Combinatorics of a game Suppose there are $n$ people sitting in a circle, with $n$ odd. The game is played in rounds until one player is left. 
Each round the remaining players point either to the person on their right or their left. If two players pick the same person they both are out.
There are $2^n$ possible opening rounds. I would like to analyse how many ways there are for $0,2,4,6, \ldots, n-1 $ people to be knocked out in the initial round for a given $n$.
edit 1:see $\textit {Bridgeburners}$ answer for $k=2$ people out in the first round.
Also I would like to work out the probability of winning if possible and if each person had a number, what possible number combinations can be found in the final stages of the game. 
Any references to this problem from before would be welcome either.  
Attempt
So I can prove that there are only $2$ ways nobody is knocked out.
Let $0$ represent a person pointing left and $1$ a person pointing right. We can then describe the state of the first round by a 'word' as follows $a_1a_2\ldots a_n  $ with each $a_i=0,1$. Two people are eliminated if the following subword occurs in the word,  $1a0$ and $a=0,1$. 
Wlog suppose $a_1=1$. Then if no people are to be eliminated, we have $a_k=1$ for all $k$ odd  $1 \leq k\leq n $. But then $a_2$ must be also else players $2$ and $n$ are eliminated. So it follows the state is $\overbrace{111\ldots 1}^n$. The proof is very similar if $a_1=0$.
I'm finding it difficult to calculate the number of ways for $2$ people. Mainly because of the condition that no-one else can be eliminated after an initial 'choice pair'.
edit 2: I've added an 'answer', which isn't an answer, but just a matlab script and some of it's outputs in a table. I forgot about this question, which was mainly for fun, and I didn't really want to clutter the question up hence it's below in an "answer".
 A: Start with $x_1 = 1$ and $x_3 = 0$, and let's find a way to make sure that players $1$ and $3$ are the only ones eliminated. Well let's assume that $x_2 = 1$. This choice would be symmetric with $x_2 = 0$ and reversing the direction of the circle.
With $x_2 = 1$ we require $x_4 = 1$, which requires $x_6 = 1$, etc, up to the last even number, $x_{n-1} = 1$. That covers the even players. We are left with the remaining $\frac{n-3}{2}$ players, $x_5, x_7, \cdots, x_n$. Note that player $n$ is safe from $2$, player $5$ is safe from $3$, and player $n-1$ is safe from $1$.
If $x_5 = 1$ that requires the remaining to be $1$. But if $x_5 = 0$ then $x_7$ can take whatever position it wants. If $x_7 = 1$ that requires the remaining to be $1$, but if $x_7 = 0$ then $x_9$ can do whatever it wants... you see the pattern? This means that in the odd set $x_5,x_7, \cdots x_n$, which consists of $k = \frac{n-3}{2}$ players, we have the leftmost $q$ players only facing left, and the rightmost $k-q$ players only facing right, with $q$ anywhere between $0$ and $k$. Thus there are $k + 1 = \frac{n-1}{2}$ configurations (with $q$ anywhere from $0$ to $k$ being $k+1$ options) for that odd set $x_5, x_7, \cdots, x_n$. 
So in the case $\{x_1,x_2,x_3\} = \{1,1,0\}$ we have $\frac{n-1}{2}$ possible configurations where everyone except players $1$ and $3$ are safe. Now let's see how symmetry can expand this solution to all configurations of two eliminations.
First $x_2 = 1$ is identical to $x_2 = 0$ and reversal of direction about player $2$. So we must also have $\frac{n-1}{2}$ configurations for $x_2=0$ and thus $n-1$ total configurations that eliminate players $1$ and $3$. 
Then, there are $n$ unique pairs that can be eliminated (1-3, 2-4, 3-5, ..., n-2). From circular symmetry, those cases are all identical to the one analyzed above by a simple re-labelling of the numbers. Thus there are $n (n-1)$ possible ways for two players to be eliminated.
I guess the number of possibilities for $k > 2$ players is yet to be solved, and maybe someone can do it and see if they recognize a pattern. It seems much more complicated. Also, TBH, I was working this out while I was typing here, so I could be wrong, and I urge you to check my logic and point out my flaws, so I can embarrassingly remove my answer in shame.
A: Not an answer but some numerics since I happened to remember I asked this question a long time ago. I also didn't want to clutter up the question, so I've put it here.
Matlab code to output the number of ways $0,2,4,\ldots,n-1$ people are out in the opening round.
n=9; %n should be odd.
states=2^n;

numbers=0:states-1;

bins=dec2bin(numbers); %numbers in binary

binlog=(bins=='1'); %binary as logical array

check=zeros(states,n); %checks who's out for each config 1 out 0 still in

numoutonernd=[[0:2:n].' zeros(floor(n/2)+1,1) ]; % output, out in round one.

for i=1:states

    for j=1:n

        temp=j+2;
        if temp>n
            temp=mod(temp,n); %periodic boundary
        end

        if binlog(i,j) && not(binlog(i,temp)) %(i,j)=1 and (i,j+2)=0

            check(i,j)=1;
            check(i,temp)=1;

        end

    end

end

for j=1:floor(n/2)+1

    numoutonernd(j,2)=length(find(sum(check,2)==numoutonernd(j,1)));

end

numoutonernd

Here's a table with some of the outputs. The first row is the number of people, $n$ above, and first column indicates the number of people eliminated.
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& \bf{3} & \bf{5} & \bf{7} & \bf{9} & \bf{11} & \bf{13} & \bf{15} \\ \hline
 \bf{0}& 2 & 2 & 2 & 2 & 2 &2 &2\\ \hline
 \bf{2}& 6 & 20 & 42 & 72 & 110 &156 &210\\ \hline
 \bf{4}&  & 10 & 70 & 252 & 660 &1430 &2730\\ \hline
 \bf{6}&  &  & 14 & 168 & 924 &3432 &10010\\ \hline
 \bf{8}&  &  &  & 18 & 330 &2574 &12870\\ \hline
 \bf{10}&  &  &  &  & 22 &572 &6006\\ \hline
\bf{12}&  &  &  &  &  &26 &910\\ \hline
\bf{14}&  &  &  &  &  & &30\\ \hline
\end{array}
The $2$ people out row is answered already by Bridgeburners.
Interestingly the pattern $10,70,252,660,1430,\ldots$ for the $4$ people out row isn't showing up on oeis. 
The diagonal patterns might be easier to analyse, e.g. it would seem the pattern $6,10,14,18\ldots$ is just $2n$, whereas $20,70,168,330,572\ldots$ is $(n-1)\cdot((n-1)^2 - 1)/3$, according to oeis. 
edit: I was under the impression the initial choice always had some sort of symmetry which would mean each entry in the table would be divisible by $n$, but that doesn't seem to be true for $n=9,15,21\ldots?$ Probably clear from the formula in the preceding paragraph.
