Probability of Following Winning Player Let's say we're playing a game with a total of $n$ people. In each round of the game, two random players $A$ and $B$ are selected. One of $A$ or $B$ is randomly chosen to be the winner, and the other the loser. Without loss of generality, let us assume $A$ wins. Then, $B$ is eliminated from the pool of $n$ contestants, and denoted as a 'follower' of $A$. Furthermore, any previous followers of $B$ also become followers of $A$. $A$ is re-entered into the pool of contestants, and this process is repeated until one person is left and wins the game. Clearly, this last person remaining will have everyone following him.
My question is: what is the probability that a player $X$, randomly selected at the beginning of the game, ends up following the overall winner $Y$, before $Y$ wins? I'll say any player is definitionally following themselves at the beginning of the game, to formalize the problem. I think the answer is $\frac{1}{2}$ by symmetry, but this also seems somewhat counter-intuitive.
If you're curious, the above question is motivated by the game Fortnite; when another player kills you, you can then watch them play. If another player kills the player who killed you, you then watch the new winner play.
 A: You're right, it is $\frac{1}{2}$.
We can prove this by induction:
Base: 
2 players. One will be the winner and the other the loser of the only game. The winner was a follower of themselves from the start, i.e. before the win took place, but the loser was not. So, the probability is indeed $\frac{1}{2}$
Step: 
Inductive Hypothesis: Suppose that with $n$ players the probability of one of those players being a follower of the eventual winner $Y$ before the last win is $\frac{1}{2}$
Now let's take $n+1$ players.  If you take a random player $X$, there are three ways this player ends up being a follower of the eventual winner $Y$ before the last win:
A. $X$ plays in the first game  (probability $\frac{2}{n}$), loses (probability $\frac{1}{2}$) to some other player $Z$ (who may or may not be $Y$), and $Z$ gets put back in the group (now consisting of $n$ players) and becomes (or already is, if $Z$ is $Y$) the follower of the eventual winner $Y$ before $Y$'s last win (for once $Z$ becomes a follower of $Y$, $X$ will too). Since $Z$ is one of $n$ players, and since $Z$'s earlier win has no effect on the outcome of any of the future games, we can treat $Z$ as a random player and thus we can apply the Inductive Hypothesis, meaning that  $Z$ being the follower of $Y$ before $Y$ last win is $\frac{1}{2}$. Thus, the probability of this happening is: 
$$\frac{2}{n} \cdot \frac{1}{2} \cdot \frac{1}{2}  = \frac{1}{2n}$$
B. $X$ plays in the first game ((probability $\frac{2}{n}$), wins (probability $\frac{1}{2}$), and gets put back in the group (now consisting of $n$ players), and becomes the follower of the eventual winner $Y$ before $Y$'s last win (with probability $\frac{1}{2}$ by Inductive Hypothesis). The probability of this happening is:
$$\frac{2}{n}  \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}  = \frac{1}{2n}$$
C. $X$ does not play the first game (probability $\frac{n-2}{n}$), but after first game (when $n$ players are left) becomes the follower of the eventual winner $Y$ before $Y$'s last win (probability $\frac{1}{2}$ by Inductive Hypothesis). The probability of this happening is: 
$$\frac{n-2}{n} \cdot \frac{1}{2}  = \frac{n-2}{2n}$$
So, the probability that $X$ ends up being a follower of the eventual winner $Y$ before the last win is:
$$\frac{1}{2n}+\frac{1}{2n}+\frac{n-2}{2n}=\frac{n}{2n}=\frac{1}{2}$$
A proof that does not use induction is as follows:
When there are two players left, one (call this player $Y1$) will have $k$ followers  and the other ($Y2$) has $n-k$ followers. There are two ways in which random player is a follower of the eventual player $Y$ before $Y$ wins the last game:
A. The player is a follower of $Y1$ (probability $\frac{k}{n}$, and $Y1$ wins the last game against $Y2$ (i.e $Y=Y1$; probability $\frac{1}{2}$). The probability of this happening is:
$$\frac{k}{n}\cdot \frac{1}{2}=\frac{k}{2n}$$
B. The player is a folower of $Y2$ probability $\frac{n-k}{n}$), and $Y2$ wins the last game against $Y1$ (i.e. $Y=Y2$; probability $\frac{1}{2}$). The probability of this happening is:
$$\frac{n-k}{n}\cdot \frac{1}{2}=\frac{n-k}{2n}$$
The probability of a random player being a follower of eventual winner $Y$ before $Y$'s last win is therefore:
$$\frac{k}{2n} + \frac{n-k}{2n}=\frac{n}{2n}=\frac{1}{2}$$
A: Consider the game at the state where only two players $X$, $Y$ are left. If your probability of following $X$ is $p$, then, the probability of following the winner at the end will be $$\frac p2+\frac{1-p}2=\boxed{\frac12},$$ depending on whether $X$ or $Y$ wins. This proves our result. $\blacksquare$
