# Accounting for Bet in Bonus Game of Slot Game

##### Question

Should the bet be separated from the payout when analysing a bonus game of a slot game (like it is done when analysing a base game)?

##### Description

When calculating the RTP (return to player) of a slot game, the bet and payout are considered separately. For example, let's say there are 20 combinations are possible, of which only 8 combinations pay (let's suppose 2x). The RTP of the base game is:

$$bet=\1$$

$$RTP_{base\ game}=\frac{payout}{total\ bet}=\frac{8\times(2\times1)}{20\times\1}=\frac{\16}{\20}=80\%$$

Note that the payout doesn't account for the bet a player pays before clicking the button. Thus, their real win for 2x (for a single combination) will be $1, not $2 (–bet + payout = –$1 + 2 ×$1).

Let's suppose when the player gets a winning combination, they are proposed to play the bonus game: coin toss. The player may take the winnings or play the bonus game. If they win the bonus game, their payout gets doubled. If they lose, they just lose their bet (that they deposited before playing the base game), the bonus game ends. To calculate the average winnings for a single game, the following distribution of the bonus games will be used:

$$n_{games\ total}=16\ games$$

# $$prize$$ $$\boldsymbol{n_{win\ \&\ continue}}$$ $$\boldsymbol{n_{win\ \&\ take}}$$ $$\boldsymbol{n_{lose}}$$
2 $$8\times bet=\8$$ 0 1 1
1 $$4\times bet=\4$$ 2 2 4
0 $$2\times bet=\2$$ 8 8 0

But since the bet is not included into the payout for the base game analysis, should the same be done for the bonus game analysis? Below, there are the 3 variants that I considered but I cannot figure out which is appropriate (if any).

$$Total\ payout=\sum_{i=0}^2 (n_{win\ \&\ take\ i}\times prize_i - n_{lose\ i}\times bet)=$$

$$=8\times\2-0\times\1+2\times\4-4\times\1+1\times\8-1\times\1=\27$$

$$Average\ payout=\27 / 16 = \1.6875$$

(looks like the player doesn't deposit their bet before clicking the button)

$$Total\ payout=\sum_{i=0}^2 (n_{win\ \&\ take\ i}\times prize_i)=$$

$$=8\times\2+2\times\4+1\times\8=\32$$

$$Average\ payout=\32 / 16 = \2$$

(equals to the base game payout which is 2x)

$$Total\ payout=\sum_{i=0}^2 (n_{win\ \&\ take\ i}\times{prize_i})-n_{games\ total}\times{bet}=$$

$$=(8\times\2+2\times\4+1\times\8)-16\times\1=\16$$

$$Average\ payout=\16 / 16 = \1$$

(pure win of the player)

Which $$RTP_{total}$$ is correct?

$$RTP_{total}=\frac{8\times\1.6875}{20\times\1}=\frac{\13.5}{\20}=67.5\%$$

$$RTP_{total}=\frac{8\times\2}{20\times\1}=\frac{\16}{\20}=80\%$$

$$RTP_{total}=\frac{8\times\1}{20\times\1}=\frac{\8}{\20}=40\%$$

The return to player is just what the player receives at the end of the game divided by the bet the player makes. What mechanics go on inside the game do not matter. There may be exit points along the way and you need to define the player's strategy at those points. You should not add together the winnings from win and continue and win and take as the player only gets one of those. I don't understand your bonus structure at all. In the original game the RTP was $$80\%$$ as you say. If the player plays each of those wins again s/he will win $$\frac 16\%$$ of the time and get $$4$$ in those cases. The RTP is then $$64\%$$
• Also, I didn't add win & continue and win & take. Commented Jun 13, 2023 at 14:09
I've found an answer: since the bet is not accounted in the base game, it should be disregarded when calculating an RTP for the bonus game too, since the total bet goes to the denominator.