Calculating Odds of a Slot Machine

English is not my best language, so please excuse the grammar and syntax of what is to follow.

I need to calculate the payout % of a slolt. Actually, any slot, if I can do it for one, I'll be fine for the others.

I decided to start with simulating a 5 x 3 slot machine, with reels of 32 positions. At the end, I'll need to calculate the payout of a slot with reels of 127 positions.

I'm not sure of the match invovled, so to validate the math, I decided to use a perl script (with 5 nested loops) to validate that the calculations are correct (hence the 32 positions, with 127, I pretty sure the script would time out).

My problem is that the script gives me 1015 ways to get 4 'A' and the math gives me 1013 ways to get 4 'A'.

The math I use to get to 4 'A' is as follow.

32 positions per reels.
reel 1: 1 A and 1 wild
reel 2: 2 A and 1 wild
reel 3: 1 A and 1 wild
reel 4: 2 A and 1 wild
reel 5: 2 A and 1 wild

The symbols have to line up from left to right.

So, the math should be (1 + 1) * (2+1)*(1+1)*(2+1)*(32-2-1) = 1044

and I should take out the times when I get 4 'W' right, 1*1*1*1*31 = 31

So 1044 - 31 = 1013.

Is that correct?

Once I confirm the odds of each result, I'll be able to calculate the payout %.

Thanks

Is there a separate pay for 4 Wild, if not, is the pay for 4A greater than the pay for anything else except 5A? If so, then 4 Wild is effectively the same thing as 4A, if A is the highest paying symbol w/ no separate pay for 4 wild.

Quote: slotcalculator

My problem is that the script gives me 1015 ways to get 4 'A' and the math gives me 1013 ways to get 4 'A'.

The math I use to get to 4 'A' is as follow.

32 positions per reels.
reel 1: 1 A and 1 wild
reel 2: 2 A and 1 wild
reel 3: 1 A and 1 wild
reel 4: 2 A and 1 wild
reel 5: 2 A and 1 wild

The symbols have to line up from left to right.

So, the math should be (1 + 1) * (2+1)*(1+1)*(2+1)*(32-2-1) = 1044

and I should take out the times when I get 4 'W' right, 1*1*1*1*3129 = 29
you cant use 31 for the 5th reel because you didn't use 31 for the 5th reel in the initial counting of 4 Aces

So 1044 - 29 = 1015.

Is that correct? now it is

Once I confirm the odds of each result, I'll be able to calculate the payout %.

Thanks

Fixed your post.

Here's what you need to do:

[(1+1)*(2+1)*(1+1)*(2+1) - 1*1*1*1]*(32-2-1) = 1015

There is a separate higher payout for 4 Wilds but ...

4 Wild might translate (for payout purposes) into 5 of a kind, based on the result of the fifth reel.

I'll have to adapt for the cases where 4 Wilds will pay more than 5 of another kind because of having 4 initial wilds.

Now that I know how to count them properly, thanks to the next posts, I have to tackle this part.

Thanks.

Quote: Mission146

Is there a separate pay for 4 Wild, if not, is the pay for 4A greater than the pay for anything else except 5A? If so, then 4 Wild is effectively the same thing as 4A, if A is the highest paying symbol w/ no separate pay for 4 wild.

After more work on my problem, I realize that what you said here is pertinent for one of the problems I am now facing. (wich I can't crack, ... yet).

For us to have a common ground, I'll comment on the Atkins Diet slot by the Wizard.


Based on the number of symbols on the reels, the number of times I can get 4 Bacon is
[ ( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) - ( 1 * 1 * 1 * 1)]*(32-3-1) = 20,972.

But the Wizard reports, as per his enumeration script, 20,832.

After looking at the payouts, I realized that in some cases, 3 Atkins pays more than 4 of a Kind (for example 4 Bacon).

Hence, I have to calculate the number of times it is possible to get AtkinsAtkinsAtkinsBacon where the fifth reel is not Atkins and not Bacon and remove that from 20,972.

So, I figured this would be 1*1*1*4(number of bacon on 5th reel)*(32-3-1) = 140

Now it works, 20,972 - 140 = 20,832.

So, all proud that I had figure this part out ... I went and did the same things for all the 4 of a kinds and can match all of the number from the wizard.

A few minuters later, moving to the 5 of kind, crash and burn. I can figure out the correct number all the way down to cheese, using this method, but not Bacon and Mayonaise.

Since 4 Atkins pays more than 5 Bacon, I need to do ( 5 Bacon ) - (5 bacon where the first 4 are Atkins)

( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) *(3+1) - ( 1 * 1 * 1 * 1*1) = 2999 ways to get 5 Bacons.

Now remove the number of time it's actually 4 of a kind (so remove AtkinsAtkinsAtkinsAtkinsBacon) is 1*1*1*1*3 = 3

Hence, I should have 2,999 - 3 = 2,996 BUT the wizard is getting 2,976.

What am I doing wrong?

Quote: slotcalculator

English is not my best language, so please excuse the grammar and syntax of what is to follow.

I need to calculate the payout % of a slolt. Actually, any slot, if I can do it for one, I'll be fine for the others.

I decided to start with simulating a 5 x 3 slot machine, with reels of 32 positions. At the end, I'll need to calculate the payout of a slot with reels of 127 positions.

I'm not sure of the match invovled, so to validate the math, I decided to use a perl script (with 5 nested loops) to validate that the calculations are correct (hence the 32 positions, with 127, I pretty sure the script would time out).

My problem is that the script gives me 1015 ways to get 4 'A' and the math gives me 1013 ways to get 4 'A'.

The math I use to get to 4 'A' is as follow.

32 positions per reels.
reel 1: 1 A and 1 wild
reel 2: 2 A and 1 wild
reel 3: 1 A and 1 wild
reel 4: 2 A and 1 wild
reel 5: 2 A and 1 wild

The symbols have to line up from left to right.

So, the math should be (1 + 1) * (2+1)*(1+1)*(2+1)*(32-2-1) = 1044

and I should take out the times when I get 4 'W' right, 1*1*1*1*31 = 31

So 1044 - 31 = 1013.

Is that correct?

Once I confirm the odds of each result, I'll be able to calculate the payout %.

Thanks

Hello slotcalculator, Sorry to bust your bubble but slots are different from back in the day. But to answer your question on your own, simply walk behind a slot tech when he is working on a machine and you will see the percentage labeled there.

Here in Canada the OLG (Ontario Lotto & Gaming Commission) has them all set to 85% Payout in most Canadian Casino's
and every machine nowadays reset RNG after every pull.

Back in the day, we were able to sit at a slot machine and crouch down just a little bit and look inside the machine showing at the top right side (Total In) (Total Out) counters and on many Blazing 7 Machines, standing in front of the machine approx eye level you notice a small 1/4" x 4" clear glass window which displayed the counters.

This allowed anyone to record the amount going in and out of the machine, seeing if paid out or ready to payout utilizing the 85% law.

They are now removed. I don't want to get into detail but something in todays machines called Co-Processors or eprom chip or something like that I don't want to mention were placed in around 1998-2000 resetting machines after every pull.

Quote: slotcalculator

A few minuters later, moving to the 5 of kind, crash and burn. I can figure out the correct number all the way down to cheese, using this method, but not Bacon and Mayonaise.

Since 4 Atkins pays more than 5 Bacon, I need to do ( 5 Bacon ) - (5 bacon where the first 4 3 are Atkins because 3 Atkins are also worth 50 as well as subtracting out 4 Atkins because they are worth 500)

What am I doing wrong?

( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) *(3+1) - ( 1 * 1 * 1 * 1*1) = 2999 ways to get 5 Bacons.

Now remove the number of times four Atkins Appear with 5th reel bacon:
1*1*1*1*3 = 3

Now remove the number of times three Atkins Appear with 4th reel bacon and 5th reel either bacon or Atkins:
1*1*1*5*(3+1) = 20

So now we have:
( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) *(3+1) - 1 - 3 - 20 = 2,976

Quote: tringlomane

( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) *(3+1) - ( 1 * 1 * 1 * 1*1) = 2999 ways to get 5 Bacons.

Now remove the number of times four Atkins Appear with 5th reel bacon:
1*1*1*1*3 = 3

Now remove the number of times three Atkins Appear with 4th reel bacon and 5th reel either bacon or Atkins:
1*1*1*5*(3+1) = 20

So now we have:
( 4 + 1 ) * ( 4 + 1 ) * ( 4 + 1 ) * ( 5 + 1 ) *(3+1) - 1 - 3 - 20 = 2,976

Damn, if I had another eye, I'd be a cyclop.

Thanks

Quote: slotcalculator

Damn, if I had another eye, I'd be a cyclop.

Thanks

When I was first learning to do this, I made many similar mistakes. Definitely can be frustrating. It's generally why it's recommended doing 5 nested loops and evaluating each combination via computer program.