Poll slots 2

[Zoozie]

I believe Zoozie's simulation is accurate @ 1:88 and not yours @ 1:75.

I just discovered we disagreed on this probabilty for Moonshine. Are you sure you used the
flask-scatter and not the policeman-scatter KK? If not you might have to recalculate all scatter-probabilities on
your website..

 

[Zoozie]

Number of symbols per reel: 35, 35, 35, 36, 40
Number of scatters per reel: 1, 1, 1, 1, 4
Which by my calculations makes the chance of getting 3 or more scatters 1 in 55.
To me that sounds way to low!
Can someone please check that for me?

I get:
Probability for getting a feature each spin:816480/61740000~0.013224489795918367~1/76
Notice that 816480/61740000 is the exact value.

Edit:

For 3+ scatters it is:~1/76
For 4+ scatters it is: ~1/1371
For 5 scatters it is :~1/63519

 

[gerilege]

It's even more simplier, though again, let's make a small simplification, let reel4 also have 35 symbols only. I haven't had a look on the reel strip for reel 5, but I assume that there are always at least two other symbols between the scatters, please let me know if this is not the case.
3 scatters:
Two cases, we have a scatter on reel5 or not.
a) We have a scatter on reel 5, then we need exactly two other scatters from the first four reels. 6 combos. So reel 5 = (12/40), other four = (3/35)*(3/35)*(32/35)*(32/35), altogether:
(12/40)*(3/35)*(3/35)*(32/35)*(32/35)*6
b) We get 3 scatters on the first four reels. 4 combos. We have 3 on the first 4 reels: (3/35)*(3/35)*(3/35)*(32/35) and no scatter on the last: (28/40). Altogether:
(3/35)*(3/35)*(3/35)*(32/35)*(28/40)*4.

4 scatters, again two cases, we have a scatter on reel5 or not.
a) we have a scatter on reel5, and do not have on one of the first four. 4 combos. Scatter on reel5 (12/40), have 3 from the first four: (3/35)*(3/35)*(3/35)*(32/35), altogether:
(12/40)*(3/35)*(3/35)*(3/35)*(32/35)*4.
b) we have 4 scatters on the first 5 reels, 1 combo. (3/35)*(3/35)*(3/35)*(3/35), but no scatter on last reel (28/40), altogether:
(3/35)*(3/35)*(3/35)*(3/35)*(28/40)
5 scatters is 3*3*3*3*12/(35*35*35*36*40)

(12/40)*(3/35)*(3/35)*(32/35)*(32/35)*6 +
(3/35)*(3/35)*(3/35)*(32/35)*(28/40)*4 +
(12/40)*(3/35)*(3/35)*(3/35)*(32/35)*4 +
(3/35)*(3/35)*(3/35)*(3/35)*(28/40) =

(3/35)*(3/35)*(1/40)*(1/35)*(1/35)*
{
(12*32*32*6 +
(3*32*28*4 +
(12*3*32*4 +
(3*3*28)
} =
804060/60025000 ~= 1:75.
Again the probability should be less or equal than 1:75 because of the small simplification I made, this (and the assumption about "scatterred enough" scatters) explains why I'm slightly off to Zoozie's result and confirm that the result from his simulation is accurate.

 

[gerilege]

KK, it would be easier to find the missing piece in your calculations if you would publish it. Actually I believe I probably know where you do not consider a particular thing, it is a common issue which people usually do not consider. Why am I saying this? Your 5 scatters probability is OK, but the feature trigger and 4 scatters are significantly off.

 

[Rusty]

Nope, lost me there.

a) We have a scatter on reel 5, then we need exactly two other scatters from the first four reels. 6 combos. So reel 5 = (12/40), other four = (3/35)*(3/35)*(32/35)*(32/35), altogether:

Can you explain why is reel 5 (12/40) thanks, and are there not 8 combos for any Two scatters on first 4 reels?
1a,2a,3 1a,2a,4 1a,2b,3 1a,2b,4 1b,2a,3 1b,2b,3 1b,2a,4 1b,2b,4

or
1,2,3, 1,2,4, 1,3,4 2,3,4 combosX2 because there are Two scatters on reels 1 and 2

Thanks for checking it out KK, I never knew dogfarther was a clone,are you sure, the reel lengths appear to be different for a start.
I will have a look.
If your layout is accurate then it makes the my results even more curious.

 

[Rusty]

My rough calculations give the chances of a feature on the dogfather slot as 1 in 80 for any of the 6 combos including reel 5
and 1 in 432 for any of the other four combos

so around 6 in 400+ = 66+ a good guesstimate would be 1 in 70.
Very low

Ahh just seen Zoozies post 1 in 76 still low.

How do you calculate the 816480 Zoozie?

 

[Zoozie]

How do you calculate the 816480 Zoozie?

I simply count them 1 by 1 from checking all 35*35*35*36*40 possible
spin results. Well, the slotanalyzer does that.

 

[Rusty]

I simply count them 1 by 1 from checking all 35*35*35*36*40 possible
spin results. Well, the slotanalyzer does that.

Thanks,is there anyway to verify the data?
I was hoping you had a formula, there must be one.

Anyway do you know what the RTP is for the paytable wins alone including scatter so we can calculate what the approx RTP is of the feature?

 

[gerilege]

Nope, lost me there.

a) We have a scatter on reel 5, then we need exactly two other scatters from the first four reels. 6 combos. So reel 5 = (12/40), other four = (3/35)*(3/35)*(32/35)*(32/35), altogether:

Can you explain why is reel 5 (12/40) thanks, and are there not 8 combos for any Two scatters on first 4 reels?
1a,2a,3 1a,2a,4 1a,2b,3 1a,2b,4 1b,2a,3 1b,2b,3 1b,2a,4 1b,2b,4

or
1,2,3, 1,2,4, 1,3,4 2,3,4 combosX2 because there are Two scatters on reels 1 and 2

Thanks for checking it out KK, I never knew dogfarther was a clone,are you sure, the reel lengths appear to be different for a start.
I will have a look.
If your layout is accurate then it makes the my results even more curious.

There are 4 scatters on that real, and each scatter is in the view 3 times (as you pointed that out to me fortunately last time ), hence 3*4 = 12, and there are 40 symbols on the reel.

Exactly 2 scatters on 4 reels:
On this second Halloweenies slot there are only 1 scatter on each of the first four reels. Scatters are on reels 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4, that's 6.
Ok now?

 

[gerilege]

As for verifying Zoozie's data, you can use my first formula with the s1, ..., l2 numbers. I used a simplified version of that in my post, if you would like to do it yourself, you have to take into consideration that only 3 reels are 35 long, one is 36, and one is 40 so it takes a bit more time to sum up the calculations.
Hint: For 3 scatters, you have 3 reels with 35 symbols, one with 36 symbols, one scatter each, and one 40-long reel with 4 scatters. You will have the following cases:
3 scatters:
-3 scatter on the 35-long reels (1 combo).
-1 scatter on the 40-long reel, 1 on the 36-long and 1 on one of the 35-longs (3 combos).
-1 scatter on the 40-long reel, 2 on the 35-longs (3 combos).
-1 scatter on the 36-long reel, 2 on the 35-longs (3 combos).
4 scatters:
You don't have a scatter on a 35-long reel (3 combos)
You don't have a scatter on the 36-long reel (1 combo)
You don't have a scatter on the 40-long reel (1 combo)
5 scatters: trivial, (3*3*3*3*12)/(35*35*35*36*40).
Please let me know if you would like work this out yourself, or you need further hints. There's an alternative way to do these kind of calculations, but let's stick to one method at the same time. When you have the results we can verify it with the other method if you wish. Keep in mind that you always have to include each reel into the product for 5 reels, so either the probability that a reel has a scatter, or the probability of it does not have.(i.e not only (3/35)*(3/35)*(3/35), but (3/35)*(3/35)*(3/35)*(33/36)*(28/40) for the first line above for 3 scatters)

 

[Rusty]

There are 4 scatters on that real, and each scatter is in the view 3 times (as you pointed that out to me fortunately last time ), hence 3*4 = 12, and there are 40 symbols on the reel.

Exactly 2 scatters on 4 reels:
On this second Halloweenies slot there are only 1 scatter on each of the first four reels. Scatters are on reels 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4, that's 6.
Ok now?

Ahh Halloweenies, thought we were still on Moonshine at that point

 

[Rusty]

As for verifying Zoozie's data, you can use my first formula with the s1, ..., l2 numbers. I used a simplified version of that in my post, if you would like to do it yourself, you have to take into consideration that only 3 reels are 35 long, one is 36, and one is 40 so it takes a bit more time to sum up the calculations.
Hint: For 3 scatters, you have 3 reels with 35 symbols, one with 36 symbols, one scatter each, and one 40-long reel with 4 scatters. You will have the following cases:
3 scatters:
-3 scatter on the 35-long reels (1 combo).
-1 scatter on the 40-long reel, 1 on the 36-long and 1 on one of the 35-longs (3 combos).
-1 scatter on the 40-long reel, 2 on the 35-longs (3 combos).
-1 scatter on the 36-long reel, 2 on the 35-longs (3 combos).
4 scatters:
You don't have a scatter on a 35-long reel (3 combos)
You don't have a scatter on the 36-long reel (1 combo)
You don't have a scatter on the 40-long reel (1 combo)
5 scatters: trivial, (3*3*3*3*12)/(35*35*35*36*40).
Please let me know if you would like work this out yourself, or you need further hints. There's an alternative way to do these kind of calculations, but let's stick to one method at the same time. When you have the results we can verify it with the other method if you wish. Keep in mind that you always have to include each reel into the product for 5 reels, so either the probability that a reel has a scatter, or the probability of it does not have.(i.e not only (3/35)*(3/35)*(3/35), but (3/35)*(3/35)*(3/35)*(33/36)*(28/40) for the first line above for 3 scatters)

I think I can simplify this further.
AS my main interest of concern is how often the feature will trigger I will neglect 4 and 5 scatter triggers as this makes no difference to the figures.

This also simplifies our calculations as we can now say;

3 scatters on the 35 symbol reels=27 possible reel positions(3*3*3)*36*40(other reel positions)*1(combos)=38880

1 scatter on 40symbol reel, 1 on 36 and 1 on any of 35=108(12*3*3)*35*35*3=396900

1 scatter on 40 symbol reel, 2 on 35=108(12*3*3)*36*35*3=408240

1 scatter on 36 symbol reel and 2 on 35=27(3*3*3)*40*35*3=113400

so all possible reel positions containing at least 3 scatters is
38880+396900+408240+113400=957420

As we know total combinations are 61740000/957420=64.48#

I think this method is simple and accurate but if you see any glaring errors let me know, which of course you will

 

[KasinoKing]

Well, it's quite late, I've been working hard all day & drinking a bit this evening and those figures are making my head spin!
I'll try to have a look in the morning.

In the meantime, here's my (much simpler I think) way of working out the probabilities (which may be seriously flawed, as Zoozie said):-

Lets take DogFather, with the reel layouts I posted above:-

There are 10 ways of hitting three scatters, the first is on reels 1, 2 & 3.
There is 1 scatter per reel, therefore there are 3 reel positions which will have a scatter showing.
So the number of ways you can hit those scatters is 3 x 3 x 3 = 27 ways.
The total number of different reel positions is 35 x 35 x 35 = 42,875
So the chances of getting 3 scatters on the first 3 reels is 42,875/27 = 1 in 1,587.96
(Or 6.30 times in 10,000 spins - you'll see why I quote that later...)

Does everyone agree with that? If not, what's wrong?

Now let's look at reels 1, 2 & 5:-
Reel 5 has 4 scatters, so there are 12 reel positions to have 1 scatter showing.
So the number of ways you can hit those scatters is 3 x 3 x 12 = 108 ways.
The total number of different reel positions is 35 x 35 x 40 = 49,000
So the chances of getting 3 scatters on these 3 reels is 49,000/108 = 1 in 453.17
(Or 22.04 times in 10,000 spins)

So I used the same calculation for all 10 reel possibilities, getting the number of times each combo should appear in 10,000 spins. Then just add them all together and divide into 10,000.
I originally did these calculations using fractions - and got exactly the same result.

Don't forget, my figure is for 3 or more scatters across the 5 reels, so also includes getting 4 or 5.

My spreadsheet:-

 

[Rusty]

Having done some further analysis of the paytable and win combinations it appears that this would account for roughly 50% RTP, leaving around 45% split between bonus and free spin features.

There are One or Two strange idiosyncrasies though.

For example there are less Glove combinations than Rabbit, knukkleD or cigar and yet it pays less by far than any other symbol?

Rabbit,knukkleD and cigar have exactly the same reel layout and yet they all have different paytable values?

The chances of hitting a Jackpot are over 15 Million to one!

Even the chance of hitting 5 Lady dogs is close to 1 Million to One.

The best paying symbols are the Cigar, Rabbit and revolver while the Glove is amongst the worse.

A lot of the common symbols are grouped so as to be nearly all in one half of a reel layout.

As we can see from some of the above the paytable is not based on the probability of outcome for each symbol.

Maybe MG have some simple answers as to why these oddities exist etc.
Then again maybe they don't.
Anyone here have any thoughts?

 

[Rusty]

Well, it's quite late, I've been working hard all day & drinking a bit this evening and those figures are making my head spin!
I'll try to have a look in the morning.

In the meantime, here's my (much simpler I think) way of working out the probabilities (which may be seriously flawed, as Zoozie said):-

Lets take DogFather, with the reel layouts I posted above:-

There are 10 ways of hitting three scatters, the first is on reels 1, 2 & 3.
There is 1 scatter per reel, therefore there are 3 reel positions which will have a scatter showing.
So the number of ways you can hit those scatters is 3 x 3 x 3 = 27 ways.
The total number of different reel positions is 35 x 35 x 35 = 42,875
So the chances of getting 3 scatters on the first 3 reels is 42,875/27 = 1 in 1,587.96
(Or 6.30 times in 10,000 spins - you'll see why I quote that later...)

Does everyone agree with that? If not, what's wrong?

Now let's look at reels 1, 2 & 5:-
Reel 5 has 4 scatters, so there are 12 reel positions to have 1 scatter showing.
So the number of ways you can hit those scatters is 3 x 3 x 12 = 108 ways.
The total number of different reel positions is 35 x 35 x 40 = 49,000
So the chances of getting 3 scatters on these 3 reels is 49,000/108 = 1 in 453.17
(Or 22.04 times in 10,000 spins)

So I used the same calculation for all 10 reel possibilities, getting the number of times each combo should appear in 10,000 spins. Then just add them all together and divide into 10,000.
I originally did these calculations using fractions - and got exactly the same result.

Don't forget, my figure is for 3 or more scatters across the 5 reels, so also includes getting 4 or 5.

My spreadsheet:-

Thanks KK that is confirmation of my results but with greater clarity.

I have not played this game much but does this seem correct when playing the game?

Now taking that figure and dividing into a full cycle of spins we get how many free spin triggers we could expect. 949846
Multiply this by 13 the amount of free spins awarded and you get
12,347,998 free spins in a full cycle but these free spins would yield another 2.5 million spins in retriggers and so on.
Basically this adds up to roughly 16 million free spins paying X3 prizes so 48 million if you like in a cycle of 62 million spins

If my paytable calculations are correct that would pay around 24 million coins around 39% RTP which when added to around 50% RTP off wins alone would leave the bonus round to about 7% RTP

If only we knew what the expected bonus RTP was.

 

[KasinoKing]

Maybe MG have some simple answers as to why these oddities exist etc.
Then again maybe they don't.
Anyone here have any thoughts?

Now you've totally lost me!

The number of win combinations v paytable does not ever have to have any connection - that's irrelevant.

Same with the layout of the reels - irrelevant.
If each reel went Glove, Glove, Glove, Glove, Revolver, Revolver, Revolver, etc... the return % would still be exactly the same.

Chance of getting 'Jackpot' of 5 wilds = 17.5 x 17.5 x 35 x 36 x 40 = 15,435,000 on 1 winline.
(2 wilds on reels 1 & 2)
Divide by the number of winlines (20) = 1 in 771,750.

5 Lady Dogs (including wilds) = 1 in 42,875

Those are the odds if playing max lines, obviously.

 

[Rusty]

Now you've totally lost me!

The number of win combinations v paytable does not ever have to have any connection - that's irrelevant.

Same with the layout of the reels - irrelevant.
If each reel went Glove, Glove, Glove, Glove, Revolver, Revolver, Revolver, etc... the return % would still be exactly the same.

Chance of getting 'Jackpot' of 5 wilds = 17.5 x 17.5 x 35 x 36 x 40 = 15,435.00 on 1 winline.
(2 wilds on reels 1 & 2)
Divide by the number of winlines (20) = 1 in 771,750.

5 Lady Dogs (including wilds) = 1 in 42,875


Those are the odds if playing max lines, obviously.

You have missed the point completely.

AS you say the symbol positions do not matter so why have them all bunched up instead of evenly scattered?
You are looking at it back to front.

More importantly you would expect a random slot paytable to corrospond with the probability of outcome, why does that not make sense to you?

Would you expect to play roulette and have number 6 pay 25/1 and number 33 45/1?

ps all my odds are based on single line I should have mentioned that, not many people play one line do they but it is still more accurate to state 1 million to one as you have to multiply your stake to achieve better odds.

 

[KasinoKing]

You have missed the point completely.

AS you say the symbol positions do not matter so why have them all bunched up instead of evenly scattered?

Because it doesn't make the slightest bit of difference.
Why would the slot designers waste time trying to spread them evenly on a video slot?
Most MG slots are laid out so you can't see the same symbol twice in one 'window' - that's all that matters (to them).
Some other softwares don't worry about this at all! (You play Grand Virtual!)

More importantly you would expect a random slot paytable to correspond with the probability of outcome, why does that not make sense to you?

I would expect that, yes. But again it makes no difference & I can understand why they might do it for various reasons.

Would you expect to play roulette and have number 6 pay 25/1 and number 33 45/1?

Only if I was betting on number 33...

 

[Rusty]

Because it doesn't make the slightest bit of difference.
Why would the slot designers waste time trying to spread them evenly on a video slot?
Most MG slots are laid out so you can't see the same symbol twice in one 'window' - that's all that matters (to them).
Some other softwares don't worry about this at all! (You play Grand Virtual!)


I would expect that, yes. But again it makes no difference & I can understand why they might do it for various reasons.


Only if I was betting on number 33...

Sorry KK you are not seeing it are you.

The question is why group the symbols not why not group them.
It is an unatural design plain and simple and I am curious as to why take the trouble that is all.

As for the second part it makes all the difference in the World.

It proves the slot is not designed around probability of outcome.
If it were you would design the paytable around these probabilities.
Why go out of your way to have a paytable that does not corrospond with probability?

You say you have some reasons as to why say a 3/1 shot pays 1/1 while 1/1 shot pays 3/1 (hypothetical example for clarity) I would like to know them because I can not for the life of me understand why.

It does not prove anything and the slot could still have expected RTP and be random but it sure as hell is a strange way to go about designing a random slot.

I need a break I don't know about you?

 

[KasinoKing]

KK, it would be easier to find the missing piece in your calculations if you would publish it. Actually I believe I probably know where you do not consider a particular thing, it is a common issue which people usually do not consider. Why am I saying this? Your 5 scatters probability is OK, but the feature trigger and 4 scatters are significantly off.

Hmmm... I'm a bit tired now, but you keep quoting exactly 1 scatter on 2 reels, or exactly no scatter on reel 3 etc...

My figures for getting 3 scatters is based on only looking at 3 of the 5 reels at a time - I'm ignoring whether a scatter lands on the other reels or not.

In other words, I'm saying (for Moonshine) the chance of getting 3 or more scatters is 1 in 75.
Now this could be wrong - but I don't see how.

Using my same formula I get the chance of the feature on Thunderstruck as 1 in 123.6 (which seems to be the accepted figure), but 1 in 55 on DogFather which seems to me to be way too low!
But I can't see the flaw in my calculations...

 

[Rusty]

Hmmm... I'm a bit tired now, but you keep quoting exactly 1 scatter on 2 reels, or exactly no scatter on reel 3 etc...

My figures for getting 3 scatters is based on only looking at 3 of the 5 reels at a time - I'm ignoring whether a scatter lands on the other reels or not.

In other words, I'm saying (for Moonshine) the chance of getting 3 or more scatters is 1 in 75.
Now this could be wrong - but I don't see how.

Using my same formula I get the chance of the feature on Thunderstruck as 1 in 123.6 (which seems to be the accepted figure), but 1 in 55 on DogFather which seems to me to be way too low!
But I can't see the flaw in my calculations...

I think you have made a typo there KK your calculations show 1 in 65 the same as mine.
It is tiring making all these manual calculations and you should see mine for the paytable which includes instances of 3,4 and 5 of everything and coins paid etc Gahhh.
It would of been quicker for me to write some code which I guess I should do because handling numbers as big as telephone numbers is bound to lead to human error.
At least it is cheaper than playing Dogfather

 

[Zoozie]

Here is how to calculate scatter probabilities by hand.

Let p1,p2,p3,p4,p5 be the probabilities for a scatter on reel 1,2,3,4,5.
Define q1 =(1-p1), q2=(1-p2), .... , q5=(1-p5). Ie. this is the probability for NOT hitting a scatter on that reel.

Then

P(excactly 5 scatters) = p1*p2*p3*p4*p5 ( I think we all agree on that one)

P(excactly 4 scatters) = p1*p2*p3*p4*q5 + p1*p2*p3*q4*p5 + p1*p2*q3*p4*p5 + p1*q2*p3*p4*p5 + q1*p2*p3*p4*p5 (notice one is not a scatter, taking all combinations where one is not scatter) There is a total of 5 combinations .

P(excactly 3 scatters) = p1*p2*p3*q4*q5 + p1*p2*q3*p4*q5 + ...... + q1*q2*p3*p4*p5 (taking all combinations where 2 reels are not scatter) There is a total of 10 combinations .

You can not use binomial distribution since probabilites for a scatter on each reel varies.

 

[jetset]

Just a (retrospective) thought....why were Cryptologic-Wagerlogic slots not included in this poll?

 

[gerilege]

I think I can simplify this further.
AS my main interest of concern is how often the feature will trigger I will neglect 4 and 5 scatter triggers as this makes no difference to the figures.

This also simplifies our calculations as we can now say;

3 scatters on the 35 symbol reels=27 possible reel positions(3*3*3)*36*40(other reel positions)*1(combos)=38880

1 scatter on 40symbol reel, 1 on 36 and 1 on any of 35=108(12*3*3)*35*35*3=396900

1 scatter on 40 symbol reel, 2 on 35=108(12*3*3)*36*35*3=408240

1 scatter on 36 symbol reel and 2 on 35=27(3*3*3)*40*35*3=113400

so all possible reel positions containing at least 3 scatters is
38880+396900+408240+113400=957420

As we know total combinations are 61740000/957420=64.48#

I think this method is simple and accurate but if you see any glaring errors let me know, which of course you will

What your are doing is a second method called logical sieve. But you can't ignore 4 or 5 scatters, when trying to find the probability for 3 or more scatters, because you will get slightly off results, I will show you why.
5 scatters award one feature, as do 4 scatters. If you calculate by the whole length of the remaining non-scatter reels, then you also include 4 or 5 scatters, which award the same feature as 3 scatters would. With other words you calculate the 5 scatter feature triggers ten times, as when you have 5 scatters, you have ten combos of 3 scatters, but it is only one trigger, but you sum up them along the 3 scatters ten times above. Similarly when you have 4 scatters, you have 4 combos of 3 scatters as well, so you also sump up the 4 scatters 4 times more then the number of features they trigger. So when you're done with the above method, you have to deduct the probability of 5 scatters ten times, and the 4 scatters 4 times.
Last step is to realize that 5 scatters are also 5 combos of 4 scatters, and in the step above you deducted them twice, so you have to add back 5 times the 5 scatters.
I will be most likely offline for the rest of the day, but I'm fully confident you can finish it now After that I will show you a third, backward logic method to get the same result.

 

[KasinoKing]

(Post #71) I think you have made a typo there KK your calculations show 1 in 65 the same as mine.

If you read my post that you were quoting again, you will see I was talking about Moonshine, not DogFather.


I missed this one last night!

I think I can simplify this further.
AS my main interest of concern is how often the feature will trigger I will neglect 4 and 5 scatter triggers as this makes no difference to the figures.

This also simplifies our calculations as we can now say;

3 scatters on the 35 symbol reels=27 possible reel positions(3*3*3)*36*40(other reel positions)*1(combos)=38880

1 scatter on 40symbol reel, 1 on 36 and 1 on any of 35=108(12*3*3)*35*35*3=396900

1 scatter on 40 symbol reel, 2 on 35=108(12*3*3)*36*35*3=408240

1 scatter on 36 symbol reel and 2 on 35=27(3*3*3)*40*35*3=113400

so all possible reel positions containing at least 3 scatters is
38880+396900+408240+113400=957420

As we know total combinations are 61740000/957420=64.48#

I think this method is simple and accurate but if you see any glaring errors let me know, which of course you will

Same result as I got (for DogFather)... well at least we agree on something!

{Edit} Missed this one too!

Sorry KK you are not seeing it are you.

The question is why group the symbols not why not group them.
It is an unatural design plain and simple and I am curious as to why take the trouble that is all.

As for the second part it makes all the difference in the World.

It proves the slot is not designed around probability of outcome.
If it were you would design the paytable around these probabilities.
Why go out of your way to have a paytable that does not corrospond with probability?

You say you have some reasons as to why say a 3/1 shot pays 1/1 while 1/1 shot pays 3/1 (hypothetical example for clarity) I would like to know them because I can not for the life of me understand why.

It does not prove anything and the slot could still have expected RTP and be random but it sure as hell is a strange way to go about designing a random slot.

I need a break I don't know about you?

Well...
A) I haven't worked out the chances of any of the lower symbols hitting - so I don't even know if what you're saying is true in the first place.
B) Even if it is true, I'm sorry but to be honest I don't give a flying toss about that. All I want to know is the correct odds of hitting the bonus features.
c) I've had a break now - but another one sounds like a good idea!

What I really want is Gerilege or Zoozie to look and respond to my post #63...
Please?