Scatter math question

Consider a 5 reel slot, with one scatter symbol on each reel. Three rows are shown on the display. The stops per reels for reels 1-5 are {11,12,11,12,11} respectively. Find the number of ways to get 5, 4, and 3 scatter symbols.

I got
5: 243
4: 4,212
3: 29,187

If you have a sexy method of doing this, like using combinations, please post.

I did this for another slot a while ago here Just use the below table instead.

Reel	1	2	3	4	5
scatter	3	3	3	3	3
other	8	9	8	9	8

5 scatters = 3*3*3*3*3 = 243
4 scatters = 8*3*3*3*3 + 3*9*3*3*3 + 3*3*8*3*3 + 3*3*3*9*3 + 3*3*3*3*8 = 3402
3 scatters =8*9*3*3*3 + 8*3*8*3*3 + ... + 3*3*3*9*8 = 19053

Shouldn't you be subtracting the scatters for the non-scatter multiple? Here's my work:

Excel Sheet

Also, anyone know why my posts are sent to the last page in the forum?

Quote: nmacgre

Shouldn't you be subtracting the scatters for the non-scatter multiple?

Yes. And that's what Miplet did.

Your math is looking for the scatter on any payline for 4 reels, but is looking for a non-scatter on the center line of the last reel, etc.

Miplet is looking for the scatter to be on any payline for 4 reels, and for it to not appear on any payline for 1 reel, etc.

Quote: nmacgre

Also, anyone know why my posts are sent to the last page in the forum?

Yeah, it's actually at the top of the unread threads - which may put it far back if there are a lot of threads you haven't read.

Just click the "Mark threads read" button at the top of the list.

Ah I see. So would I multiply the non-scatters by 3?

Thanks for the forum tip by the way.

Not quite. The question always boils down to "how many ways can the symbols I care about show up on this reel to yield the award I care about?"

If you want to count the 4-symbol scatter that ends in a non-scatter symbol, you need to have no scatter symbols on *any* of the visible stops on the 5th reel. If there are 11 stops total and 1 scatter, there are only 8 stops which meet that criterion.

Ok, makes perfect sense. Thanks.

Hi there,

glad that i've found this forum via the contact system from the wizard homepage :).
I thought i could place my question in this thread, because i had the same question for days and couldn't find an answer (so i don't have to create a new thread). Thank you very much for explaining everything.

I have two questions:
1.) Is it possible to have one formula for the math that's posted here? Because writing down all possible combinations is very, how to say... you could forget only one combination and the complete math is wrong. So is there an easier way to determine how many combinations are possible.

2.) This question keeps me crazy... I tried to figure out, how Wizard calculated the combinations in his Atkins slot machine. In the moment i try to figure out the steak. What i've done so far:
There're 3 on reel 1 (3 stops), 2 on 2ns, 2 on 3rd, 2 on 4th and 3 on the 5th reel. Makes a total of 12 steak symbols in the game. Because the Atkins symbol is wild, it could replace the steaks. To make the math easier i made a group of steaks and wild symbol. A wild symbol is found only once on each reel, makes 5 in total.
So my new steak group is: 4, 3, 3, 3, 4 -> total of 17 symbols.

Now i tried to calculate the combinations for 4 steaks in a row. My math was:
4x3x3x3x28 = 3,024
(28 because: The last reel should have all symbols except wild and steak (to prevent the 5-in-a-row). I merged Wild and steak, so 4 symbols shouldn't show up to make the 4-in-a-row perfect).
Now the result of Wizard was: 2,996 exactly 28 less then my result.
My question is... where the hell did this 28 came from??? I already substracted 28 for the 5th reel. I'm not seeing the wood for the trees.

Any help is appreciated. Thank you in advance.

Best whises
Frank

You are counting four wilds in the first 4 reels. This will pay more (500) than four steaks(200).

Hi miplet,

thanks for the answer. But i already thought about it. But for the 4 Wild, why substracting 28? When only 1 wild-symbol exist on every reel, a steak-wild group is created, the only thing i have to do is to substract the amount of 4 wild-combinations, and that's 1x1x1x1 not 28???

I'm heavily confused now... Because to make the 4 steaks-in-a-row i need 3 wilds and one steak symbol. If i would use 4 wilds and a steak i'd have a 5-in-a-row of steaks. So 4 wilds is for sure 4-wilds and not 4 steaks. 4 steaks could be created with 3 wilds...
Oh man i've no idea :).

Could you give me a math example like in your 2nd post where you explained the math for the scatter? Maybe i'll understand then. In the moment i can't imagine the trick behind the math.

Best regards
Frank

EDIT: Wait a minute... Maybe i got it. I think you meant that i have a steak-wild group. I combined the two to make the math easier right? But the little thing i missed is... as i made this merged group, not only 4 steaks are included but also 4 wilds-in-a-row. And that's what i've missed and that's what you answered me. But why 28?

(4x3x3x3)=108 ways for the first 4 reels minus the 1 way to have 4 wilds is 107. Then multiply by 28.

Thank you very much for the example miplet, now i understand. My thinking was i the wrong direction, but now it's clear!
Thanks again!

Any idea for the 1st question, how to simplify the 'combination'-finding to determine the scatter-combinations? Writing all possible combinations down in an excel sheet seems very time consuming and error-prone.

Best regards
Frank

Hi there,

it's me again. I have a new problem with the calculations. I try to understand it. The examples given hier work well for the 4-in-a-row.
But what to do with the Wild-Symbols.

In the PDF-file i linked before, there are 513 combinations for 3 Atkins (Wild-Symbol). I took care of the combinations which are higher than this 3 in a row (4 in-a-row with other symbols). 4 symbols match that criteria, where the 4 in-a-row is more worth than 3 Atkins-symbols.

So i have the following:
Reel #4:
Steak: 2x, Ham: 3x, Drumsticks: 2x, Sausage: 4x

Reel #5:
Steak: 3x, Ham: 4x, Drumsticks: 3x, Sausage: 4x

So... i calculated:
3 Atkins symbol in a row (only 1 symbol per reel):
1x1x1

Then i substracted 11 from the 4th reel => 32-11 = 21 and the 4th wild symbol = 21 - 1 = 20
Next step: Substracted 14 from the 5th reel => 32-14 = 18 and the 5th wild symbol = 18 - 1 = 17

So my math was: 1x1x1x20x17 = 340

Wizard has as result in his PDF: 513 combinations...
What did i wrong this time??? :(

Thanks for any help.

Best regards
Frank

Read this thread. Basically you want to loop through all 32^5 combinations and score them. Doing it by hand is very error prone. Anyways doing it the hard way:
The first 3 have to be wilds (1*1*1) which you already have.
With a 4th reel is a scale (1) the 5th can be anything (32) 1*32=32
With a 4th reel of mayo(4) the 5th can be anything (32) (wild wild wild mayo wild , and wild wild wild mayo mayo can be counted as 5 mayos for 50 but the Wiz counts it as 3 wilds) 4*32 = 128
With a 4th reel of Bacon(5) the 5th can be anything (32) (Same thing as mayo) 5*32 = 160
With a 4th reel of Cheese(4) the 5th can be anything but Cheese(3) or wild(1) = (32-3-1=28)( wild wild wild cheese [non wild or cheese] can be counted as 4 cheeses for 50 but the Wiz counts it as 3 wilds) 4*28 = 112
With a 4th reel of Butte3(3) the 5th can be anything but Butter(4) or wild(1) = (32-4-1=27) (Same thing as cheese) 3*27=81

32 + 128 + 160 +112 +81 = 513

I assume this is a history lesson as well? and a RNG is not being used !

Hi miplet,

first at all, i'm very thankfull for your patience and the "hard-way" explanation you gave me. Now i understand. Error prone is the exact phrase to describe the problem.

As you see, i forgot so many thing. I took a look in that thread and i clapped my hands. I'm able to write programs in VB.net and didn't see the option to write a program which has 5 nested for-loops.

I made this a couple minutes ago. First thing i did, was to check if he cycle through all 32^5 combinations. It did perfectly.
Now the big question is... how do i code the counter which count the corresponding combinations correctly.
My first try was to group wild and normal symbols into one group, which makes it easier. But to 'teach' the program which combination is valid and which not, is extremly difficult (or i think to difficult), because i can't code 20.000 possible combinations which the program should count, that's very error-prone.
No idea so far :).
I think i'll post in that thread and simply ask ;).

Again, thank you very much for your patience and help.

Best regards
Frank