If we assume everyone is a perfect logician the expected value of the game is strictly non-positive for all players; no agent can expect to gain from playing the game. I think this is a reasonable prediction from the model because I’ve literally never seen this game played in real life.
For those who are unfamiliar, `Big bank take little bank’ is a game in which two people compare wallets and whoever has more cash in their wallet takes the cash of the poorer wallet. If I have $10 in my wallet and my friend Bill has $20, Bill would take my $10.
I first became interested in the game in 2017 while Rap Geniusing “Swang” by Rae Sremmurd. In his verse, Swae Lee sings “Big Bank take little bank // Every day pouring up drank” (Lee & Jxmmi, 2017). After learning the rules, I was curious at what the optimal entry point would be for someone looking to play the game. That is, given some distributional assumptions, at what point is the expected value of playing the game positive.
It is clear that if you are in a room of people and you think that you have the least money in the room, you should never approach someone to play the game since your expected value would always be negative (you would always lose). In contrast, if you believe that you have the most money in the room, then you should always play the game (you would always win). My question is at what point does your decision to play switch. I assume that your utility is given by
\[ U(Game) = E(Winnings) + \theta\]
where \(E(Winnings)\) is the expected amount of winnings from playing the game and \(\theta\) is some pleasure parameter that signifies how much enjoyment you get from playing the game. If you are a risk-seeker then \(\theta\) is positive whereas if you are risk-averse \(\theta\) is negative. Before moving into calculating the optimal entry point I state the assupmtions of the model
Anyone you ask to play the game will agree to play the game. I will discuss at the end what happens when this assumption is relaxed, but for now it is simpler not to have to consider your oponents decision to play or not.
There are infinite players. Again, this is mainly to simplify the model and allow us to use continuous probability distribution functions. I will discuss at the end how strategy changes in the finite case. The general gist is that I don’t think it changes too much.
You have knowledge of the distribution of cash in people’s wallets. This sounds like a stronger assumption than it is. You could reasonably conduct a survey and experimentally determine this distribution for the population at large.
Notationally, I use \(f(x)\) to denote the probability distribution function of cash in people’s wallets and \[F(x) = \int_{-\infty}^x f(x) dx\] to denote the cumulative distribution function. \(F(x)\) represents the percentage of people who have less than \(\$x\) in their wallets. In the next section I calculate the expected value of the game and use this to calculate the optimal entry point.
Given that you have knowledge of the distribution of cash in people’s wallets, anyone you ask to play the game will play, and you know you have \(\$x\) in your wallet, expected value is given
\[\begin{align*} E(\text{Play the game}| \text{Cash} = x) &= p(\text{winning} | \text{Cash} = x) \cdot E(\text{winning} | \text{Cash} = x) - (1 - p(\text{winning} | \text{Cash} = x))\cdot x\\ &= \int_{-\infty}^x f(x) dx \cdot \int_{-\infty}^x xf(x) dx - x(1 - \int_{-\infty}^x f(x) dx)\\ &= F(x)\int_{-\infty}^x xf(x) dx - x + xF(x) \\ \end{align*}\]The condition for playing the game is then
\[\begin{align} U(Game) &\geq 0 \\ 0 &\leq F(x)\int_{-\infty}^x xf(x) dx - x + xF(x) + \theta \\ x &\leq \frac{F(x) \int_{-\infty}^x xf(x) dx + \theta}{1 - F(x)} \end{align}\]This last inequality gives us the general answer for any cash distribution! Finding out the specific entry point for the game would just be a matter of solving the integrals and isolating \(x\). The cumulative distribution functioncan easily be found online for any common distribution, so solving this inequality wouldn’t requite more than a few minutes of algebra and some simple integration. As an example I solve below under the assumption that cash in peoples wallet is distributed under a uniform distribution with minimum 0 and maximum 1. This is the case where everyone has between 0 and 1 dollar in their wallet and each amount in between the two is equally likely.
In this case we have that \[ f(x) = \{\begin{array}{lr} 1, & \text{for } 0\leq x\leq 1\\ 0, & \text{otherwise }\\ \end{array} \]
so that \(F(x) = x\) for \(0 \leq x \leq 1\) and \(\int_{-\infty}^x xf(x) dx = \frac{x^2}{2}\). In this case our entry condition from the last section becomes
\[\begin{align*} 0 &\leq \frac{x^3}{2} - (1 - x)x + \theta \\ &\leq x^3 + 2x^2 - 2x + 2\theta \end{align*}\]I ignore \(\theta\) for now and apply the quadratic formula to get that the part dependent on \(x\) (\(x^3 + 2x^2 - 2x\)) is greater than \(0\) when \(x \geq \sqrt{3} - 1 \approx 0.7321\). This means that our full entry condition is approximately
\[x + \theta_1 \geq 0.7321\]
where \(\theta_1 = 2\theta\). If you are risk-nuetral this means that (under our assumptions) you have a positive expected value whenever you have more than 0.7321 dollars in your wallet. This may sound like a useless result, but given that all uniform distributions are equivalent under linear transformation, we can generalize and say that if you assume that the distribution of cash in people’s wallets is described by a uniform distribution, your expected value of playing the game is approximately positive if you are in the 73rd percentile of income!
In sections 3 and 4 I offer a general solution for when you should play the game big bank take little bank. Equation (3) tells us that the expected value is generally positive when
\[x \leq \frac{F(x) \int_{-\infty}^x xf(x) dx + \theta}{1 - F(x)}\]
whereas section 4 applies this result to the uniform distribution to show that the expected value of the game is positive in this case if you believe you have more cash in your wallet than \(73.21\%\) of the population. I’ll now spend a little time discussing what happens when the model assumptions are relaxed.
My first assumption was that anyone you ask to play the game will agree to play the game. This was important in deriving the results above since the other person agreeing to play the game didn’t give us any new information about the amount of cash in their wallet. If we relax this assumption and assume that the all players are perfect logicians, that means that if you ask someone to play the game and they agree, you know that they also (in the uniform case) have an income above the 73rd percentile. This reduces your probability of winning and your expected value and so the entry point is higher. The other person also incorporates this and also has a higher entry point and this cycle presumably continues until the only people with non-negative expected value are the people who have the most money. They would then only play with themselves and no one would ever win any money. I think then, that if we assume everyone is a perfect logician the pure expected value of the game (ignoring the ‘fun’ parameter \(\theta\)) is strictly non-positive for all players and no one would ever play the game. I think this is a reasonable prediction from the model because I’ve literally never seen this game played in real life.
My second assumption was that there are infinite players. If we have finite players, I would assume you could approximate at the optimal entry point with the solution from the infinite player case. For a specific answer you could probably use order statistics to approximate at what rank you are in the income distribution and then use that to calculate the expected value of playing the game. This is a question for the future.
Thanks for sticking around through this post. If you find an error here feel free to let me know. This is an initial run at a problem and it’s possible I’ve made a mistake.