jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 7th, 2018 at 7:49:54 AM permalink
I would like to know the maths for working out the probability of getting free spins on an EGM... using the following as an example but most are generally the same

5 reels - 3 symbols stopped each spin on each reel.

Any 3 from 5 scattered free spin symbols to trigger free games (pretty standard)

Reels 32 long and 1 free spin symbol per reel.

hope this is enough info!

thanks in advance
rsactuary
rsactuary
  • Threads: 29
  • Posts: 2315
Joined: Sep 6, 2014
Thanked by
jimny48
September 7th, 2018 at 8:13:52 AM permalink
Quote: jimny48

I would like to know the maths for working out the probability of getting free spins on an EGM... using the following as an example but most are generally the same

5 reels - 3 symbols stopped each spin on each reel.

Any 3 from 5 scattered free spin symbols to trigger free games (pretty standard)

Reels 32 long and 1 free spin symbol per reel.

hope this is enough info!

thanks in advance



Probability of getting a bonus game is = pr (3 symbols) + pr (4 symbols) + pr (5 symbols)
= 5!/(3!x2!)x((3/32)^3)x(29/32)^2 + 5!/4!x((3/32)^4)x(29/32) + (3/32)^5
=0.013892

Approximately once every 72 spins. I'm sure someone will check my math
ThatDonGuy
ThatDonGuy
  • Threads: 118
  • Posts: 6465
Joined: Jun 22, 2011
Thanked by
jimny48
September 7th, 2018 at 3:55:03 PM permalink
Quote: rsactuary

Probability of getting a bonus game is = pr (3 symbols) + pr (4 symbols) + pr (5 symbols)
= 5!/(3!x2!)x((3/32)^3)x(29/32)^2 + 5!/4!x((3/32)^4)x(29/32) + (3/32)^5
=0.013892

Approximately once every 72 spins. I'm sure someone will check my math


I heard that...
Of the 325 = 33,554,432 ways the wheels can stop, there are:
33 x 292 x 10 = 227,070 ways of getting three symbols
34 x 29 x 5 = 11,745 ways of getting four symbols
35 = 243 ways of getting five symbols
or a total of 239,058 winning combinations

The probability is 239,058 / 33,554,432 = about 1 / 140

rsactuary - your equation is correct, but you may have entered it into your calculator or computer wrong.
rsactuary
rsactuary
  • Threads: 29
  • Posts: 2315
Joined: Sep 6, 2014
September 7th, 2018 at 6:18:22 PM permalink
Quote: ThatDonGuy

rsactuary - your equation is correct, but you may have entered it into your calculator or computer wrong.



Indeed. I think when I copied formulas down , I didn't anchor some. Thanks for the correction.
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 8th, 2018 at 1:23:32 AM permalink
Thank you both heaps!

if its not too much to ask can you explain the maths a bit? or even algebra it in the long format. I'm not a mathematician but should be able to change it up for different features once I get the logic behind each step, as I would like to use it for varying machines.


:-)
ThatDonGuy
ThatDonGuy
  • Threads: 118
  • Posts: 6465
Joined: Jun 22, 2011
September 8th, 2018 at 3:20:46 PM permalink
Quote: jimny48

Thank you both heaps!

if its not too much to ask can you explain the maths a bit? or even algebra it in the long format. I'm not a mathematician but should be able to change it up for different features once I get the logic behind each step, as I would like to use it for varying machines.


I'll put it in a Spoiler Box so people who don't want to see it won't have to scroll through the whole thing.

Each reel has 32 symbols on it. Since we're only interested in the wild symbols, let's assume that the symbols are in this order:
Wild, 1, 2, 3, 4, ..., 28, 29, 30, 31
There are 32 different positions each reel can stop in, and three of them (Wild / 1 / 2, 31 / Wild / 1, and 30 / 31 / Wild) will show the Wild symbol in that reel.

Here's rsactuary's method:

The total probability of having 3 or more Wilds = the probability of exactly 3 wilds + the probability of exactly 4 wilds + the probability of exactly 5 wilds.

The probability of exactly 3 wilds = (the probability that a reel will show a wild)3 x (the probability that a reel will not show a wild)2 x (the number of ways you can have 3 reels with wilds and 2 without wilds).
The probability that a reel will show a wild in this case is 3/32.
The probability that a reel will not show a wild in this case is 29/32.
The number of ways to select the 3 reels with wilds out of 5 is usually called "the number of combinations of 5 things taken 3 at a time" (note "combinations", not "permutations" - the difference is, for example, {1, 2, 3} and {2, 3, 1} are the same combination of 3 items, but two different permutations as the order is different). This is 5! / (3! x (5-3)!) = 10.
Thus, the probability of exactly 3 reels with wilds = (3/32)3 x (29/32)2 x 10 = 227,070 / 33,554,432.

Similarly, the probability of exactly 4 reels with wilds = (3/32)4 x (29/32) x (5! / (4! (5-4)!) = 11,745 / 33,554,432
and the probability of exactly 5 reels with wilds = (3/32)5 = 243 / 33,554,432

The total probability = (227,070 + 11,745 + 243) / 33,554,432 = 239,058 / 33,554,432 = about 1 / 140.36.

My method is slightly different in that I counted the total number of ways to get 3, then 4, then 5, and added those up, then divided by the 325 possible ways the 5 reels could stop - in other words, I counted the 227,070, the 11,745, and the 243 first, then added them and divided by 33,554,432, rather than getting three fractions and adding them.

jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 11th, 2018 at 3:54:35 AM permalink
Thank you so much! both of those methods will help me heaps!
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 12th, 2018 at 5:15:41 AM permalink
Ok so I finally looked at it. I made the example simple so I had to do some work and learn but I cant get a formula to come into my brain if the reel lengths are different and the number of wilds on each are different.

I can step it out with what you have given me but that's not neat. Essentially I am stuck on the fact that I need to add all 5 (combination) sums for 4 wilds out of 5 and all 10 for 3.

Is there a way considering for 4 combinations the number of stopping positions with a wild is multiplied by the others on 4 occasions each with 1 being stopping positions with none? and for 3 wilds its 3 and 2?

I made it hard for myself so I had to use some of my brain but im stumped if I can figure out a formula solution on my own! Hope im not too far off haha
rsactuary
rsactuary
  • Threads: 29
  • Posts: 2315
Joined: Sep 6, 2014
September 12th, 2018 at 8:48:40 AM permalink
Yes there is. If no one beats me to it, I will attempt it tonight. It definitely gets messy.

Edited to add: I went back and tried to understand what you were asking and I can't figure it out. Can you state a specific example that you'd like to figure out? Hopefully you can generalize from there.
Last edited by: rsactuary on Sep 12, 2018
ThatDonGuy
ThatDonGuy
  • Threads: 118
  • Posts: 6465
Joined: Jun 22, 2011
September 12th, 2018 at 6:27:18 PM permalink
Quote: rsactuary

Yes there is. If no one beats me to it, I will attempt it tonight. It definitely gets messy.

Edited to add: I went back and tried to understand what you were asking and I can't figure it out. Can you state a specific example that you'd like to figure out? Hopefully you can generalize from there.


Do you think there's an "easy way" if the number of stops on each reel, and the number of wilds on each reel, is different? I think he's stuck with a Brute Force solution.

In other words, yes, you probably do have to check each of the 10 different sets of 3 reels that can have wilds separately for the 3-reel number, and the 5 different sets of 4 reels that can have wilds separately for the 4-reel number.

One thing makes it easier: the denominator - the total number of possible combinations - is always the same; it is the product of the number of stops on each reel.

Let S1, S2, S3, S4, and S5 be the number of stops on reels 1-5, respectively, and W1, W2, W3, W4, and W5 be the number of wilds on each reel.
The total number of positions of the five reels is S1 x S2 x S3 x S4 x S5
The ten ways to get 3 wilds:
1,2,3: W1 x W2 x W3 x (S4 - W4) x (S5 - W5)
1,2,4: W1 x W2 x (S3 - W3) x W4 x (S5 - W5)
1,2,5: W1 x W2 x (S3 - W3) x (S4 - W4) x W5
1,3,4: W1 x (S2 - W2) x W3 x W4 x (S5 - W5)
1,3,5: W1 x (S2 - W2) x W3 x (S4 - W4) x W5
1,4,5: W1 x (S2 - W2) x (S3 - W3) x W4 x W5
2,3,4: (S1 - W1) x W2 x W3 x W4 x (S5 - W5)
2,3,5: (S1 - W1) x W2 x W3 x (S4 - W4) x W5
2,4,5: (S1 - W1) x W2 x (S3 - W3) x W4 x W5
2,4,5: (S1 - W1) x (S2 - W2) x W3 x W4 x W5
The five ways to get 4 wilds:
1,2,3,4: W1 x W2 x W3 x W4 x (S5 - W5)
1,2,3,5: W1 x W2 x W3 x (S4 - W4) x W5
1,2,4,5: W1 x W2 x (S3 - W3) x W4 x W5
1,3,4,5: W1 x (S2 - W2) x W3 x W4 x W5
2,3,4,5: (S1 - W1) x W2 x W3 x W4 x W5
And the one way to get 5 wilds:
1,2,3,4,5: W1 x W2 x W3 x W4 x W5
Add those 16 numbers up, and divide by (S1 x S2 x S3 x S4 x S5) to get the total probability.
CrystalMath
CrystalMath
  • Threads: 8
  • Posts: 1911
Joined: May 10, 2011
September 12th, 2018 at 7:01:31 PM permalink
Quote: ThatDonGuy

Do you think there's an "easy way" if the number of stops on each reel, and the number of wilds on each reel, is different? I think he's stuck with a Brute Force solution.

In other words, yes, you probably do have to check each of the 10 different sets of 3 reels that can have wilds separately for the 3-reel number, and the 5 different sets of 4 reels that can have wilds separately for the 4-reel number.

One thing makes it easier: the denominator - the total number of possible combinations - is always the same; it is the product of the number of stops on each reel.

Let S1, S2, S3, S4, and S5 be the number of stops on reels 1-5, respectively, and W1, W2, W3, W4, and W5 be the number of wilds on each reel.
The total number of positions of the five reels is S1 x S2 x S3 x S4 x S5
The ten ways to get 3 wilds:
1,2,3: W1 x W2 x W3 x (S4 - W4) x (S5 - W5)
1,2,4: W1 x W2 x (S3 - W3) x W4 x (S5 - W5)
1,2,5: W1 x W2 x (S3 - W3) x (S4 - W4) x W5
1,3,4: W1 x (S2 - W2) x W3 x W4 x (S5 - W5)
1,3,5: W1 x (S2 - W2) x W3 x (S4 - W4) x W5
1,4,5: W1 x (S2 - W2) x (S3 - W3) x W4 x W5
2,3,4: (S1 - W1) x W2 x W3 x W4 x (S5 - W5)
2,3,5: (S1 - W1) x W2 x W3 x (S4 - W4) x W5
2,4,5: (S1 - W1) x W2 x (S3 - W3) x W4 x W5
2,4,5: (S1 - W1) x (S2 - W2) x W3 x W4 x W5
The five ways to get 4 wilds:
1,2,3,4: W1 x W2 x W3 x W4 x (S5 - W5)
1,2,3,5: W1 x W2 x W3 x (S4 - W4) x W5
1,2,4,5: W1 x W2 x (S3 - W3) x W4 x W5
1,3,4,5: W1 x (S2 - W2) x W3 x W4 x W5
2,3,4,5: (S1 - W1) x W2 x W3 x W4 x W5
And the one way to get 5 wilds:
1,2,3,4,5: W1 x W2 x W3 x W4 x W5
Add those 16 numbers up, and divide by (S1 x S2 x S3 x S4 x S5) to get the total probability.



You also need to multiply W1, W2, ... by the visible height of that reel. This also assumes that two wilds will not be visible on the same reel at the same time.
I heart Crystal Math.
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 13th, 2018 at 12:52:46 AM permalink
Yeah! that's exactly what I did, I set the stepped out combinations in excel once then just did a set of if statements so I could drag it to save time, then you can change the numbers around.

But is there an "elegant" way to do that? as I want to throw piles of shit at it!


So for example rsactuary


Reel 1 is 50 long with 2 wild

Reel 2 is 80 long with 3 wild

Reel 3 is 52 long with 2 wild

Reel 4 is 50 long with 2 wild

Reel 5 is 51 long with 2 wild
unJon
unJon
  • Threads: 14
  • Posts: 4661
Joined: Jul 1, 2018
September 13th, 2018 at 4:36:00 AM permalink
Quote: jimny48

Yeah! that's exactly what I did, I set the stepped out combinations in excel once then just did a set of if statements so I could drag it to save time, then you can change the numbers around.

But is there an "elegant" way to do that? as I want to throw piles of shit at it!


So for example rsactuary


Reel 1 is 50 long with 2 wild

Reel 2 is 80 long with 3 wild

Reel 3 is 52 long with 2 wild

Reel 4 is 50 long with 2 wild

Reel 5 is 51 long with 2 wild




Set up cells in Excel for W1-5 and S1-5. Then have your dragged formulas with fixed references to those cells. Then you can just change the values in those cells to whatever you want and the dragged formulas will spit out the new numbers.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 13th, 2018 at 5:36:04 AM permalink
thats not a maths solution. its what ive done? or am I missing something?
unJon
unJon
  • Threads: 14
  • Posts: 4661
Joined: Jul 1, 2018
September 13th, 2018 at 5:41:10 AM permalink
Quote: jimny48

thats not a maths solution. its what ive done? or am I missing something?



I thought you hard coded the W1-5 and S1-5 into your formulas as numbers so it wasn’t easy for you to change them to “throw piles of shit” at them. I was suggesting you pull them out and use cell references so you can easily change them to let you “throw piles of shit” at them. If you want elegant for elegance sake, then I dunno.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 13th, 2018 at 4:29:16 PM permalink
Apologies that did look rude what i wrote. I was just hoping there was a maths way so you could change the stopping window size to 4 on each reel or even add an additional reel (6) as some features do. Also for when wilds arent seperate and are in blocks on a reel. Which i can step out but wondered if there was a formula way of doing it. I can't think of one but I'm not a maths person i just love excel haha
rsactuary
rsactuary
  • Threads: 29
  • Posts: 2315
Joined: Sep 6, 2014
September 13th, 2018 at 4:44:53 PM permalink
Quote: jimny48

Apologies that did look rude what i wrote. I was just hoping there was a maths way so you could change the stopping window size to 4 on each reel or even add an additional reel (6) as some features do. Also for when wilds arent seperate and are in blocks on a reel. Which i can step out but wondered if there was a formula way of doing it. I can't think of one but I'm not a maths person i just love excel haha



When you start varying things like this, you do end up getting close to brute force.
jimny48
jimny48
  • Threads: 1
  • Posts: 8
Joined: Sep 7, 2018
September 13th, 2018 at 8:24:18 PM permalink
I had to look that up haha... earlier in the year i was building a spreadsheet for a steel fabrication company and i was trying to streamline ordering to make it as cost efficient as possible. I nearly lost my mind after an hour of trying to figure out how to do it. I actually had to get up and go out haha. When i came back i googled and found it actually had a name "the cutting stock problem"

Is that what you mean? Too many variables makes it close to impossible?

This has been awesome too! Thanks!
  • Jump to: