kkwong77
Joined: Mar 27, 2019
• Posts: 4
March 27th, 2019 at 11:23:10 PM permalink
Hi all,
I saw a slot game that is a 243 way game, called Alaksan Fishing, it seems like not a normal 243 ways game since it supports WILD pay. This WILD is quite confusing and I don't know how it works. I do understand the math of way game but the only with WILD pay is quite confusing. Anyway, here is what I have today

Let's define the symbol (W is WILD) & paytable

W C F DF B E P BT S T
5x
100
70
60
40
20
20
16
16
12
12
4x
20
12
10
8
4
4
3
3
2
2
3x
12
6
4
3
1.4
1.4
1
1
1
0.6

The screen I saw is
 W E E C C W W W S T W W W B E

From the screen, here is what I get
27 ways of 3xE each pay 1.4 = 37.8
12 ways of 4xS each pay 2 = 12*2 = 24
12 ways of 4xB each pay 4 = 12*4 = 48
12 ways of 5xC each pay 5 = 12*70 = 840

Since WILD substitute all symbols except scatter and bonus, we also find the following pay comb from WILD only:
12 ways of 3xF each pay 4 = 48
12 ways of 3xDF each pay 3 = 36
12 ways of 3xP each pay 1 = 12
12 ways of 3xBT each pay 1 = 12
12 ways of 3xT each pay 0.6 = 7.2

Very last one, since the wild pay as well in the way game, we should count that as well
it will be 12 ways of 3xWI = 12x12 = 114

sum up all those pays, we gain 1209 but the game only shows the winning of that screen as 1077. I seems that the slot does not count WILD win 12 ways, it just counts 1-way WILD win to pay 12 only, in that case, the total win is

37.8 + 24 + 48 + 840 + 48 + 36 + 12 + 12 + 7.2 + 12 (1-way WILD pay only) = 1077

It looks odd to me. If this is something wrong in the house pay in WILD-pay way game or mistake in my computation, any comment is welcome. Thanks
Last edited by: kkwong77 on Mar 28, 2019
charliepatrick
Joined: Jun 17, 2011
• Posts: 2818
March 28th, 2019 at 2:12:13 AM permalink
I think you may have done some double counting. If WWW didn't pay 12 the way I understand it is you would get...
12 ways to get 5 Cs
12 ways to get 4 S's (note that while there are 3 of 243 lines that score this WWWSC WWWSBi WWWSE only one counts)
12 ways to get 4 B's
15 ways to get 3E's (WEE WWE WEW)

However as WWW pays 12 then they may argue you get 12 ways to get 5 C's as that is a bigger winline followed by some 3E's using only 1 or 2 wilds. (Note this gets 1050 + 15*1.4 = 1071). However this does seem wrong logic.
kkwong77
Joined: Mar 27, 2019
• Posts: 4
March 28th, 2019 at 9:22:16 AM permalink
Quote: charliepatrick

I think you may have done some double counting. If WWW didn't pay 12 the way I understand it is you would get...
12 ways to get 5 Cs
12 ways to get 4 S's (note that while there are 3 of 243 lines that score this WWWSC WWWSBi WWWSE only one counts)
12 ways to get 4 B's
15 ways to get 3E's (WEE WWE WEW)

However as WWW pays 12 then they may argue you get 12 ways to get 5 C's as that is a bigger winline followed by some 3E's using only 1 or 2 wilds. (Note this gets 1050 + 15*1.4 = 1071). However this does seem wrong logic.

My bad, I took a picture of the win screen and paytable but many typo in the first post. I fix all of them in the top thread but it still wrong. It seems that they only pay 1-way WILD itself instead of 12 ways. I don't know why. But 1077 is what I get from the screen while it should be 1209 in my computation.
CrystalMath
Joined: May 10, 2011
• Posts: 1907
March 28th, 2019 at 11:12:06 AM permalink
It's tough to figure out what's going on here.

If I were designing the game, I would pay the following pays:
27 ways of 3xE each pay 1.4 = 37.8
12 ways of 4xS each pay 2 = 12*2 = 24
12 ways of 4xB each pay 4 = 12*4 = 48
12 ways of 5xC each pay 5 = 12*70 = 840
12 ways of 3xW each pay 12 = 12*12 = 144

This is what I think it paid:
15 ways of 3xE each pay 1.4 = 15*1.4 = 21 (the other 12 ways were paid as 3x wilds)
12 ways of 4xS each pay 2 = 12*2 = 24
12 ways of 4xB each pay 4 = 12*4 = 48
12 ways of 5xC each pay 5 = 12*70 = 840
12 ways of 3xW each pay 12 = 12*12 = 144
Total = 1077
I heart Crystal Math.
kkwong77
Joined: Mar 27, 2019
• Posts: 4
March 28th, 2019 at 12:32:40 PM permalink
Quote: CrystalMath

It's tough to figure out what's going on here.

If I were designing the game, I would pay the following pays:
27 ways of 3xE each pay 1.4 = 37.8
12 ways of 4xS each pay 2 = 12*2 = 24
12 ways of 4xB each pay 4 = 12*4 = 48
12 ways of 5xC each pay 5 = 12*70 = 840
12 ways of 3xW each pay 12 = 12*12 = 144

This is what I think it paid:
15 ways of 3xE each pay 1.4 = 15*1.4 = 21 (the other 12 ways were paid as 3x wilds)
12 ways of 4xS each pay 2 = 12*2 = 24
12 ways of 4xB each pay 4 = 12*4 = 48
12 ways of 5xC each pay 5 = 12*70 = 840
12 ways of 3xW each pay 12 = 12*12 = 144
Total = 1077

Thanks for your reply. The number looks correct but could you please share your idea why you would think the pay is 1077. I know your computation gives consistent pay as I got but I just there is not mentioned anything from the game rule how the WILD should pay. Different from line game, there will say the highest pay will be given per line win, it is confusing how they handle WILD in way game.

Also, I am trying to follow your idea but I don't get how to obtain 15 ways to get 3xE instead of 27
CrystalMath
Joined: May 10, 2011
• Posts: 1907
March 28th, 2019 at 1:29:57 PM permalink
Quote: kkwong77

Thanks for your reply. The number looks correct but could you please share your idea why you would think the pay is 1077. I know your computation gives consistent pay as I got but I just there is not mentioned anything from the game rule how the WILD should pay. Different from line game, there will say the highest pay will be given per line win, it is confusing how they handle WILD in way game.

Also, I am trying to follow your idea but I don't get how to obtain 15 ways to get 3xE instead of 27

In this case, I'm almost certain that they are paying all 27 ways on the three of a kind, but they are paying the highest amount only for each way. On 12 of those ways, the highest pay is 3 wilds. On the other 15 ways, the highest pay is 3 E.

I read the rules online, and they are not clear on this.
I heart Crystal Math.
kkwong77
Joined: Mar 27, 2019
• Posts: 4
March 28th, 2019 at 6:24:47 PM permalink
Quote: CrystalMath

In this case, I'm almost certain that they are paying all 27 ways on the three of a kind, but they are paying the highest amount only for each way. On 12 of those ways, the highest pay is 3 wilds. On the other 15 ways, the highest pay is 3 E.

I read the rules online, and they are not clear on this.

Yes, it is quite confusing regarding the "way" pay. If I understand it correctly, there are 243-way game since from any symbol on the first reel, there are total 243 ways to get to any end symbols on the last reel. But if I look at all the 3 of a kind way, i.e. the last symbol of pay end in the middle reel, for the screen I gave, all combinations similar to W W E will pay 3xE and all W W W will pay 3xW and all W E W will pay 3xE as well. If we pick one specific path to look at, says all WILDS on the middle row, there will be W W W, however, if we count two more symbols at the last 2 reels (i.e. C & C), then the "way" sharing W W W but with higher pay is W W W C C, so does it mean when W W W C C is paid for highest reward, all 3-of-a-kind W W W should not be paid?

It is too confusing how to define a way and what does it mean by "highest pay per way". To me, it looks like that a way is how to get to 2-of-a-kind (if any), 3-of-a-kind, 4-of-a-kind, 5-of-a-kind pay, so my understanding on "highest pay per way" is to return the highest win within all 3-of-a-kind win when WILD substitution happens but we don't compare 3-of-a-kind way to 4-of-a-kind way. So the following with "X" are two different ways of winning

 A B C D E X X X F G H I J K L

 A B C X X X X X D E F G H I J

and even the last one share some path with the previous one, they should pay at the same time because they are two different way to win, the last one does not count as the higher pay than the previous one.
Last edited by: kkwong77 on Mar 28, 2019
CrystalMath
Joined: May 10, 2011