Squid Game conditional probabilities

Difficulty:   ★★★☆☆   undergraduate

Earlier we analysed the probabilities for the bridge-crossing scenario in the Squid Game episode “VIPs”, which has “deadly high stakes” according to the Netflix blurb for the series. 🙂 So far, we made the assumption of no foreknowledge. This means our results for the players’ progress describe their chances as they stand before the game begins. Equivalently, if the game has started, our results assume the analyst knows nothing about prior contestants, and cannot view the state of the bridge.

But now, suppose we are told only that a specific player numbered i died on step number n. (That is, they stood safely on column n – 1, but chose wrongly amongst the next pair of glass panels on column n, breaking a pane and plummeting downwards.) Then the next player is definitely safe on step n, but has no information about later steps, so the game is essentially reset from that point. Hence the “conditional probability” that player I > i is still alive on step N > n is simply:

    \[P(I,N|i\textrm{ died on }n) = a_{I-i,N-n}.\]

Recall a_{i',n'} \equiv P(i',n') is the chance player i′ is alive on step n′ (given no information nor conditions). We labelled as b_{i',n'} = \binom{n'-1}{i'-1}2^{-n'} the chance they died on step n′ specifically, so analogously:

    \[P(I\textrm{ dies on }N|i\textrm{ died on }n) = b_{I-i,N-n}.\]

Now, suppose we are told only that a specific player I will die on step N. What is the probability for an earlier player’s progress? Bayes’ theorem says that given two events A and B, the conditional probabilities are related by P(A|B) = P(B|A)P(A)/P(B), which in our case is:

    \[\begin{aligned}c_{i,n} &:= P(i\textrm{ died on }n|I\textrm{ dies on }N) \\ &= \frac{b_{I-i,N-n}b_{i,n}}{b_{I,N}} \\ &= \frac{\binom{N-n-1}{I-i-1}\binom{n-1}{i-1}}{\binom{N-1}{I-1}}.\end{aligned}\]

The powers of 2 cancelled. The Table below shows some example numbers.

Table: Probability player i died on step n, given player I = 5 will die on step N = 8

n = 1

2 3 4 5 6 7 8

i = 1

4/7 2/7 4/35 1/35 0 0 0 0
2 0 2/7 12/35 9/35 4/35 0 0 0
3 0 0 4/35 9/35 12/35 2/7 0 0
4 0 0 0 1/35 4/35 2/7 4/7 0
5 0 0 0 0 0 0 0 1

In general, on any given row (fixed player i) the entries are nonzero only for n between i and N – I + i inclusive. This forms a diamond shape. For the row sum \sum_{n=i}^{N-I+i}c_{i,n} computer algebra returns a hypergeometric function times two binomial coefficients, which appears to simplify to 1 (for integer parameters) as expected, since player i must die somewhere. On any given column \sum_i c_{i,n} = (I-1)/(N-1) which is independent of n, meaning each step has equal chance that some player will die there. In particular the first entry itself takes this value: c_{1,1} = (I-1)/(N-1).

We examine other properties and special cases. By construction the last row and column are zeroes apart from c_{I,N} := 1; our general formula does not apply for n = N. If we are told where the second player I = 2 died, then player i = 1 has an equal chance 1/(N – 1) of dying on any earlier step. Also from the definition it is clear:

    \[c_{i,n} \equiv c_{I-i,N-n},\]

so the table is symmetric about its central point. The ratio of adjacent entries follows from the binomial coefficients:

    \[\begin{aligned}\frac{c_{i-1,n}}{c_{i,n}} &= \frac{(i-1)(N-I-n+i)}{(I-i)(n-i+1)}, \\ \frac{c_{i,n-1}}{c_{i,n}} &= \frac{(N-n)(n-i)}{(n-1)(N-I-n+i+1)}.\end{aligned}\]

It follows that at step n, player i = (I – 1)n/N and the subsequent player have the same “fail” chance. Presumably the maximum lies within this range. Physically we require the indices i and n to be integers. For the chosen Table parameters above, the relation just given is simply in/2, so every second column contains an adjacent pair of equal values. For the steps (columns) on the other hand, on n = (i – 1)(N – 1) / (I – 2) and the following step the “elimination” chance is equal. Note these special index values are linear functions of the other index (i or n respectively), where we regard I and N as fixed.

By rearranging terms we can write equivalent expressions for the chance to be eliminated, such as:

    \[c_{i,n} \equiv \frac{i\binom{N-I}{n-i}\binom{I-1}{i}}{n\binom{N-1}{n}}.\]

conditional probability in Squid Game
The probability Squid Game bridge contestants will expire on a given step, given the condition: player I = 9 will die on step N = 25. It forms a sort of boat- or saddle-shape. The blue dots are for integer index values, which are physical. In previous results the “ridge line” of high probability was at roughly n = 2i, but for the conditional probability it is spread out between the endpoint events, so in this case is roughly n = 3i.

For suitably large parameters, the probability resembles a gaussian curve. We can apply the de Moivre-Laplace approximation (with parameter p := ½ say) to the binomial coefficients. This gives a gaussian for a fixed step number n, as a function of the player number. I omit the height, but its centre and width are determined from the exponent which is:


The spread is maximum at n = N/2, in this approximation. Now to obtain a gaussian approximation for a fixed player i, apply the results of the previous blog post using the substitutions x \rightarrow n-1, a \rightarrow i-1, b \rightarrow I-i-1, and X \rightarrow N-2. The centre is n_0 := x_0 + 1 = (i-1)(N-1)/(I-2)+1/2. One option for the height of the gaussians — when looking for a simple expression — is to use the sums 1 and (I – 1)/(N – 1) determined before. Recall for a normalised gaussian, the height 1/\sqrt{2\pi}\sigma is inversely proportional to the standard deviation.

There are other conditional probability questions one could pose. Suppose we are given a window, bounded by the events that player J died on column L, and later player K dies on column M? Inside this window, the probabilities reduce to our above analysis: the chance i dies on n is just c_{i-J,n-L}, where we also substitute I \rightarrow K-J and N \rightarrow M-L. As another possible scenario to analyse, we might be informed that player I is alive on step N. Then we would not know how far they progressed, just that it was at least that far. Or, we might be told player I died on or before step N.

A concluding thought: Bayes’ theorem is deceptively simple-looking. I tried harder ways beforehand, trying to puzzle through the subtlety of conditional probability on my own. But with Bayes, the main result followed easily from our previous work.

🡐 asymptotics | Squid Game bridge | ⸻ 🠦

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