1
$\begingroup$

Suppose you are presented with the following game:

Game Interface

How the game works:

  1. Player places bets on one or more symbols.
  2. The wheel is spun.
  3. If the wheel lands on a symbol that the player has placed a bet on then the player wins. Winnings are calculated by multiplying the bet of the symbol that landed by the multiplier value displayed together with the symbol on the wheel.

What I'm having difficulty with is developing a formula that relates the means of each symbol so that the overall mean can be calculated. Why are the means related: Because the wheel can only fall on a single symbol in a round but bets can be placed on each symbol separately.

What I've come up with so far is as follows:

Suppose we only consider 1 symbol at a time and treat all other symbols as loosing positions, we can compute its mean (aka it's theoretical mean or return to player) as follows:

$$ \overline{x}=\frac{1}{m}\sum f_ix_i$$

Where $\overline{x}$ is the mean, m is the number of possible positions, f is the number of position for a specific symbol (its probability) with multiplier of x (its 'weight').

Thus the mean for each symbol individually would be:

  • $\overline{x_{apple}} = \frac{1}{16} (3\times6 + 3\times3) = \frac{27}{16} = 1.6875$

  • $\overline{x_{banana}} = \frac{1}{16} (1\times20 + 1\times5 + 2\times3) = \frac{31}{16} = 1.9375$

  • $\overline{x_{banana}} = \frac{1}{16} (1\times15 + 2\times5 + 3\times3) = \frac{17}{8} = 2.125$

I've simulated the above 3 scenarios with 10000000 rounds using the formula below to calculate the actual mean:

$$ \overline{x}=\frac{1}{n}\sum \frac{w_i}{b_i}$$

Where n is the total number of rounds, w is the amount won in the round and b is the amount that was bet for the round.

Two things came of this:

  1. The actual mean showed convergence to the theoretical mean.
  2. The value of the bet unintuitively didn't play a role in determining the outcome as many runs of the simulations were done for each symbol with different bet values. Some runs were even done with random bet values using a RNG.

Note: The game presented above is a simplified model of game that I'm working on, however the solution may provide insight for others on how related means affect the overall mean.

Please help I've been stuck on this for days now. Any guidance ideas or formulas will be highly appreciated.

$\endgroup$
  • $\begingroup$ In your calculations it is not clear what you identify as "the bet of the symbol" and what you identify as "the multiplier value" for the symbol. In the picture I see four banana slices with different numbers (weight? bet?) on each slice. It is also not clear how the weight/bet placed on one banana slice is relevant if the wheel lands on another banana slice. In that sense, why do we care if the picture shows a banana at all? $\endgroup$ – Michael Jan 22 at 0:38
  • $\begingroup$ Thank you @José Carlos Santos I've updated the question with the proper formatting. $\endgroup$ – MSD Jan 22 at 0:46
  • $\begingroup$ @Michael The different number in the slice on the wheel is the weight(multiplier) of landing in that slice for that symbol. The bet isn't placed on a specific slice in the wheel, the bet is placed on the symbol and if the wheel lands on a slice with the symbol that has a bet greater than 0 then the winning value will be the bet on the symbol multiplied by the weight in the slice. Eg. You bet 1 on Apple and 2 on Banana, the wheel in spun and lands on the slice with banana and a multiplier of 20, you win 2*20 = 40 $\endgroup$ – MSD Jan 22 at 1:31
  • $\begingroup$ Then if we place our bets and if $V$ is the random amount we win, we get $E[V] = \frac{1}{16}\sum_{i=1}^{16} b_i w_i$ where $b_i$ is the bet established for the fruit of location $i$, and $w_i$ is the multiplier of location $i \in \{1, ..., 16\}$. So if locations $i$ and $j$ both belong to banana then $b_i=b_j = \mbox{bet}_{banana}$. You could write $$ E[V] = bet_{banana}\frac{1}{16}\sum_{i \in B} w_i + bet_{grape}\frac{1}{16}\sum_{i\in G} w_i + bet_{apple}\frac{1}{16}\sum_{i\in A} w_i$$ where $B,G,A$ are the sets defining locations for Banana,Grape,Apple $\endgroup$ – Michael Jan 22 at 1:36
  • $\begingroup$ @Michael The formula of $E[V] = \frac{1}{16}\sum_{i=1}^{16} b_i w_i$ yields inconsistent values of $E[V]$ with different bet values. E.g with $bet_{banana} = 1$, $bet_{grape} = 0$, $bet_{apple} = 0$, $E[V] = 2.125$ but for $bet_{banana} = 2$, $bet_{grape} = 0$, $bet_{apple} = 0$, $E[V] = 4.25$ $\endgroup$ – MSD Jan 22 at 1:59
0
$\begingroup$

This formalizes my comments above. Let $(b_B, b_G, b_A)$ be the bet value placed on [B]anana, [G]rape, [A]pple. We assume $b_B, b_G, b_A$ are nonnegative and not all zero. Let $V$ be the total we win. Then $$ E[V] = \frac{1}{16}\sum_{i=1}^{16} b_i w_i $$ where $w_i$ is the weight of location $i \in \{1, ..., 16\}$; $b_i$ is the bet value of the fruit associated with location $i$. This can be rewritten: $$ E[V] = b_B\underbrace{\left(\frac{1}{16}\sum_{i\in B}w_i\right)}_{\theta_B}+ b_G\underbrace{\left(\frac{1}{16}\sum_{i \in G}w_i\right)}_{\theta_G} + b_A\underbrace{\left(\frac{1}{16}\sum_{i \in A}w_i\right)}_{\theta_A}$$ where $B, G, A$ denote the subsets of locations $i \in \{1, ..., 16\}$ associated with Banana, Grape, Apple, respectively. Using $\theta_B, \theta_G, \theta_A$ values as defined by the above underbraces, we get $E[V] = b_B\theta_B + b_G\theta_G + b_A\theta_A$. The ratio of expected return to total bet is then: $$ \frac{E[V]}{b_B+b_G+b_A} = \left(\frac{b_B}{b_B+b_G+b_A}\right)\theta_B + \left(\frac{b_G}{b_B+b_G+b_A}\right)\theta_G + \left(\frac{b_A}{b_B+b_G+b_A}\right)\theta_A$$ In this case we have \begin{align} \theta_B &= \frac{20 + 5 + 3 + 3 }{16} = 1.9375\\ \theta_G &= \frac{3+3+5+15+3+5}{16} = 2.125\\ \theta_A &= \frac{6+3+6+6+3+3}{16} = 1.6875 \end{align} If you want to use $x_B, x_G, x_A$ as the fraction of our bet that we place on each fruit, so for example $x_B = b_B/(b_B+b_G+b_A)$, then $$\frac{E[V]}{b_B+b_G+b_A}=x_B\theta_B + x_G\theta_G+x_A\theta_A$$


Max expectation solution:

To maximize the expectation we should put the full portion of our bet on the fruit that maximizes the sum of its weights, that is we choose the max of $\theta_B, \theta_G,\theta_A$. In this case we should place our full bet on grape, so $(x_B,x_G,x_A)=(0,1,0)$, for a maximized expected payout-per-bet of 2.125.

Max worst-case solution:

The worst-case weight of each fruit is 3. So if we allocate our bet equally over all fruit, with fraction vector $(x_B,x_G,x_A)=(1/3, 1/3, 1/3)$, we get in the worst case: $$ \frac{V}{b_B+b_G+b_A} \geq (3)(1/3)=1 \quad \mbox{always} $$ This means we never loose any money. It is not difficult to see that this equal allocation maximizes the worst-case payout-per-bet. Under this equal allocation we get an expected payout-per-bet of: $$ \frac{E[V]}{b_B+b_G+b_A} = \frac{\theta_B}{3}+\frac{\theta_G}{3}+\frac{\theta_A}{3}\approx 1.91666666 $$ This is not as high as the expected payout-per-bet associated with placing everything on grape. However, placing everything on grape means we may loose it all (nonzero risk), whereas this equal-allocation scheme guarantees we never loose money (zero risk), and furthermore we are expected to gain money. So, everyone would want to play this game, but the sponsor of this game would quickly go bankrupt.

$\endgroup$
  • $\begingroup$ Findings: 1. To maintain a predictable (more stable) $E[V]$ value over sufficient trials without consideration of each symbols specific bet value its useful to set the individual weights or the probabilities of each symbol with a specific weight such that $\theta_B = \theta_G = \theta_A$. Lets call this value $\theta_k$ then the equation would simplify to be $\frac{E[V]}{b_B+b_G+b_A} = \theta_k$ $\endgroup$ – MSD Jan 23 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.