Squid Game bridge probabilities

Difficulty:   ★★★☆☆   undergraduate

In the popular Korean series Squid Game, one episode features a bridge-crossing game, whose probabilities are a fun challenge to calculate. (Warning: partial spoilers ahead.) In this cruel fictional scenario, called “Glass Stepping Stones” in the English subtitles, glass panels are suspended above a long fall. Contestants must leap between them. At each step the leading player is forced to choose left or right, knowing only that one panel is made of ordinary glass which will shatter, and the other is strengthened (“tempered”) glass which will hold. Later contestants cross the same bridge, and watch all previous attempts, so can learn the successes and failures.

The odds are simple for the first player. On each leap forward, there is a 50% chance they will fall to their death. Hence the chance of surviving N steps is 1/2^N, an exponential decrease. In the show (~30 minute mark), one player actually calculates this: 15 untested steps remain ahead of him, for a horrifyingly low 1/32768 chance of survival from that point. (Actually this is the third player, but more on that later.)

Squid Game bridge
The third contestant on the Squid Game bridge accurately calculates his chances as 1 in 32768

But suppose we do not know the outcome of earlier players. At the start, before anyone has moved, what is the probability a_{i,n} say, that player number i will still be alive on step number n? We showed a_{1,n} = 1/2^n. For player 2, it is certain they will survive step 1, by copying the first player if they were successful, or switching to the opposite pane if not. By extension player i is certain to survive the first i – 1 steps, hence a_{i,n} = 1 for all n \le i - 1.

In general, we set up a recurrence relation. But consider firstly the case i = 2. What is the chance they are alive at step n? If the first player died on step 1 (I mean, they leaped from the starting platform to an ordinary glass panel at step 1), then their successor must guess n – 1 tiles to reach step n successfully (I mean, to still be alive on panel n, and not fall through it). The probability of this combination of events is (1 - a_{1,1})/2^{n-1}. Similar reasoning applies to any step up to n – 1. However if the first player is still alive on n – 1, their follower is guaranteed to reach step n successfully. (Any later performance of the first player is irrelevant to their successors at step n.) The overall probability is the sum over these possibilities, which for an arbitrary player is:

    \[a_{i,n} = a_{i-1,n-1} + \sum_{k=1}^{n-1} ( a_{i-1,k-1} - a_{i-1,k} ) / 2^{n-k}.\]

This gives the probability in terms of the previous player. (Note the term in parentheses is the chance the previous player will die on step k precisely.) Hence starting from the initial conditions given earlier, we may build up an array of values using a spreadsheet, computer program, or computer algebra system. The latter choice preserves exact fractions, which feels very satisfying. Also we define a_{i,0} = 1 for convenience, where “step 0” may be interpreted as the ledge contestants safely start from. The Table below gives the first few values.

Table: Probability player i is still alive by step n, given no foreknowledge

n = 1

2 3 4 5 6 7 8

i = 1

1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256
2 1 3/4 1/2 5/16 3/16 7/64 1/16 9/256
3 1 1 7/8 11/16 1/2 11/32 29/128 37/256
4 1 1 1 15/16 13/16 21/32 1/2 93/256
5 1 1 1 1 31/32 57/64 99/128 163/256
6 1 1 1 1 1 63/64 15/16 219/256

In Squid Game the bridge has 18 (pairs of) steps. The probability of crossing the entire bridge is the probability of being alive on step 18, as the next leap is to safety. In theory the 9th player has nearly even odds of making it: a_{9,18} = 53381/131072 \approx 0.41, and the next player likely will: a_{10,18} = 77691/131072 \approx 0.59. In the show, 16 players compete in this challenge, so the last player has excellent odds, supposedly: a_{16,18} = 65493/65536 \approx 0.9993.  However our analysis does not account for human behaviour! In the show, time pressure, rivalries, and imperfect memory compete with logical decision making and the interests of the group as a whole. On the other hand, some players claim to distinguish the glass types by sight or sound, which would give an advantage. These make interesting plot elements, but would spoil the simplicity and purity of a mathematical analysis.

Returning to the recurrence relation, it simplifies to:

    \[a_{i,n} = \frac{a_{i,n-1} + a_{i-1,n-1}}{2}.\]

Hence each term is just the average of two previous terms. However I wanted to derive this via direct physical interpretation, not algebraic manipulation alone. This is intuitively satisfying. With the end result in mind, we relate P(i,n) \equiv a_{i,n} to the previous step and previous player. [Update, 21st April: A simpler way is to consider step 1. If player 1 guesses breaks it, there are i – 1 remaining players for the next n – 1 steps. If player 1 instead guesses correctly, there are i players for the next n – 1 steps. This gives the recurrence relation.] Consider the three cases for player i – 1: they (A) died before step n – 1, (B) died on step n – 1, or (C) made it safely to step n – 1 or further. The total probability is the sum of these cases:

    \[P(i,n) = P(i,n|A)P(A) + P(i,n|B)P(B) + P(i,n|C)P(C).\]

The first term for example is the “conditional probability” that i is alive at step n, given that case A occurred; times the probability of case A itself occurring. There is a similar decomposition to the above for P(i,n-1). Now most parts of the expression are straightforward. If i – 1 died at step n – 1, then the next player is definitely safe at that step, but may only guess at the following step, so P(i,n-1|B) = 1 and P(i,n|B) = 1/2. If i – 1 was safe at step n – 1 or further, then the next player is safe for an extra step: P(i,n-1|C) = 1 = P(i,n|C). For case A the conditional probabilities are more difficult, but we do not need to calculate them. Observe that if the previous player died before n – 1, then steps n – 1 and n are uncharted territory. Hence the chance the following player makes it to n safely, is half of whatever it was for them to reach n – 1 safely: P(i,n|A) = \frac{1}{2}P(i,n-1|A). Hence the decomposition becomes:

    \[P(i,n) = \frac{1}{2}P(i,n-1|A)P(A) + \frac{1}{2}P(B) + P(C).\]

But this is just \frac{1}{2}P(i,n-1) apart from the C term, as seen from expanding out the conditional cases. Now P(C) = P(i-1,n-1) \equiv a_{i-1,n-1}. It follows P(i,n) = \frac{1}{2}(P(i,n-1) + P(i-1,n-1)) as before. We did not need to evaluate P(A) or P(B), though this is straightforward.

Now that the reader (and author!) have more experience with conditional probability, let’s return to the third player in the Squid Game episode. Before anyone moved, he had chance a_{3,18} = 43/65536 \approx 1/1500 of surviving the bridge. This would seem to contradict the earlier calculation, which gave a lower chance by a factor of 21½, a surprising contrast! The black-masked “Front Man” said to the VIP observers, “I believe this next game will exceed your expectations” (~12:30 mark), but in this sense it did not 😆 . The distinction is the information learned. Conditional probability is a subtle and beautiful thing. If we know nothing about the previous attempts, nor the state of the bridge, the probabilities are our variables a_{i,n}. But if we are given the information player I died on step N for instance, then the following player has no information about later steps, and the bridge scenario is essentially reset from that point onwards. Hence P(3,18|2\textrm{ died on }3) = a_{3-2,18-3} = 1/2^{15}.

This scenario has been a valuable learning experience, as I had not worked with conditional probabilities before. Probability is very important in physics, particularly quantum physics where it is intrinsic (it is usually assumed). I originally came up with an incorrect recurrence relation, but realised this upon comparison with an article in Medium , which uses an elegant combinatorial argument. The scenario had already captured my attention, but realising my flaw drove further my need to understand. A related article  is also helpful; I recommend these if you find my discussion hard to follow. There is even a draft paper  on the Squid Game bridge probabilities! Presumably this is all little more than a specific application of textbook combinatorics. Still, it is fun to rediscover things for oneself.

🡐 ⸻ | Squid Game bridge | general solution 🠦