In the previous post we argued that volatility in a slot economy is determined in part by the casino operator when they set the terms of free, earned and purchased credits. In particular, it is important to understand how much time and effort managers put into the earned credits component, and to put into clear terms why they are doing it. It turns out they are correct to do so, and this article will begin to build a theory that can help organize how they go about this.
Previously when we looked at the earned credits component of the economy accounting identity, we assumed that the difficulty to complete was low or that players would complete the challenge. This simplification helped us focus on the main result that although there are many combinations of reward and difficulty that result in the same return to player for the individual from that feature, there was only one combination consistent with equilibrium given casino rake. The obvious reality though is that only a portion of the players will actually complete a feature. Predicting and measuring this proportion is our aim. We begin with an explicit form of earned credits.
EARNEDt(Bt,Bt-1) = k*Bt-1*I(Bt>=(1+c)*Bt-1), (1)
We are writing out the word EARNED because we are going to also use the mathematical expectation operator, E[...], and we we want to avoid confusion. I(Bt>=(1+c)*Bt-1) is the indicator function which is equal to 1 when Bt >= (1+c)*Bt-1 and is 0 otherwise. In words it guarantees that earned credits are 0 unless the player bets (1+c) times yesterdays bets, in which case they are awarded with k times yesterday's bets.
So what on average do we expect the players to earn (i.e. we take the expectation E[...])?
E[ EARNEDt(Bt,Bt-1) ] = E[ k*Bt-1*I(Bt>=(1+c)*Bt-1) ]. (2)
And by linearity of the expectation operator,
E[ EARNEDt(Bt,Bt-1) ] = k*Bt-1*E[I(Bt>=(1+c)*Bt-1) ]. (3)
Now we apply one more trick, which is probably one of the deeper and more useful results from probability and measure theory. The expectation of an indicator function is identical to the probability of that event.
So,
E[ EARNEDt(Bt,Bt-1) ] = k*Bt-1*P(Bt>=(1+c)*Bt-1). (4)
Our basic algebra combined with a little probability theory gives us a precise statement. We actually do need to understand the chance that players complete the challenge in order to manage the economy and keep it in equilibrium. Arguing as before, we fix return to player of the feature at 2%,
k/(1+c) = 2%, (5)
and combine it with the economy accounting identity in equilibrium to get,
k*P(Bt>=(1+c)*Bt-1) = c * rake. (6)
(See "How to Tune a Feature" to make these steps clearer).
Instead of two equations in two unknowns, k and c, to determine equilibrium and return to player as we did when we assumed the player always completed, now we must estimate the probability that the player will or can bet that much. Plugging in different guesses for P shows how substantially different the answers you might get are when compared to the naïve assumption of completion.
Luckily we have some ideas about P(Bt>=(1+c)*Bt-1). The first is that it is clearly a function of c. Intuitively as c gets larger it is harder to complete the feature and P decreases. Graphically we might assume something like the following.
The average player is pretty much guaranteed to complete so long as expected free and paid coins can cover the ask of (1+c) times yesterday's bets, Bt-1, after which point it begins to drop off. But how fast? and how do we account for different size bets? Those are the interesting and difficult questions which require brave and concrete models. Let's show how to proceed with the relevant variables and make a very strong assumption to show what is at the heart of the question.
P(Bt>=(1+c)*Bt-1) = P(St*avg(Bt)>=(1+c)*St-1*avg(Bt-1)), (7)