Glass Stepping Stones in SQUID GAME: Mathematical Analysis (Probability Theory)

0:00 Introduction
0:51 Explaining the Rules of the Game
1:44 Probability of Player 1 Surviving
2:47 Probability of Player 2 Surviving
5:45 Probability of Player 3 Surviving
7:43 GENERALIZATION. Probability of Player k Surviving (Result at 11:53)
11:59 Calculating Each Player's Probability & Drawing the Graph
12:38 EXPECTED VALUE of Number of Survivors? (Result at 14:20)
In the famous Emmy-winning Netflix show Squid Game, there exists a Glass Stepping Stones game, where 16 players have to cross the glass bridge made of 18 pairs of glass panels. Can we actually calculate the exact probability of each player surviving the game? What would be the average number of survivors in this game? Here, I am going to provide mathematical analyses to this game and calculate such probabilities and expected number of survivors, and throughout the video you will encounter some basic topics in probability theory and statistics.
#SquidGame #BinomialDistribution #GlassSteppingStones #probability #ProbabilityDistribution #statistics
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[@sehr.geheim]

Very good, I especially liked how you also calculated the expected number of survivors

[@CornerstonesOfMath]

Thanks. Although that meant extra work for me, I'm glad that I did it :)

[@Smourbiff24]

I know it's been a while but I don't understand why at 15:00 the fact of adding all the probabilities gives you the average number of survivors...

[@CornerstonesOfMath]

Well, I'll try my best. I will use the same notations as in the video, where p_k = probability of player k surviving, Nfail = number of fails, p(Nfail=i) = probability where the number of fails is i, X = number of players surviving, and p(X=k) = probability of exactly k players surviving. It is explained in the previous part of the video that p_k = Σ(i=0 to k-1) p(Nfail=i) and p(X=k) = p(Nfail=16-k), which can be also thought as p(Nfail=i) = p(X=16-i). As a first explanation, you can just start from the sum of probabilities p_1 + p_2 + p_3 + ... + p_16 and show that the expression can be rearranged into the expected value of X: p_1 + p_2 + p_3 + ... + p_16 = p(Nfail=0) + [ p(Nfail=0) + p(Nfail=1) ] + [ p(Nfail=0) + p(Nfail=1) + p(Nfail=2) ] + ... + [ p(Nfail=0) + p(Nfail=1) + p(Nfail=2) + ... + p(Nfail = 15) ] = p(Nfail=15) + 2*p(Nfail=14) + 3*p(Nfail=13) + ... + 16*p(Nfail=0) = p(X=1) + 2*p(X=2) + 3*p(X=3) + ... + 16*p(X=16) = E(X) We can write the same thing more elegantly if we know how to use the double sigma notation (if you don't understand, it's just the same thing as above written differently). Since p_k = Σ(i=0 to k-1) p(Nfail=i), p_1 + p_2 + p_3 + ... + p_16 = Σ(k=1 to 16) p_k = Σ(k=1 to 16)Σ(i=0 to k-1) p(Nfail=i) (double sigma notation) = Σ(i=0 to 15)Σ(k=i+1 to 16) p(Nfail=i) (changing the order of sigmas) = Σ(i=0 to 15)Σ(k=i+1 to 16) p(X=16-i) (number of fails being i means number of survivors being 16-i) = Σ(i=0 to 15) (16-i)*p(X=16-i) (inner sigma simply gives (16-i) because p(X=16-i) doesn't depend on k) = E(X) Another way to explain it is by being less precise about the calculation and care more about the concepts. Here, the probability of total k players surviving p(X=k) has the property of probability mass function (PMF) which is discussed in statistics textbooks. But the probability of Player k surviving, p_k, does NOT have the same property (or dimension/unit) as PMF. p_k is expressed as the SUM of such probabilities, more precisely, p_k = Σ(i=0 to k-1) p(Nfail=i), or p_k = Σ(i=0 to k-1) p(X=16-i) hence p_k actually has the equivalent property (or dimension/unit) of the cumulative distribution function (CDF), which is the sum of probability mass function (CDF = Σ PMF). Not exactly the same as the CDF defined from p(X=k) (which is Σ(i=1 to k) p(X=i)), but quite "similar" to that. Therefore, if we add up p_k, it's equivalent to adding up CDFs, which can be thought as Σp_k = Σ CDF = Σ Σ PMF and one of two sigmas simply gives the term having the dimension/unit of people, so Σp_k = Σ PMF*(some variable having the dimension/unit of people, like X) = E(X). This is the best I can do. Hope it helps at least to some degree.

[@jwang3417]

Thanks for the excellent content!

[@CornerstonesOfMath]

Thanks! Amazing to see that my old content still gets attention.

[@mik4466]

Epic video!

[@CornerstonesOfMath]

Thanks! 👍

[@xxbendover6995]

nice video 👍

[@CornerstonesOfMath]

Thank you! Glad you enjoyed it :)