Best queue position on a glass tile bridge

Ten players stand in a fixed order to cross a bridge of $10$ steps. Each step consists of a left and a right panel, exactly one of which holds weight; which one is a fair coin flip, independent across steps, and the panels look identical. The player at the front chooses a panel at each step, continuing until they either fall through (and are out) or reach the end. Whenever the leader falls, the next player in line takes over from that point, already knowing every panel revealed so far. The first player to reach the far side wins. If $p_i$ is the probability that the $i$th player in line wins, what is $\max_i p_i$?

$\tfrac{63}{256}$. Each step is guessed blind exactly once, by whoever first reaches it, and that guess fails with probability $\tfrac12$. So the total number of eliminations is $\text{Bin}(10,\tfrac12)$, and player $i$ wins exactly when there are $i-1$ eliminations: $p_i=\binom{10}{i-1}/2^{10}$. The maximum is at $i=6$: $\tfrac{252}{1024}=\tfrac{63}{256}$.