Pascal's Wager and Squid Game (TW: Math)
Hello,
I find that the core themes of the Squid Game franchise revolves around a fundamental philosophical question: is it better to be content with a certain, short-term gain, or is it worth taking a major risk for a much-less likely long-term gain? This is deconstructed the best in the first episode with the Bread and Lottery scene: each homeless person is given the choice between a certain small gain (bread), or a chance at a large gain (lottery ticket). Every single one of them (save one) chose the latter, despite the overwhelming odds stacked against them. The Squid Game itself is a more complicated version of this same idea: people can walk away with their own lives spared (or with some extra cash per S2 rules), or they can take the risk of getting 45.5 billion Won (about 32 million dollars), despite the overwhelming odds that it will get them killed.
This philosophical question can be debated back and forth across many different dimensions. But in terms of mathematics, there is a very straightforward and objective algorithm to make this kind of decision from a utilitarian perspective, called Pascal's Wager. Pascal's Wager is hardly a new concept, but it was invented by French mathematician Blaise Pascal in the 1600s.
The idea of Pascal's Wager is that you take the potential reward from each possible decision, then multiply it by the probability of achieving that reward. That is known as the expected outcome. Then, the most optimal choice is the one that yields the highest expected outcome.
For example, suppose you are given a choice between two envelopes: A and B. Envelope A has a 10% chance of having $1,000, and envelope B has a 50% chance of having $500. The expected outcomes from each decision goes as follows:
- Envelope A: $1,000 * 0.1 = $100
- Envelope B: $500 * 0.5 = $250
Because $250 < $100, then envelope B is the correct choice.
Now let's apply this algorithm to the Bread and Lottery game: Google tells us that the probability of getting a winning lottery ticket is 1 out of 292 million. The price of a loaf of bread in Korea is about $15. Since the bread has no catch to it, then the probability is 1 (100% chance).
- Lottery ticket: $1,000,000 * (1/292 million) = $0.0034 (or 0.34c)
- Bread: $15 * 1 = $15
So the bread is the obvious choice. It doesn't matter that there is a chance of getting 1 million dollars, the expected outcome of the lottery ticket is less than a penny, due to how low the probability is.
Now let's apply this logic to the Squid Game itself. It would take a long time to calculate what the probability is for surviving all six games, but we can imagine it must be extremely small. Take the glass bridge for example: you have to make a 50-50 guess to cross each pane of glass, meaning the probability of surviving is 1/2. But since there are 18 panes of glass in total, then the probability of getting all of them right is 1/(2^18) = 0.0000038. That's not very good odds!
Using this probability against the final prize money, the expected outcome comes out to: 45.5 billion Won * (1/2^18) = 173,000 Won (or about $120). Not exactly worth risking your life over. And remember, This is only the probability for one out of six total games in the show, which makes the probability way smaller.
However, this basic analysis assumes that the players don't lose anything once they choose to walk out of the show, and we know that isn't the case. The game specifically recruited people who were deep in insurmountable debt, typically ranging from 2-6 billion Won. This makes them much more desperate, and argues that they would rather risk their life making money rather than going back to face their loan sharks. But is that really true?
The amount of debt is drastically different between different players, as is their varying degrees of holding value to their life. We see some contestants like Oh Il-Nam and Ji-Yeong are willing to throw their life away for the greater good, while others cling to it desperately. Thus, for a more precise analysis we should consider these factors as variables.
Let:
- D = amount of debt the player starts the game with
- X = monetary value of the player's life
- P0 = probability of the player being eliminated in any of the six games
Then the player should keep playing Squid Game as long as the following expression is true:
- 1) -D < (45.5 billion Won)*(1-P0) - X*P0
In other words, the player should keep playing as long as the expected outcome from the prize money is greater than their debt, after deducting the expected loss from risking their life (X*P0).
Since we expect that winning the game is very unlikely, then we can assume that P0 approaches 1. This would reduce the expression to:
- 2) -D < -X
In other words, it wouldn't really matter how high the prize money is. If the player's debt (D) is greater than the value of their life (X), then they are literally worth more dead than alive. But barring cases of existential nihilism (like Oh Il-Nam, since he has a brain tumor and has nothing left to lose), we can assume that shouldn't be the case.
But what if we factor in S2 rules? In this case, the players don't leave empty-handed if they choose to stop the game, but rather they get to keep all the winnings accumulated up to that point. Since this prize money will increase every round, then we should change our analysis to consider whether the player should quit per round, rather than the game as a whole.
Let:
- D = amount of debt the player starts the game with
- X = monetary value of the player's life
- P(n) = probability of the player being eliminated in the nth round
- M(n) = gross winnings of the player if they quit in the nth round
Then, in the nth round, the player should keep playing Squid Game as long as the following expression is true:
- 3) M(n) - D < M(n+1)*(1-P(n+1)) - X*P(n+1)
With some basic assumptions, we can refine this expression considerably. The rules of the game clearly defines what M(n) is going to be, given the number of players who have been eliminated. For each player eliminated, the prize money increases by 100 million Won (about $70,000), which would then be evenly distributed to all the remaining players. Knowing that there are 456 players to start with, then the equation for M(n) can be given as follows:
Let:
- C(n) = cumulative number of players that have been eliminated as of the nth round
- M(n) = gross winnings of the player if they quit in the nth round
Then:
- 4) M(n) = (C(n) * (100 million Won))/(456-C(n))
Next, we need to find a more refined way to define C(n). Let us assume that there is an even probability distribution across all the players, which means that every player has about the same probability of being eliminated in each round. In reality, some players have a significance advantage over others in different challenges, as they vary considerably by age, gender, height, weight, etc. But just for the sake of argument, let's just assume that the probability of being eliminated is about the same for everyone on average.
This would mean that the probability of any individual player being eliminated in the nth round is equal to P(n) (the same probability that we, as a player, can be eliminated). Thus, the expected number of players that are eliminated in the nth round is equal to (456-C(n-1))*P(n). With all that information, a recursive definition for C(n) can be constructed like this:
Let:
- C(n) = cumulative number of players that have been eliminated as of the nth round
- P(n) = average probability that any individual player is eliminated in the nth round
Then:
- 5) C(n) = C(n-1) + (456 - C(n-1)) * P(n) [NOTE: C(0) = 0]
Using this definition, we can combine it with equation (4) to create a new definition for M(n+1):
Let:
- C(n) = cumulative number of players that have been eliminated as of the nth round
- M(n) = gross winnings of the player if they quit in the nth round
- P(n) = average probability that any individual player is eliminated in the nth round
Then:
- 6) M(n+1) = ((C(n) + (456 - C(n)) * P(n+1)) * 100 million Won) / (456 - C(n) - (456 - C(n)) * P(n+1))
That seems like a monstrous equation, but we can reduce it with a bit of algebra. Remember that we can use the definition in equation (4) to substitute in:
- 7) M(n+1) = ( M(n) + (P(n+1) * 100 million Won) ) * (1/(1-P(n+1)))
Now we can substitute this equation into expression (3). Remember that (1/(1-P(n+1))) cancels out when multiplied by (1-P(n+1).
Let:
- D = amount of debt the player starts the game with
- X = monetary value of the player's life
- P(n) = average probability that any individual player is eliminated in the nth round
- M(n) = gross winnings of the player if they quit in the nth round
Then, in the nth round, the player should keep playing Squid Game as long as the following expression is true:
- 8) M(n) - D < M(n) + P(n+1) * (100 million Won) - X * P(n+1)
Now we see that M(n) is on both sides of the expression, so it can be canceled out. The expression then reduces to:
- 9) - D < ((100 million Won) - X) * P(n+1)
Now similar to expression (2), we can assume that the games are so difficult to beat that P(n+1) approaches 1. This now reduces the expression to:
- 10) -D < (100 million Won) - X
In a rather surprising turn of events, we stumble across the true reason why players are so incentivized to keep playing in S2, much more than in S1. This seems really counterintuitive, since the players are now allowed to keep some money when they leave, so one would assume it would be incentive for them to cash out early. But because the games are so deadly, then it becomes extremely likely that some players are going to be eliminated, which pretty much guarantees that the pot of winnings will keep increasing. As long as the risk on the player's life isn't too high (X*P(n+1)), then they always have some incentive to keep playing (100 million Won * P(n+1)).
But this is where the confidentiality of Squid Game is so insidious. Because the players don't know what the next game is, then they have no idea what the value of P(n+1) is going to be, thus making it impossible to make an informed decision based on the equations above.