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[GameLink] Amazon Queen (WMS)
ReelSize: 3x5
Paylines: 20 Line
Provider: WMS
Features: Both Way
Another new part of the model is free game, which usually first calculates the expected value of one free game, then multiplies it by the average number of free games to get the average payment of the free game, and then adds it to the scatter symbol payment that triggers the free game.
Compared with the previous spreadsheet, there is one more H1 4-of-a-kind and 3-of-a-kind from right to left. Basically the algorithm does not need to modify any special formulas. Here is a place to pay attention to in the design. Consider the connection of H2-WWW-H1. This line is regarded as the 4-of-a-kind of H2 from the left and the 4-of-a-kind of H1 from the right. Some games will give two payments. Some games will only give the one with the higher payment, and you must deduct the repeated connections. This game is a design that gives both payments.
After getting the expectation of free game, then find the average number of free spins.
First calculate the probabilities of retrigger (the probability of 3/4/5 CF), and then put it into the formula of the average number of games. It is recommended that readers can guide this formula by themselves, or use a different algorithm. Let’s take the remaining $N$ free game and the probability of $p3/p4/p5$ to trigger another 10/25/100 games as an example.
Let $E_N$ = The average spins of remaining $N$ spins, then
$E_N = N * E_1$ (The average spins of remaining 20 spins is equivalent of the average spins of remaining 1 spins * 20)
$E_1=(1-p3-p4-p5)*1+p3*(1+E_{10})+p4*(1+E_{25})+p5*(1+E_{100})$
$=1+p3*10*E_1+p4*25*E_1+p5*100*E_1$
(There are 4 possibilities for the remaining 1 game, respectively, the probability of $1-p3-p4-p5$ will only play the remaining 1 game, plus the $p3$ probability to play the remaining 1 game plus 10 games to trigger again The number of games,..., plus $p5$ probability to play the remaining 1 game plus another 100 games to trigger)
$E_1 = 1/(1-p3*10-p4*25-p5*100)$ (After moving to finishing, you can get the average spins of the remaining 1 game)
$E_N = N/(1-p3*10-p4*25-p5*100)$
Then multiply the expectation of a single free spin by the average number of free spins to get the average payment for triggering the free game.
Finally add the average payment of the free game to the scatter symbol payment of base game to get the RTP of this game.
Note: The reel strips used in the par sheet may not be the real strips of the game.
Parsheet_04 - Amazon Queen (WMS)
ReelSize: 3x5
Paylines: 20 Line
Provider: WMS
Features: Both Way
Game Instructions
This game can be regarded as a regular slot game, consisting of a base game and a free game. The game feature is that H1 (orangutan) has both ways payouts. Both ways means that, in addition to the continuous connection from the leftmost reel to the right, the continuous connection from the rightmost reel to the left is also considered a connection, which means that there is twice the connection opportunity, but usually 5-of-a-kind only count once.Another new part of the model is free game, which usually first calculates the expected value of one free game, then multiplies it by the average number of free games to get the average payment of the free game, and then adds it to the scatter symbol payment that triggers the free game.
Model Descriptions
Free Game
Let's start with free game.Compared with the previous spreadsheet, there is one more H1 4-of-a-kind and 3-of-a-kind from right to left. Basically the algorithm does not need to modify any special formulas. Here is a place to pay attention to in the design. Consider the connection of H2-WWW-H1. This line is regarded as the 4-of-a-kind of H2 from the left and the 4-of-a-kind of H1 from the right. Some games will give two payments. Some games will only give the one with the higher payment, and you must deduct the repeated connections. This game is a design that gives both payments.
After getting the expectation of free game, then find the average number of free spins.
First calculate the probabilities of retrigger (the probability of 3/4/5 CF), and then put it into the formula of the average number of games. It is recommended that readers can guide this formula by themselves, or use a different algorithm. Let’s take the remaining $N$ free game and the probability of $p3/p4/p5$ to trigger another 10/25/100 games as an example.
Let $E_N$ = The average spins of remaining $N$ spins, then
$E_N = N * E_1$ (The average spins of remaining 20 spins is equivalent of the average spins of remaining 1 spins * 20)
$E_1=(1-p3-p4-p5)*1+p3*(1+E_{10})+p4*(1+E_{25})+p5*(1+E_{100})$
$=1+p3*10*E_1+p4*25*E_1+p5*100*E_1$
(There are 4 possibilities for the remaining 1 game, respectively, the probability of $1-p3-p4-p5$ will only play the remaining 1 game, plus the $p3$ probability to play the remaining 1 game plus 10 games to trigger again The number of games,..., plus $p5$ probability to play the remaining 1 game plus another 100 games to trigger)
$E_1 = 1/(1-p3*10-p4*25-p5*100)$ (After moving to finishing, you can get the average spins of the remaining 1 game)
$E_N = N/(1-p3*10-p4*25-p5*100)$
Then multiply the expectation of a single free spin by the average number of free spins to get the average payment for triggering the free game.
Finally add the average payment of the free game to the scatter symbol payment of base game to get the RTP of this game.
Simulation Result
File Download
Here is the par sheet of this game, if you are interested, you can download the par sheet file for research.Note: The reel strips used in the par sheet may not be the real strips of the game.
Parsheet_04 - Amazon Queen (WMS)
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