Info wanted to improve an Autofruit emulation

[beromat]

I don't think I ever played the Autofruit, or one of its siblings, back in the days when penny arcades really were 1D arcades as I preferred its bigger cousins, the horse racing and roulette type of game as I found those more visually interesting. So when I joined this forum a few months back, seeing pictures and videos of Autofruits was a bit of an "oh, don't remember that" moment. It was also, as a retired programmer, an "I bet it wouldn't be too difficult to write a program that does that" moment. So that is what I did, I wrote a "fun money" program and I've put it on my website as freeware along with other stuff I've written. However, that is possibly just a first stage in development, and in order to do more I need to ask you guys who actually have these machines a few questions as the inner workings are not completely obvious if you have no access to them.

To give my questions a context I'd better briefly digress into what I have already done. I have made an emulation of the original Lemon pays 8D machine. Why an emulation and not a simulation? That is down to my ability with graphics which a Scot might describe as being "like a coo wi' a gun"! Why the 8D version? Because simple arithmetic, combined with videos of the 6D machine's "Losses" switch says it has to be simpler.

It seemed obvious that the innards of the machine must have some analogue to a roulette wheel. The questions were, how many "slots" and how many of those "slots" per fruit. So looking at the 8D machine payouts, if there was 1 position for the Bell then for the Lemon there had to be 12/8 positions for the Lemon. Clearly that doesn't work, so try again with 2 positions for the Bell, that gives (12 / 8) x 2 = 3 positions for the Lemon. Nya ha! Progress! In a similar vein it soon becomes apparent how many slots the other fruits need. That gives a total of 35 which looked uncomfortable to divide into 360 degrees so I figured there was probably a Lose position, giving 36 nicely sized 10 degree slots. I then saw the videos, counted the positions on the rotary switch and bingo! Thus encouraged I proceeded with the program. I even remembered not to stop on Lose when there are no bets!

So to the questions for further development.

1) I would like to make my program as accurate as possible, so whilst I have the correct number of each fruit in a list which the program steps through and looks reasonably realistic. I would actually like to have the same order as the real machine. Can anyone supply that please? Being circular, it doesn't matter where you start with the list. This would be an utterly trivial change to make.

2) I would like to then add an option to make it emulate both versions of the machine, but, as a first stage, without the Losses switch. This involves slightly more work, so before I commit myself to that it would be nice to know if the sequence of connections from the rotor to the fruits is still the same as the 8D machine, but with the fixed Loss position changed to a Lemon? If you work out how many slots are needed to give the correct payouts as before it comes to 18 including 2 for Lemon, so double that to fill 36 slots. So, if anyone knows the 6D machine sequence I would appreciate it if that information could be shared. (Incidentally, that disproves the myth that having Lemon = 8D is more generous than 6D as 3 slots paying out 8D comes to exactly the same 24D as 4 slots paying out 6D! (I cannot feel smug, I fell for that myth too!)

3) The final stage will be to emulate the Losses switch. But it is not clear how that works. What is clear is that all 36 positions of the rotor must connect to fruits otherwise the display whilst "playing" would not be one light per clunk of the rotor. Hence, and this is why I initially said the 6D machine must be more complex, some rotor positions must be overrideable to a Loss state. Equally clearly, 3 of these dual purpose positions must be selectable by setting the switch to 3, but which? Are they sequential or spread around the rotor. Is it the same 3 positions plus 3 more when the switch is at 6, or 6 different positions. And obviously the same questions arise for the 9 position. Can anyone enlighten me please?

B.T.W. #1 : It is highly unlikely that the machines were rigged to hit Lose more if you played all 5 symbols at once. If you play the same 1 symbol 36 times and each stop on the rotor is hit you will lose 12D regardless of which is your selection. If you play more coins in a fixed selection you simply lose 12D times the number of coins in the same 36 play scenario. So why would anyone want to discourage you from losing money 5 times faster than necessary by imposing extra Lose results? Surely they would want to encourage you to keep throwing money away at a high rate?

B.T.W #2 : The download link is at http://racingdog.hostfree.pw/Freeware.html right at the bottom (Note the capital F)

So what do I think of the machine now I have had a chance to play it, albeit in emulated form? Well the basic machines are a rip off, but that is true of all our beloved machines. However, if the Lose selector is set to 3 it becomes a complete rip off, to 6 and it is a total rip off, to 9 and it is a complete and total rip off! Basically, if you see a Lose light illuminate on a 6D machine, find something else to play!

Thanks in advance,
Derek

[brigham]

I have the contact positions written-out somewhere, for both 'no-bets' and 'coin in' sequences. I'll have a look for them.
Have you grasped the notion that there is no random element?

[beromat]

That would be great ta.

My ability with my wristwatch's stop watch may be exceedingly suspect, but I got the impression the length of time the rotor moves was not exact from one game to the next, so I had that tagged as a possible source of randomness. As I said, I'm just guessing about everything else from limited information, but I'd convinced myself that no other aspects were random as it just sounded too complicated for the technology available.

If nothing is random, doesn't that make outcomes predictable?

Cheers,
Derek

[rippyspennyarcade]

The length of time the rotor moves is different each game thanks to an interrupter stop/start unit. Every time the contact drops in to a notch it stops the main cam unit for a few seconds. This prevents any recurring sequence.

[brigham]

The 'variator', as that rotor is known, delays the start of the selector by a different amount each game, which gives the APPEARANCE of a random element.
However, the variator itself runs continuously, and in step with the main control drum, by virtue of both being synchronized to the same mains supply; so it too plays a known and repeated part in the game cycle.
It's a long cycle, and I was at one time going to count how many games it takes to repeat, but more interesting things intervened.
Like the ENIGMA, it has a weak spot. The giveaway is the 1/- win. (Elvis Presley, on my version). There is only ONE position on the selector which is connected to the 1/- payout, when coins are in play. When Elvis comes up, we KNOW where the selector is resting.

[pennymachines]

And then there's the hornswoggle...

[brigham]

I've found my notes.
I'll post them tonight, after work.
They relate to 'Top Ten', but as far as I'm aware, they are all just cosmetic variations on the same machine.

[brigham]

Right, here we go!

These notes refer to the Jamieson's 'Top Ten' of 1962.
There are 36 positions on the selector, which I have numbered from 1 to 36, from an arbitrary position 1 at the extreme rear of the cabinet, and running consecutively in the direction of rotation of the wiper.
There are five possible 'win' positions on the front glass, which I have numbered 1-5 from left to right.
These pay out 4, 2, 12, 2, 6, respectively.
Position 1 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 2 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
Position 3 connects to 'Win' 5 when no bet placed, and to 'Win' 5 when a bet is made.
Position 4 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 5 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 6 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 7 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.
Position 8 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 9 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 10 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
Position 11 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 12 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 13 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.
Position 14 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 16 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
Position 17 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.
Position 18 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 19 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 20 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
Position 21 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 22 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 23 connects to 'Win' 5 when no bet placed, and to 'Win' 5 when a bet is made.
Position 24 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 25 connects to 'Win' 3 when no bet placed, and to 'Win' 3 when a bet is made.
Position 26 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 27 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.
Position 28 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
Position 29 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 30 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 31 connects to 'Win' 4 when no bet placed, and to 'Win' 4 when a bet is made.
Position 32 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.
Position 33 connects to 'Win' 1 when no bet placed, and to 'Win' 1 when a bet is made.
Position 34 connects to 'Win' 2 when no bet placed, and to 'Win' 2 when a bet is made.
Position 35 connects to 'Win' 3 when no bet placed, and to 'Win' 4 when a bet is made.
And finally,
Position 36 connects to 'Win' 5 when no bet placed, and to 'Win' 2 when a bet is made.

These figures could be transferred to a simple table, by someone with the necessary computer skills, and the pattern of results would become evident.
The 'give-away' position is No. 25. It is the ONLY contact on the selector which is ALWAYS connected to the 1/- payout.
When the machine pays out twelve, the selector IS resting on No. 25.

This machine has an 'odds' adjuster, with positions 0,3,6,9, which correspond to the number of 'LOSE' indications introduced to the game in place of possible winning positions.
I have the figures for these, and the changes they effect, if you need them. I must try and find a less cumbersome way to post a table of figures.

Please feel free to ask, if anything is still not clear.

[beromat]

Yes I see! Thanks all, but please keep an eye on this thread (well for a little while at least!) as I may eventually need the Lose effect data.

I'm a bit busy for the next few days, but I will be analysing all that asap. Then I'll figure how much more I want to do.

Thanks again,
Derek

[beromat]

I had few minutes just to look at the figures and I was doing my head in because they didn't add up. Then I realised Position 15 isn't in the list!

Go on, be a sport, give me a clue! Please.

Cheers,
Derek

[13rebel]

I feel sure that it will be a 2d position as the Top Ten has one 12d payout position, two 6d payouts and six 4d wins.This is in line with Jamiesons 'Top Dog' payouts, although Top dog has 37 pins on the rotor and no switch for lose positions, not the 36 mentioned on the Top Ten machine. On the Top Ten machine this results in a payout of 56.7% with the switch set to zero lose positions. This is assuming randomness which appears to be questionable according to one post. It will be interesting to find out the effect of the switch in the various positions e.g. does it only affect the 2d wins etc.?

[beromat]

OK, so I assumed the missing values were 5 and 4.

I upgraded the program to the first stage as outlined above and updated the download file, so at least the lights flash the way the real thing does. I haven't used the "real" payout mapping as I didn't see the point of making a fun money computer game less attractive to play by making it meaner!

I'm not convinced about going any further with that program as the next step would be to add a Lemon = 6D option, but that wouldn't be right as it would be only using setting (0) of the "little man" switch. That would inevitably lead to the subsequent adding of an emulation of that switch to make a proper job, which would make the game even meaner, and, as I said, turning people off playing by making the game even meaner just doesn't seem like a good idea. With the 8D machine I have more carte blanche to go my own way with the payouts as most people have never used one, so won't be disappointed by that.

How mean is the game? Well there is a web site run by a guy I know in Vegas under the byline of The Wizard of Odds https://wizardofodds.com. He consults for the casinos about the odds they should set for new games and does all sorts of other statistical analysis for them. But, on the flip side, he shares his knowledge and advice for the punters via the web site. So you can take it he knows what he is talking about. Looking on his site at several slot machines in the US casinos, the payout is typically between 85 and 95% though he reports one or two outside that range (yes some really are higher!).

What of our little game? My version of the payouts, which I figured before I knew how the machine works, were based on the idea of the total payouts for each symbol would be the same over that statisticians favourite 36 plays with each possible outcome happening just once. Do that and you end up with getting back 24D for 36D paid on any one symbol. That is a payout of 66.66...% So if you do the hypothetical experiment for each symbol, that takes 180D and returns 120D. If we try that with the actual mappings (assuming the missing value was window 4 as stated above) then the return on 180D is only 106D, a percentage of 58.88...%. Assuming the 9 Losses setting only knocked out 2D wins (ho ho ho!) that is another 18D off, which is an 88D return, so that return will actually be worse than 48.88...%. Not exactly encouraging is it? But to be fair, these are amusements not genuine gambling devices and many others also have returns that a gambler would find appalling.

The unbalanced spread of symbols actually used in the real machine introduces another Achilles heel. On the payouts I was proposing there is no optimal strategy. Whatever you do loses the same amount on average, your only choice is how fast you lose it (by playing more than one coin at a time). On the actual machine, you do have an optimal strategy to minimise your losses which is to always play the Cherries as there are more of them than Oranges and those are the only two symbols over represented.

Cheers,
Derek

[13rebel]

beromat,you seem to acknowledge in your opening line of position 4 and 5 are missing that there are 37 pin positions. this being the case then the calculation base is 185 [not 180].The missing positions are both 2d not 2d and 6d,this leads to a calculation of 104/185 = 56.2%. This is based on both the Autofruit [6d version] and Top Ten machine as a friend has both these machines.This is also the case with Top DOG machine. If the switch is set to 9 lose at the 7x 2d ,positions,1 x 6d and 1x4d positions for example then the return to player is 43.2%. Surely an operator would not set the return to player amount at this measly amount would they?

[brigham]

OOOOPS!
I can't find how to edit the original list, so please accept my apologies, and insert the following information:

Pin 15 is connected to 'Win' 2 in both cases.

The 'Odds' adjuster simply replaces a number of 'Win' positions with the same number of 'Lose'.
(The position number of the selector IS the number of 'Lose' positions inserted. I have the figures if you want them).

[beromat]

13Rebel, no the two guesses were one for each list, not two for the same list. Brigham's post contained two lists in one table, one for no bet and one for bet.

Brigham, thanks for the update and all the previous info, most helpful. I wrote another program to try to spot sequences, as the thing is supposed to be non-random. This new info will simplify it and its results. It's not doing me much good though! I suspect it's more complicated than that as I've only been able to find short sequences. I'd stand more chance if I could see some slightly longer sequences than those on the videos, but unless I come across one in an arcade on a day when I have my old and slow tablet PC with me I doubt that will happen.

[brigham]

Beromat,
If you are using the 'No Bets' sequence instead of the actual payout one, you aren't emulating the gameplay.
There are SEVEN 1/- payouts in that sequence. You'll soon have the tube empty.
As for the sequence, it isn't relevant. It is neither planned nor controlled. It is no different from using a 'random numbers' table, or the second thousand places of π. My only interest in it is because it is there, and is therefore a challenge.

[badpenny]

Ensuring a constant reoccurrence of memory regarding the technical preferences as utilised in the 1960s, don't forget ….

* The electricians from then recommended responsive incremental applications.
* Forward-looking innovators preferred front edge end user innovation.
* They were keen to revamp and reutilise their front facing reciprocal enhancement.
* Before launching a new machine, they realised the urgency to get on-message about the total asset innovation.
* The upgraded model would now offer systemised reciprocal alignment guaranteed to baffle small boys.
* Repurposing integrated next-generation protocols while listening to the harmonics of The Kinks always worked.
* However maximizing informational empowerment abstractions was too basic and fooled nobody.


BP

[beromat]

Nope!

In my emulation I am using the real no bet list for display and my own (fairer) payout list.

In my analysis tool I am using the real payout list on the videos of real machines.

So I am making use of all the information, but not in the same program!

So far I can find sequences of 6 symbols with the 7th not matching if a LOSE has occurred. Or sequences of 5 with the 6th not matching if there was no LOSE. The difference is that LOSE seems to cause a 1 position shift, which when I allowed for it produced the longer sequences.

If anyone is interested the increments I'm checking for are 1,5,7,11,13,17,19,23,25,29,31,35. I left out all multiples of 2 and 3 as those numbers would cause many positions to never be hit. I'm also not looking at higher or negative increments because, if you renumber the positions to be 0 to 35, all those other possible values map onto to the ones listed by using modulo 36 arithmetic, which is custom made for coping with circular repeating items such as the rotor.

Cheers,
Derek

[brigham]

Looks like I'll have to ascertain the length of the sequence after all.
You could do it mathematically. There are only three variables, so either LCM or HCF would do it. (I can't remember which. 11+ was too long ago).
Or you could do it empirically, thus:
Wait until ELVIS pays 1/-.
Stop the machine, and mark the position of all three rotors with Marking Blue.
Start the machine, and count how many games occur before all three marks are back in the same position.
To 'prove' your result, ELVIS will pay 1/-.
This will be the LONGEST run of games before the cycle repeats. If the notches on the VARIATOR disc are symmetrical, a number of shorter sequences, or 'sub-multiples' may exist.

[beromat]

That sounds too much like hard work for something that started out as just curiosity! You'd better leave it, unless you're one of those people who can't. I'm a bit like that, but not totally!

Yes school was a long time ago, so I mistakenly thought casually mentioning modular arithmetic would be understood by all. But I was just checking the school syllabus and I see it doesn't come in until A level! So apologies to all who never did maths at A level, but you can find many explanations on the web. I'm not sure why it comes that late in the curriculum, because it really is trivial. I'm sure junior school kids could hack it if they had to!

Cheers,
Derek