wins $P_2$, $P_2$ wins $P_3$, $\ldots$, $P_{k - 1}$ wins $P_k$, $P_k$ wins $P_{k + 1}$, $P_{k + 2}$ wins $P_{k + 1}$, $P_{k + 3}$ wins $P_{k + 2}$, $\ldots$, $P_{\ell - 1}$ wins $P_{\ell - 2}$, $P_{\ell}$ wins $P_{\ell - 1}$, $P_{\ell}$ wins $P_{\ell + 1}$, game ends after $\ell$ matches. (It is possible that $k = 0$. Note that $\ell \ge k + 2 \ge 2$.) Probability of this event for given $k$ is obviously $\frac{1}{2^{\ell}}$. And $k$ can be $0, 1, \ldots, \ell - 2$, so we have $\ell - 1$ ways to choose $k$. Then $p_{\ell} = \frac{\ell - 1}{2^{\ell}}$.
Now it's time to compute the desired probability of $r$th player's win:
$$P\{\omega = r\} = p_r + p_{r + m} + p_{r + 2m} + \ldots\\
= \sum_{i = 0}^{+\infty} p_{r + i\cdot m} = \sum_{i = 0}^{+\infty} \frac{r + i\cdot m - 1}{2^{r + i\cdot m}}\\
= \frac{m}{2^r}\sum_{i = 0}^{+\infty} \frac{i}{(2^m)^i} + \frac{r - 1}{2^r}\sum_{i = 0}^{+\infty} \frac{1}{(2^m)^i}\\
= \frac{m}{2^r}\sum_{i = 0}^{+\infty} \sum_{j = 0}^{i - 1} \frac{1}{(2^m)^i} + \frac{r - 1}{2^r}\cdot\frac{1}{1 - \frac{1}{2^m}}\\
= \frac{m}{2^r}\sum_{0 \le j < i} \frac{1}{(2^m)^i} + \frac{r - 1}{2^r}\cdot\frac{2^m}{2^m - 1}\\
= \frac{m}{2^r}\sum_{j = 0}^{+\infty} \sum_{i = j + 1}^{+\infty} \frac{1}{(2^m)^i} + \frac{2^m(r - 1)}{2^r(2^m - 1)}\\
= \frac{m}{2^r}\sum_{j = 0}^{+\infty} \frac{1}{(2^m)^{j + 1}}\cdot\frac{1}{1 - \frac{1}{2^m}} + \frac{2^{m - r}(r - 1)}{2^m - 1}\\
= \frac{m}{2^r}\cdot\left(\frac{1}{1 - \frac{1}{2^m}}\right)^2\cdot\frac{1}{2^m} + \frac{2^{m - r}(r - 1)}{(2^m - 1)}\\
= \frac{m}{2^r}\cdot\frac{2^m}{\left(2^m - 1\right)^2} + \frac{2^{m - r}(r - 1)}{2^m - 1}.$$
(Note that $P_1$ can't end game after the first match and we should exclude summand $p_1$ from $P\{\omega = 1\}$, but $p_1 = 0$ so its exclusion wouldn't change anything.)
A: I'll change the matches sequence slightly, to hopefully make the formulae simpler. So I'll assume the matches go:
$Pm-P1,P1-P2,P2-P3,P3-P4,...,Pm-P1$
That is, the first player that may win is $P1$.
Let's denote the probability that the $j$th player wins in the first round, by $p_j$, and the total probability that the $j$th player wins by $P_j$
We'll further code the match winner as $0$ if the first player wins, and $1$ if the second player wins.
So the tournament may be represented by a string of bits, and it ends when we have a sequence of $[01]$, example (first row is the bit index, second row is the bits sequence, a case with the 5th player winnig):
 0123 45
[0011 10]

Note that, except for the last $0$, every $0$ must be preceded by a $0$, otherwise the string would stop. It is clearer if we look at the first few cases.
So that $i=1$ wins (in the first round) there is one only sequence
10

we have $p_1=\frac{1}{2^2}$
So that $i=2$ wins (in the first round) there are two
0 10
1 10

we have $p_2=\frac{2}{2^3}$
So that $i=3$ wins (in the first round) there are three
00 10
01 10
11 10

we have $p_3=\frac{3}{2^4}$
The pattern is clear, so:
$p_j=\frac{j}{2^{j+1}}$
Noting that
$\sum_{i=1}^{n-1} \frac{i}{2^i} = 2-\frac{n+1}{2^{n-1}}$ (from wikipedia)
we may confirm that
$\sum_{i=1}^{\infty}p_i= \frac12 \sum_{i=1}^{\infty} \frac{i}{2^i} =1$
I was tempted to say that winning on any successive round is directly proportional to $p_j$ on the first round, so we could get the total probability by normalising the first round's, but it is not so.
The best I can think of is doing the summation, for all rounds, long and boring, but not really hard (uses another formula from the same source, for the second summation). 
$P_j = \sum_{r=0}^\infty p_{j+m\times r}$
$P_j = \sum_{r=0}^\infty \frac{j}{2^{j+m\times r +1}} + \sum_{r=0}^\infty \frac{m\times r}{2^{j+m\times r +1}}$
$P_j = \frac{1}{2^{j+1}}\left( \frac{j}{1-\frac1{2^m}} + m \frac{\frac1{2^m}}{ \left(1-\frac1{2^m}\right)^2 }\right)$
$P_j = \frac{1}{2^{j+1}}\left(\frac{ 2^m(2^m-1)j + 2^m m}{ \left(2^m-1\right)^2 }\right)$
Which does add up to one (so hopefully, now it is correct). Examples,
For 3 players

j    Pj
-------
1  0.408
2  0.347
3  0.245


For 7 players

j   Pj
-------
1  0.266
2  0.259
3  0.192
4  0.128
5  0.080
6  0.048
7  0.028

