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Author Topic: Calculating the odds of EM reel machines  (Read 8140 times)
Neonkiss
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« on: September 17, 2008, 02:42:40 PM »

A re-post from Oldreno.

Calculating the odds of EM reel machines.

The population of a set of reels is how many total unique combinations it has.
Population is the number of stops on reel #1 times the stops on #2, #3, etc.
If there are 20 stops on your reels, the population is 20*20*20=8,000. In theory, all of your reel combinations could hit within 8,000 handle pulls, but that is very very unlikely.
Bally had 20 stop reels (8000), 22 stop reels (10,648) and 25 stop reels (15,625).  That's for 3 reel machines.
Adding a 4th reel to the 20 stop bumps the population up to 160,000, the 22 stop up to 234,256, and the 25 stop to 390,625.
After you figure the population, then you calculate all the winners for one cycle of the machine.  This is the payout per cycle.  Total payout divided by population equals the payout percentage.  That minus 100% equals the hold of the machine.

Figuring total pays.
Here is a sample of a 7's/bars bally 25 stop reel strip.
Reel 1.  7, blank, 1bar, blank, 2bar, blank 3bar, blank, 1bar, blank, 2bar, blank, 1bar, blank, 1bar, blank, 2bar, blank, 3bar, blank, 1bar, blank, 2bar, blank, 1bar.
Reel #2. 7, blank, 1bar, blank, 3bar, blank, 1bar, blank, 2bar, blank, 1bar, blank, 7, blank, 1bar, blank, 2bar, blank, 1bar, blank, 3bar, blank, 1bar, blank, blank.
Reel #3. 7, blank, blank, 2bar, blank, 1bar, blank, blank, 3bar, blank, 1bar, blank, blank, 7, blank, 1bar, blank, blank, 2bar, blank, 1bar, blank, blank, 2bar, blank.
Write down the reels on a piece of paper (as you see them in order), or use a spreadsheet.  Always recount your symbols (including blanks) to make sure you have the correct number for each reel.

Next write down the number 7's for each reel, the number of 3bars, 2bars, and 1bars.
Then you take the ways for 7's to hit and calculate them out.
1-7 on 1st reel times 2-7's on 2nd reel times 2-7's on 3rd reel gives a total of 4 hits per cycle.
2-3bars on 1st times 2-3bars on 2nd, times 1-3bar on 3rd equals 4 hits per cycle.
4-2bars on 1st times 2-2bars on 2nd, times 3-2bars on 3rd equals 24 hits for 2bars.
6-1bars on 1st times 6-1bars on 2nd times 4-1bars on 3rd reels equals 144 hits for 1bars.
These 3 different kinds of bar pays are the 'matching' bars pays.  There are 172 of them.
Now you calculate the TOTAL bars hits, which includes matching bars and mixed bars.
There are 12 total bars on reel #1, 10 total bars on reel #2, and 8 total bars on reel #3.  This is 960 bars hits.
Now, we must subtract the matching bar pays from the total bars, and by subtracting 172 match bars from 960 total bars, we get 788 mixed bar pays per cycle.
Now take each kind of hit and multiply it by the payoff for that hit.
7's thus pays 800 coins per cycle (at 200 coins per hit). 3bars pays 400 coins per cycle (100 per hit), 2bars pays 1200 (at 50 each), 1bars pays 2880 (at 20 each), and finally mixed bars pay 7880 per cycle at 10 coins per hit.
This should be a total of 13160 coins paid out per cycle.
13,160 coins paid out, divided by 15,625 coins in (our population) gives a payout percentage of 84.224%.  The hold is 100-82.224, or 17.776%
Pretty darned tight.
Hit frequency is 6.162%

Now wasn't that easy?

If anyone is interested, I can post the reel strips for an 809 fruit reel multiplier, and we could all figure out that one. We even had an 8-reel keno machine at one time.
It hit and was pulled from the floor....

oldreno
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« Reply #1 on: September 17, 2008, 02:44:39 PM »

A re-post from Statfreak...

hmmm. Got up late today and missed this whole discussion. oldreno describes the method quite well and does it exactly as I have been doing it on paper for years.

As you get your hands around this method you will see that games with the nudge feature are exactly the same as games without, only they will map some of the stops for the symbol to the nudge-blank. So instead of having 3B with 4 virtual stops and the next door blank with 12, they will move the 12 stops to a different non-nudge blank and then set the 3B with 1 stop and the nudge-blank with 3. So there is no reason that a nudge game needs to payout "less" frequently than a non-nudge game.

The Haywire takes this to a new level where there are more types of "symbols" than are represented on the physical strips. These hidden "symbols" have different values and create the higher payouts. There are actually 2 of each symbol except the 7 and the HW. These special, hidden extra blanks, 1Bars, 2Bars, and 3Bars add multiplier value to the hit, They also used the concept of the Red White Blue machines where symbol order mattered (think red 1bar, white 2bar, blue 3bar, paid 50 for 1 instead of 5 for 1 as a mixed bar win). And to make it more complicated, sometimes a non special symbol is part of the mix.

For example:

   special blank,    ANYTHING,            HW                            pays 3x2 (Single HW paying 2 coins with 2 haywires)   then

   special blank,    1bar(not special),    HW                            pays 5x2 (Single HW paying 2 coins with 4 haywires)   and
   HW,                1bar(not special),    special blank                pays 5x2 (Single HW paying 2 coins with 4 haywires)   BUT

   HW,                Special blank,         ANYTHING (incl 1bar)    pays only 2x2 (Single HW paying 2 coins with 1 haywire)  and
   HW,                Special blank,         special 2bar                  pays the elusive 10x2 (single HW paying 2 coins with 9 haywires)

This list is not complete for these positions. There are other symbols that can fill in the "anything" for some other higher pays.


That's why the Harwire Deluxe was such a bugger to solve. It was the only game I never could figure out on my own until I got the PAR sheet. I got close, but no cigar.
« Last Edit: January 24, 2009, 11:48:56 AM by Neonkiss » Logged

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« Reply #2 on: January 23, 2009, 04:00:13 PM »

JRR4,

Since John0813 has pointed you to this topic, here is a copy of a PAR sheet for the Haywire Deluxe to go along with my explanation above.
I hope this helps you with the program for your class.

Stat  garfield

* SS6413.Haywire Deluxe.2cm.txt (14.88 KB - downloaded 405 times.)
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« Reply #3 on: January 23, 2009, 04:18:22 PM »

Great stuff...Thanks
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JRR4
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« Reply #4 on: January 26, 2009, 04:03:38 PM »

I have a question for you mathmeticians.  Given a PAR sheet like the one linked above, how would one compute the percentage of plays (spins) that would generate a particular event? For example, what would be the percentage of plays (spins) that are haywires?  I took the total number of hits for all haywires (about 10000) and divided by the total number of hits at the bottom of the par sheet (41412) and got close to 1/4.  I know that can't be right.  I'm thinking it should be closer to 1/40.  Any ideas?  Thanks.   
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JRR4
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« Reply #5 on: January 26, 2009, 04:23:48 PM »

Never mind - i figured it out.  The percentage of spins that will result in a haywire is the sum of 1/(pulls/hit) for all haywire combinations. It came out to be approx .04 or 1 in 25 spins.
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StatFreak
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« Reply #6 on: January 26, 2009, 08:10:11 PM »

I have a question for you mathmeticians.  Given a PAR sheet like the one linked above, how would one compute the percentage of plays (spins) that would generate a particular event? For example, what would be the percentage of plays (spins) that are haywires?  I took the total number of hits for all haywires (about 10000) and divided by the total number of hits at the bottom of the par sheet (41412) and got close to 1/4.  I know that can't be right.  I'm thinking it should be closer to 1/40.  Any ideas?  Thanks.  

 Scratch Head  41412?  Scratch Head I searched the document and can't find that figure anywhere..

Never mind - i figured it out.  The percentage of spins that will result in a haywire is the sum of 1/(pulls/hit) for all haywire combinations. It came out to be approx .04 or 1 in 25 spins.

Using the pulls per hit works, but it is not the correct way to calculate the odds as it can introduce errors; your first method was actually the correct one. Best practice is to add up the number of combinations that haywire (12,479) and divide into either the total number of permutations (64^3 or 262,144) or into the total number of winning combinations (38,750), depending on the figure you are looking for.

Your figure is close. The odds of any spin haywiring, including spins that produce no winning combination, are 1 in 21.01. The odds that a winning spin will haywire are 1 in 3.1052.
53.95% of the haywires are wins of a single haywire symbol haywiring one time. Those wins are represented in the PAR as "2x2" and pay 8 coins for a 2 coin bet.
« Last Edit: January 26, 2009, 08:17:45 PM by StatFreak » Logged

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JRR4
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« Reply #7 on: January 26, 2009, 09:51:18 PM »

Thank you very much. JR
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JRR4
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« Reply #8 on: January 28, 2009, 04:22:01 PM »

StatFreak,  Have you figured out how the spin-til-you-win type machines work?  Does the machine know what the winning reel combination will be prior to the spin-til-you-win and it just teases the player with the spin-til-you-win action or is it more complicated?
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« Reply #9 on: January 29, 2009, 12:58:31 PM »

StatFreak,  Have you figured out how the spin-til-you-win type machines work?  Does the machine know what the winning reel combination will be prior to the spin-til-you-win and it just teases the player with the spin-til-you-win action or is it more complicated?

It spins using the same RNG and the same paytable that are used in normal play. The odds of hitting any particular outcome are the same as if you were playing credits. However, since all losing spins are discarded (spun away), the total number of final outcomes possible is limited to the total number of winning combinations for that particular SS chip. In effect, you are playing a shortened set of virtual strips that has only the winning combinations (but the machine doesn't pick anything in advace. It just plays the game normally until a win is spun).
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JRR4
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« Reply #10 on: January 29, 2009, 01:00:08 PM »

I see, thank you.
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