Suppose we have a coin flipping game involving $n$ players. In each round everyone still playing flips a fair coin, and the players whose coin comes up tails are eliminated. The game continues until at most one player is still alive, and they are declared the winner.
Now, it is possible that the game does not end with a winner (e.g. if $n=2$ and both players get tails on their first flip). Let $f(n)$ denote the probability that a game with $n$ players has a winner. We have $f(0)=0$ and $f(1)=1$. For $n>1$ it follows from considering the binomial distribution that
$$f(n) = \sum_{k=0}^{n}\frac{\binom{n}{k}}{2^{n}} f(k) $$
(Here $\binom{n}{k}/(2^n)$ represents the probability $k$ players survive the current round), which can be rearranged as
$$f(n) = \sum_{k=0}^{n-1} \frac{\binom{n}{k}}{2^n-1} f(k)$$
Using this formula we can compute $f(n)$ recursively.
$$\begin{array}{cc} n & f(n) \\ 0 & 0 \\ 1 & 1 \\ 2 & 2/3 \\ 3 & 5/7 \\ 4 & 76/105 \\ 5 & 157/217 \\
\vdots & \vdots \\
20 & 0.7213 \end{array}$$
The sequence of numerators doesn't seem to be in OEIS, nor does the sequence $a_n=f(n)(2^n-1)(2^{n-1}-1)\dots(3)(1)$ from clearing all the denominators in the recursion.
Is there a way of analytically determine the limit (if it exists) of $f(n)$ as $n$ goes to infinity? It seems from calculation to be about $0.7213$, though I'm not confident in digits beyond that due to error propagation as the recursion continues.
...represents the probability 𝑘 players survive the current round...
it seems that you confused the round of flipping with the "round" of recurrence regarding a change in the remaining number of players. $\endgroup$ – Lee David Chung Lin Aug 15 '19 at 4:13