Glass Bridge Game in Squid Game (Episode 7/ Game 5) Spoilers for Squid Game.
In the show, there is a game where people try to cross a bridge, made of $n=18$ rows of 2 side by side glass panes, which they must cross one row at a time. One glass pane can support a person while the other  will break, causing the person to fall and get eliminated. Each person must select which glass pane to jump onto, from one row to the next, and try to reach the other side without falling through. In the show, some people cross the same bridge later than others, so they can tell which of the steps already crossed are sturdy or not.
Assume that the selection of the sturdy and weak glass panes are random, that later players take the same steps that previous players took up to the point that they fall through (i.e. no forgetting which pane is sturdy or guessing again on an already solved pane). Ignore all human elements, like people trying to force others to fail to figure out future panes, or being able to tell the difference between tempered sturdy panes and weaker panes (i.e. guessing is random).
Given $n$ rows of glass panes, how many players would it take until there is a player with a $>50\% $ chance of crossing the bridge?
In the show there are 16 players and $n=18$ rows of glass, so what is the most likely outcome, in terms of number of people being able to cross the bridge?
 A: Observe that the person $k$ in line has a total advancement in the bridge distributed as $S_k+k$, where $S_k\sim \mbox{NB}(k,p)$, where $p=1/2$ is the correct tile selection probability. Now, let $n=16$ be the total number of people and $m=18$ the bridge length.
Then the probability of the $k$-th person traversing is
\begin{align*}
\mathbb{P}(S_k+k> m)= 0.407,\, 0.593,\quad \mbox{for}\quad k=9,\,10,\:\: \mbox{respectively},
\end{align*}
and thus player number $10$ is the first to have more than $50\%$ chance of traversing.
Now, define the random variables $$D=\mbox{number of dead people},\quad S=\mbox{number of survivors}.$$ Then observing that $S_{k+1}$ is the sum of $S_k$ and an independent geometric random variable, say $G$, we obtain the pmf of $D$ as
\begin{align*}
p_D(k)=\mathbb{P}(D=k)&=\mathbb{P}(S_k+k\le m,S_{k}+k +G>m)\\
&=\sum_{n=1}^\infty \left(F_{S_k}(m-k)-F_{S_{k}}(m-k-n)\right)p(1-p)^{n-1}.
\end{align*}
Finally, we get that for the above parameters,
$$\mathbb{E}(S)=n-\mathbb{E}(D)=n-\sum_{k=1}^n kp_D(k)=7.$$
A: You can think of the number of new rows that can be revealed (where revealing a new row is defined as 100% demonstrating the safe pane of a row) by a single player as a geometric distribution. For any player, the expected number of new rows to be revealed is $\frac{1}{\frac{1}{2}}=2$.
Using linearity of expectation, we can see that if $k$ people cross the bridge one by one, the expected number of rows to be revealed is $2k$.
Ignoring some edge cases, the average optimal number of players it takes will be about $\frac{18}{2}$. So we can say that approximately $7$ people will survive.
