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bolverk
bolverk
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February 13th, 2017 at 7:53:56 PM permalink
Hi, everyone, Nice to meet you.
I have difficulty in calculating shifting wilds.
When The wilds appear on the reel, they will stick on the row , and go from right to left.
I can use Markov chain and apply it to one line.
But to multi-line, things are getting complicated.
Because wilds will go from right to left, that means wilds will disappear in some pay line.
Can I still apply the one line method to calculating multi-line?

Please forgive me if I make any errors; my English is a little weak.
Thanks)
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teliot
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February 14th, 2017 at 7:30:38 AM permalink
Quote: bolverk

Hi, everyone, Nice to meet you.
I have difficulty in calculating shifting wilds.
When The wilds appear on the reel, they will stick on the row , and go from right to left.
I can use Markov chain and apply it to one line.
But to multi-line, things are getting complicated.
Because wilds will go from right to left, that means wilds will disappear in some pay line.
Can I still apply the one line method to calculating multi-line?

I am not familiar with the slots you are discussing, but based on your description, the answer to your question is "yes." Back that up with a simulation (which you should be doing anyway). You can simulate just the base game, ignore the mini-games/free spins, and double check that matches the RTP for the base game in the PAR.
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CrystalMath
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February 14th, 2017 at 8:15:27 AM permalink
Quote: teliot

I am not familiar with the slots you are discussing, but based on your description, the answer to your question is "yes." Back that up with a simulation (which you should be doing anyway). You can simulate just the base game, ignore the mini-games/free spins, and double check that matches the RTP for the base game in the PAR.



It sounds a lot like Jack and the Beanstalk from NetEnt.
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Wizard
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February 14th, 2017 at 8:26:50 AM permalink
I did a slot with that feature before. I think what I did was that if that feature was invoked I scored the game with separate reel stripping that has two fully wild reels. It can be cumbersome to do it in a spreadsheet, so I would suggest cycling through every combination in the programming language of your choice.
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bolverk
bolverk
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February 14th, 2017 at 6:19:16 PM permalink
Quote: teliot

I am not familiar with the slots you are discussing, but based on your description, the answer to your question is "yes." Back that up with a simulation (which you should be doing anyway). You can simulate just the base game, ignore the mini-games/free spins, and double check that matches the RTP for the base game in the PAR.



Hi, teliot.
Thank you for your reply,
It helps a lot.
To build the par sheet of base game, I should find the steady state of the 32*32 markov chain matrix applied to one line, right?
Last edited by: bolverk on Feb 14, 2017
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bolverk
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February 14th, 2017 at 6:29:38 PM permalink
Quote: CrystalMath

It sounds a lot like Jack and the Beanstalk from NetEnt.



Hi, CrystalMath.
Thank you for your reply.

Yes, the wilds behave just like that, but this game is more complex.
It will respin until the wilds disappear when the feature invoked.
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bolverk
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February 14th, 2017 at 6:45:28 PM permalink
Quote: Wizard

I did a slot with that feature before. I think what I did was that if that feature was invoked I scored the game with separate reel stripping that has two fully wild reels. It can be cumbersome to do it in a spreadsheet, so I would suggest cycling through every combination in the programming language of your choice.



Hi, Wizard,
Thank you for your reply,
It means when the feature was invoked, you'll replace the normal reels to another reels which only two of them have wilds?
Please forgive me if I misunderstand what you mean.
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teliot
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February 14th, 2017 at 9:20:41 PM permalink
Quote: bolverk

Hi, teliot.
Thank you for your reply,
It helps a lot.
To build the par sheet of base game, I should find the steady state of the 32*32 markov chain matrix applied to one line, right?

Like I said above, I am not familiar with the slots you are discussing. That said, I think a Markov chain is overkill.
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bolverk
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February 15th, 2017 at 6:14:30 PM permalink
Thank you,
I'll do the simulation first.
I'm trying to calculate the shifting feature of Dolphin Cash made by Playtech.
I would be very grateful if you could help me.
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teliot
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February 16th, 2017 at 5:30:43 AM permalink
Quote: bolverk

Thank you,
I'll do the simulation first.
I'm trying to calculate the shifting feature of Dolphin Cash made by Playtech.
I would be very grateful if you could help me.

I've freelanced for Playtech as a mathematician as have some of my slot-mathematician friends. Are you their slot mathematician now? They create some very complex slots.
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Wizard
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February 16th, 2017 at 6:42:40 AM permalink
Quote: bolverk

Hi, Wizard,
Thank you for your reply,
It means when the feature was invoked, you'll replace the normal reels to another reels which only two of them have wilds?
Please forgive me if I misunderstand what you mean.



No, two reels will be ALL wilds. The rest will be the same as for the initial spin.
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CrystalMath
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February 16th, 2017 at 8:07:49 AM permalink
This game seems much less complicated than Jack and the Beanstalk. I also think you are over thinking it.

The wild only appears on the 5th reel, then shifts to the left each spin. In Jack and the Beanstalk, wilds appear on reels 2 through 5. Also, the player pays for each spin, unlike Jack and the Beanstalk.

There are 16 possible starting configurations for a line, where a 1 represents a wild which has shifted to the left. You can see that the fifth reel never starts with a wild.

00000, 00010, 00100, 00110, 01000, 01010, 01100, 01110, 10000, 10010, 10100, 10110, 11000, 11010, 11100, 11110

For each configuration, re-calculate the game with full wild reels for each reel with a 1.

To calculate the probability of each configuration, just calculate the probability of each of the prior 4 games having/not having wilds.. For instance, to calculate the probability of 01010, this means that the prior game had a wild in that position, two games prior did not have a wild, three games prior had a wild, and four games prior did not. Now, all you need is the probability of a wild on the last reel.

p(01010) = W2(1-W)2, where W is the probability of a wild on the last reel.
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bolverk
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February 16th, 2017 at 10:00:05 PM permalink
Quote: teliot

I've freelanced for Playtech as a mathematician as have some of my slot-mathematician friends. Are you their slot mathematician now? They create some very complex slots.


No, I'm working in a small company developing mobile apps.
but I'm aiming to be a slot mathematician.
I have learned a lot of things about slot math from here.
appreciated greatly your help.
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bolverk
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February 16th, 2017 at 10:00:46 PM permalink
Quote: Wizard

No, two reels will be ALL wilds. The rest will be the same as for the initial spin.


Thank you, Wizard.
I understand it.
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bolverk
bolverk
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February 16th, 2017 at 10:05:27 PM permalink
Quote: CrystalMath

This game seems much less complicated than Jack and the Beanstalk. I also think you are over thinking it.

The wild only appears on the 5th reel, then shifts to the left each spin. In Jack and the Beanstalk, wilds appear on reels 2 through 5. Also, the player pays for each spin, unlike Jack and the Beanstalk.

There are 16 possible starting configurations for a line, where a 1 represents a wild which has shifted to the left. You can see that the fifth reel never starts with a wild.

00000, 00010, 00100, 00110, 01000, 01010, 01100, 01110, 10000, 10010, 10100, 10110, 11000, 11010, 11100, 11110

For each configuration, re-calculate the game with full wild reels for each reel with a 1.

To calculate the probability of each configuration, just calculate the probability of each of the prior 4 games having/not having wilds.. For instance, to calculate the probability of 01010, this means that the prior game had a wild in that position, two games prior did not have a wild, three games prior had a wild, and four games prior did not. Now, all you need is the probability of a wild on the last reel.

p(01010) = W2(1-W)2, where W is the probability of a wild on the last reel.



so grateful for your help, CrystalMath.
It makes me think more clearly.

In more general case, like wilds appear on reels 2 through 5.

Let A be the markov matrix of shifting wild.

The probability of each configuration will be the top row of A^4, because of all configurations will come from state 00000 after 4 times shifting by A, right?

Shifting Wild Calculations
I make a excel file about the calculation.
I hope I don't miscalculate.
Last edited by: bolverk on Feb 17, 2017
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CrystalMath
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February 17th, 2017 at 3:13:09 PM permalink
You're welcome. For the more general case, I initially thought a Markov matrix was suitable, but I now realize that it is unnecessary and is very similar to the simpler case with wilds only on the 5th reel.

Let's assume the probability of wilds on reels 2 to 5 are w2, w3, w4, w5.

For any given game, the probability that you start the game with a wild on reel 1 is the probability of getting a wild on reel 2 in the previous game OR a wild on reel 3 of two games prior OR a wild on reel 4 three games prior OR a wild on reel 5 four games prior = 1-(1-w2)(1-w3)(1-w4)(1-w5). Then do a similar calculation for a wild on reel 2 = 1-(1-w3)(1-w4)(1-w5), and so on for reels 3 and 4.

Then, make the 16 wild configurations based on those probabilities.
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bolverk
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February 19th, 2017 at 5:40:37 PM permalink
Quote: CrystalMath

You're welcome. For the more general case, I initially thought a Markov matrix was suitable, but I now realize that it is unnecessary and is very similar to the simpler case with wilds only on the 5th reel.

Let's assume the probability of wilds on reels 2 to 5 are w2, w3, w4, w5.

For any given game, the probability that you start the game with a wild on reel 1 is the probability of getting a wild on reel 2 in the previous game OR a wild on reel 3 of two games prior OR a wild on reel 4 three games prior OR a wild on reel 5 four games prior = 1-(1-w2)(1-w3)(1-w4)(1-w5). Then do a similar calculation for a wild on reel 2 = 1-(1-w3)(1-w4)(1-w5), and so on for reels 3 and 4.

Then, make the 16 wild configurations based on those probabilities.


This way is really easy to understand and calculate.

The calculation of this method exactly matched the calculation of Markov Matrix.

Thanks for broadening my perspective of thinking.

It's really a big help to me.
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