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I'm trying to figure out the right way of calculating an expected win (or loss) for a slot base game spin with a bonus game (guess the colour).

Let's say a player makes a $1 bet and gets a combination that pays $10 (еру base game). The player can either take their winnings ($10) or play the bonus game. In the bonus game the player has to guess the colour (red or black). If the player guesses it right, the winnings will be doubled (2 × $10 = $20). Otherwise, they lose the winnings ($0) and the bonus game stops. Again, the player can either take their winnings ($20) or play the next round of the bonus game. There are maximum of 5 bonus game rounds allowed.

Thus, there are the following possible outcomes:

  • \$10 (the player decides not to play the bonus game. Just takes the prize they win in the base game);
  • \$20, \$40, \$80, \$160, or \$320 (the player plays 1, 2, 3, 4, or 5 rounds of the bonus game and takes that prize);
  • \$0 (the player plays 1, 2, 3, 4, or 5 rounds of the bonus game and loses).

Here is what I have tried so far:

For the win/loss table, I think the initial bet of $1 should be used. Or, maybe, the potential prize from the previous round/game should be listed in the Loss column?

Game Round Win, \$ Loss, \$
Base - 10 0
Bonus 1 20 1
Bonus 2 40 1
Bonus 3 80 1
Bonus 4 160 1
Bonus 5 320 1

Probability of loss is $ p_{loss} = 1 - p_{win} $. But I'm not sure what probabilities of win should be used:

  • 0.5 for all cases or
  • 0.5 for the base game (probability the player will decide to play the bonus game) and $ 0.5^{round} $ for the bonus games (probability of getting all played rounds won) or
  • something else, since not all combinations are possible (e.g., for 3 rounds, combination win, loss, win is impossible, since the bonus game ends after the loss). This looks like 0.5 should be used for all cases.
Game Round pwin ploss
Base - 0.5 0.5
Bonus 1 0.51 = 0.5 1 – 0.51 = 0.5
Bonus 2 0.52 = 0.25 1 – 0.52 = 0.75
Bonus 3 0.53 = 0.125 1 – 0.53 = 0.875
Bonus 4 0.54 = 0.0625 1 – 0.54 = 0.9375
Bonus 5 0.55 = 0.03125 1 – 0.55 = 0.96875

Then, I think, the average of the expected values for the base and bonus games should be calculated as followsNOTE:

$$ E[game] = \frac{E[spin]+\sum_{i=1}^5 E[round_i]}{n_{spins}+n_{rounds}} = \frac{(p_{win}\times win-p_{loss}\times loss)_{spin}+\sum_{i=1}^5 (p_{win}\times win-p_{loss}\times loss)_{i}}{1+5} $$

So, the question is: what's the correct way to calculate the expected win for this case?


NOTE I'm not 100% sure that the formula is written correctly in terms of notation, since I'm not very familiar with the probability and statistics.

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  • $\begingroup$ You'd realistically lose money, as a 1 dollar entry fee with a 50% of winning 10 dollars, means that everyone who only plays (only) the first round would make on average 4 dollars. And a lot of people would figure that out, (you should probably raise the entry fee) $\endgroup$
    – Sejr
    May 29 at 13:44
  • $\begingroup$ Sejr, thank you for your comment, but I want to figure out how to calculate that. $\endgroup$
    – ENIAC
    May 29 at 13:50
  • $\begingroup$ I think I understand the quistion better now, so is it this? ( where B_win is the winning chance on base round, which seems to be different than the other bonus rounds. And P_stop is the chance of the player stopping every round ) (The player on average earns around 3.42): $\frac{1}{1-P_{stop}^{1+R_{rounds}}}\left(\left(P_{stop}\right)\left(\left(B_{win}\right)\left(E_{earning}\right)-F_{fee}\right)+\left(B_{win}\right)\sum_{n=1}^{R_{rounds}}\left(P_{stop}^{1+n}\left(\left(P_{win}\right)^{\left(\frac{n^{2}+n}{2}\right)}\cdot\left(2\right)^{n}\cdot E_{earning}-F_{fee}\right)\right)\right)$ $\endgroup$
    – Sejr
    May 29 at 14:49
  • $\begingroup$ -So you want an entry fee on 5.53 or higher, if you want to earn money. $\endgroup$
    – Sejr
    May 29 at 14:51
  • $\begingroup$ Sejr, correct me, if I'm wrong. The bonus game can be triggered only if the base game gives a win. Thus, $ B_{win} = 1 $. But there is a chance that the player will not play the bonus game (I believe it is 0.5). I don't know what entry fee you mean. And I don't need to calculate it. I just need to calculate the expected win or loss with the given bet and prizes. $\endgroup$
    – ENIAC
    May 29 at 15:14

1 Answer 1

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I'm fairly sure that the equation below is what you are lookin for, I would recommend playing around with it in Desmos (https://www.desmos.com/calculator/sq5gkraui7)

$\text{Average Earning after 1 game}=-\frac{1}{1-P_{stop}^{1+R_{rounds}}}\left(\left(P_{stop}\right)\left(\left(B_{win}\right)\left(E_{earning}\right)-F_{fee}\right)+\\ \left(B_{win}\right)\sum_{n=1}^{R_{rounds}}\left(P_{stop}^{1+n}\left(\left(P_{win}\right)^{\left(\frac{n^{2}+n}{2}\right)}\cdot\left(2\right)^{n}\cdot E_{earning}-F_{fee}\right)\right)\right)$

(Where $B_{win}$ is the winning chance on base round, which seems to be different than the other bonus rounds. And $P_{stop}$ is the chance of the player stopping every round, the rest should be self explanatory )

The player on average earns around 3.42, and if you want to make money, the entry fee has to be around 5.53 or higher.

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