Glass Bridge Game in Squid Game (Episode 7/ Game 5)

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, $$\mathbb{E}(S)=n-\mathbb{E}(D)=n-\sum_{k=1}^n kp_D(k)=7.$$

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.