What is the Optimal Strategy for the Glass Game in the Squid Game?

This Article includes spoilers for the Squid Game Series

After watching the 7th episode of the Series Squid Game, I couldn’t help but try to think of an optimal strategy they could have used to maximize the number of survivors.

The Game played in this episode is the Glass Game

Players were asked to pick an apron with a number on them before the game, from 1–16, it was not revealed until a few minutes after picking that the numbers represent the order in which the players would pass through the bridge. At the start of the game, each of the sixteen remaining players stands at the opposite end of a gigantic room suspended several hundred feet above the ground. Between the entrance and the exit of the room are two “bridges” of side-by-side glass panels, each with 18 panels across. In order to win, players must cross the bridge to the other side of the room within 16 minutes. The catch: each of the panels between the two bridges is made of one of two types of glass: tempered glass, which can withstand the weight of at least two players, and regular glass, which will shatter on impact and send the player plummeting to their death. To make matters worse, even if you somehow manage to make it across to the opposite side, you’ll be pelted by a shower of splintered glass shards that will explode once the time limit is reached.

At the end of the game, only 3 of them survived with one fatally injured. The aim of this article is to try to find a reasonable optimal strategy they could have employed to save the highest number of people. The solution has to be reasonably straightforward such that they could have thought about it without any need for a strong statistics background. We will also account for the 50/50 choice at the beginning. So, given the parameters, let’s dive in.

The Correct pattern of Tempered Glass in the Game was:
L R R R R L R R L R R R L L L R R L

Strategy 1: Switch the last correct position

If the Last Tempered Glass was R, the next choice will be L.

First choice is L:

L R L¹
___ R L²
_____ R L³
_______ R L R L⁴
____________ R L R L⁵
__________________R L⁶
____________________R L R⁷
________________________L R⁸
__________________________L R L⁹
______________________________R L

In this case, 9 people die. 7 of 16 survive.

First Choice is R:

R¹
L R L²
___ R L³
_____ R L⁴
_______ R L R L⁵
____________ R L R L⁶
__________________R L⁷
____________________R L R⁸
________________________L R⁹
__________________________L R L¹⁰
______________________________R L

In this case, 10 people die. 6 of 16 survive.

Strategy 2: Stick to the last Correct option

If the Last Tempered Glass was R, the next choice will be R until it is wrong.

First Choice is L:

L L¹
__R R R R R²
__________L L³
____________R R R⁴
________________L L⁵
__________________R R R R⁶
________________________L L L L⁷
_____________________________R R R⁸
_________________________________L

In this case, 8 people die. 8 of 16 survive.

First Choice is R:

R¹
L L²
__R R R R R³
__________L L⁴
____________R R R⁵
________________L L⁶
__________________R R R R⁷
________________________L L L L⁸
_____________________________R R R⁹
_________________________________L

In this case, 9 people die. 7 of 16 survive.

Conclusion:

The most important Strategy is to not panic though. The strategies above make some assumptions about the mental state of the players and the memory. In Real Life, things like players refusing to go forward would definitely change any strategies you have.

Other variants would be to try to keep an equal number of both options. So if there are already 3 Rs and 2 Ls, you’d go with L. Or just sticking with an Option throughout the game, while this would yield a great result if your chose right, they’d be in trouble if they chose wrong.

What do you think, any other good strategies?