Poll slots 2

[zap987]

This combination should account for 30% of possible feature triggering combinations and yet it is much lower than this.
These anomalous results alone strongly suggest to me weighting or grouping and/or the use of algorithms even though there may still be random element.

He already gathered the data, just look at the reels for Moonshine and it's obvious why you will get 2 scatters on the first 2 reels and then nothing.

Both the first 2 reels have 35 stops and 2 scatter bottles, while the last 3 reels have 38,39 and 40 stops with 1 bottle on each reel. Unless I fail at math you have about 22% chance to actually hit a 3rd scatter starting with 2 on the first 2 reels which you will get 1/34 spins.

 

[Zoozie]

Zap987 allready gave a good answer why you often will see patterns
with 2 or 3 scatters. Moonshine is an good example. On RTG slots the last reel if often extremely long with only 1 scatter so this will also be noticable here.

Yes, some casinos has cheated. But have MG etc. ever been taken in
obvious cheating? There is no point in cheating with the slots really (except a progressive jackpot). BJ on the other hand can have a very small house egde and it would be much much better for the casino to cheat with.

 

[KasinoKing]

He already gathered the data, just look at the reels for Moonshine and it's obvious why you will get 2 scatters on the first 2 reels and then nothing.

Both the first 2 reels have 35 stops and 2 scatter bottles, while the last 3 reels have 38,39 and 40 stops with 1 bottle on each reel. Unless I fail at math you have about 22% chance to actually hit a 3rd scatter starting with 2 on the first 2 reels which you will get 1/34 spins.

This reminds me I have a question for anyone who knows, (Zoozie?)

What's the formula for working out the chance of getting a scatter on 2 or more reels?

e.g. If you had 2 reels, both 30 symbols with 1 scatter, the chance of getting a scatter on one reel is 1 in 10.
Therefore I assume the chance of getting a scatter on either reel is 1 in 5. (?)

But how do you work it out if say 1 reel is 1 in 10 & the other 1 in 15?
Or the chance of hitting one scatter across 3 reels, all with different numbers of symbols?

Anyone know?
Thanks!

 

[gerilege]

This reminds me I have a question for anyone who knows, (Zoozie?)

What's the formula for working out the chance of getting a scatter on 2 or more reels?

e.g. If you had 2 reels, both 30 symbols with 1 scatter, the chance of getting a scatter on one reel is 1 in 10.
Therefore I assume the chance of getting a scatter on either reel is 1 in 5. (?)

But how do you work it out if say 1 reel is 1 in 10 & the other 1 in 15?
Or the chance of hitting one scatter across 3 reels, all with different numbers of symbols?

Anyone know?
Thanks!

Normally it is easier to calculate the chance of not getting any (or 1 or more, 2 or more, etc) scatters, and subtract the probability from 1. In your case you have two reels, 30 symbols on each reel and 1 scatter. The chance of not getting a scatter on reel1 is 9/10, and it is the same on reel2. Therefore the probability of you won't get a scatter on any of the two reels is (9/10)*(9/10), which is 81/100. So the chance of getting at least one scatter on any of the two reels is 19/100. Now the chance of getting exactly two scatters is 1/10*1/10 = 1/100, combine the two results, and the chance of getting exactly one scatter on the two reels is 19/100-1/100=18/100. This is how it usually works, it gets a bit complicated with more reels, but the principle is the same. It can be generalized to reels with different length and scatter symbols, to be continued.

 

[winbig]

Normally it is easier to calculate the chance of not getting any (or 1 or more, 2 or more, etc) scatters, and subtract the probability from 1. In your case you have two reels, 30 symbols on each reel and 1 scatter. The chance of not getting a scatter on reel1 is 9/10, and it is the same on reel2. Therefore the probability of you won't get a scatter on any of the two reels is (9/10)*(9/10), which is 81/100. So the chance of getting at least one scatter on any of the two reels is 19/100. Now the chance of getting exactly two scatters is 1/10*1/10 = 1/100, combine the two results, and the chance of getting exactly one scatter on the two reels is 19/100-1/100=18/100. This is how it usually works, it gets a bit complicated with more reels, but the principle is the same. It can be generalized to reels with different length and scatter symbols, to be continued.

The problem is cracking the reels to figure out how many of each symbol & total symbols are on each one. If you don't have that information, you'll never figure out the percentages.

 

[Rusty]

You have almost answered your own question by working out the answer in your first example.
If you are getting different fractions as in your second example then you can find a common denominator.
2X15=30 > 3X10=30 > 2+3=5 the chances are 5 in 30 or 1 in 6

 

[gerilege]

You have almost answered your own question by working out the answer in your first example.
If you are getting different fractions as in your second example then you can find a common denominator.
2X15=30 > 3X10=30 > 2+3=5 the chances are 5 in 30 or 1 in 6

Sorry, this result is a slightly off. Actually getting one or two scatters with 1/15 and 1/10 chances is 1-((27/30)*(28/30)) = 4/25, getting exactly 1 scatter is 1-((27/30)*(28/30)) - (1/15*1/10).

The method I described with binomial distribution in my previous post works quite well with uniform reel lenghts. For different reel lenghts and scatter frequency you need to work a bit more.

 

[Rusty]

Moonshine reels and scatter odds;

I have just some quick calculations based on what Zapp says the reel layout is for Moonshine.

Simplifying things that means that the chances of a scatter on reel One and Two are roughly 1 in 6 and 1 in 13 on the other reels.

So there are 3 X 1/468 combos and 6 X 1/1014 combos and 1 X 1/2197 combo.

This would mean the feature triggering at an average of less than 90 spins?

Do you have the full calculations for this Zoozie?

Also as regards the frequency of the first Two scatters;

If the reel layout is as described then this would indeed account in part for the effect I describe but the chances of the first Two scatters are 1 in 36 and the chances of Two scatters appearing on any other 2 reels are not dissimilar as there are 6 combinations as opposed to just one.
Is that what we see happen?

The chances of triggering the feature after hitting the first Two scatters is roughly 1 in 5 but is that what we see happen?

Is the reel layout for Halloweenies similar to Moonshine?

 

[Rusty]

Sorry, this result is a slightly off. Actually getting one or two scatters with 1/15 and 1/10 chances is 1-((27/30)*(28/30)) = 4/25, getting exactly 1 scatter is 1-((27/30)*(28/30)) - (1/15*1/10).

The method I described with binomial distribution in my previous post works quite well with uniform reel lenghts. For different reel lenghts and scatter frequency you need to work a bit more.

yup, thanks, you are going to love my rough workings above then

 

[gerilege]

If you have a different number of scatters on each reel and different reel lenght, then it's a bit more complicated. Let's say you have now 5 reels, the length of the reels is l1, l2, l3, l4, l5, and let's assume that each scatter is at least two other symbols between any scatters on the same reel and let's denote (the number of scatters)*3 on each reels as s1, s2, s3, s4, s5 respectively (i.e. if you have 1 scatter on reel4, then s4=3).
Now the formulas:
Exactly 0 scatter on all 5 reels: (((l1-s1)/l1)*((l2-s2)/l2)*...*((l5-s5)/l5))
Exactly 5 scatters: s1/l1*s2/l2*...*s5/l5

Chances of exactly one scatter on reel1 = chances of a scatter on reel1 and no scatter on other reels = (s1/l1)*(((l2-s2)/l2)*...*((l5-s5)/l5))
Chances of exactly one scatter on reel2 = chances of a scatter on reel2 and no scatter on other reels = (s2/l2)*(((l1-s1)/l2)*((l3-s3)/l3)...*((l5-s5)/l5))
and so on to reel5.
Chances of exactly 1 scatter on all reels = sum the above probabilities for all 5 reels.

Chances of exactly 4 scatters = chance that there is exactly one reel without a scatter.
Similarly as above, chance of reel1 doesn't have a scatter but all other reels do = ((l1-s1)/l1)*(s2/l2*...*s5/l5)
Chance of reel2 doesn't have a scatter but all other reels do = ((l2-s2)/l2)*(s1/l1*s3/l3...*s5/l5)
and so on for the remaining 3 reels.
Chances of exactly 4 scatters = sum the above probabilities for all 5 reels.

Chance of exactly 2 scatters = there are exactly two reels with a scatter and three without a scatter:
There are 10 combinations, scatters are on reel 1&2 1&3 1&4 1&5 2&3 2&4 2&5 3&4 3&5 4&5.
Chance of scatters only on reel 2 and 5 for instance = (reel 2 and 5 have a scatter) * (reel 1,3,4 do not have) = (s2/l2*s3/l3)*(((l1-s1)/l1)*((l3-s3)/l3)*((l4-s4)/l4))
Chance of scatters only on reel 3 and 4 for instance = (s3/l3*s4/l4)*(((l1-s1)/l2)*((l2-s2)/l2)*((l5-s5)/l5))
and so on with the other 8 combinations.
Chance of exactly 2 scatters = sum the above ten cases up.

Chance of exactly 3 scatters = there are exactly two reels without a scatter and three with a scatter:
There are 10 combinations, there are no scatters on reel 1&2 1&3 1&4 1&5 2&3 2&4 2&5 3&4 3&5 4&5, and there are on all the other reels.
Chance of no scatters on reel 2 and 5 and scatters on all the other reels for instance = (reel 2 and 5 do not have a scatter) * (reel 1,3,4 do have) = (((l2-s2)/l2)*((l5-s5)/l5))*(s1/l1*s3/l3*s4/l4).
Chance of no scatters on reel 1 and 3 and scatters on all the other reels for instance = (reel 1 and 3 do not have a scatter) * (reel 2,4,5 do have) = (((l1-s1)/l1)*((l3-s3)/l3))*(s2/l2*s4/l4*s5/l5).
And so on for the remaining 8 combinations.
Chance of exactly 3 scatters = sum up the above ten probabilities.

Basically that's all cases I guess. With the above info you can solve the Moonshine problem, but there is an easier way to do that in that special case, I will have another look on that.

 

[gerilege]

Let's see the above formulas in action for the Moonshine problem. We have fixed the first two reels, so we have 3 reels remaining. As it was written: The last 3 reels have 38,39 and 40 stops with 1 bottle on each reel. Therefore s3=s4=s5=3, while l3=38, l4=39, l5=40. We want to know what is the chance that there is no scatter on reel3, reel4 and reel5. That is (35/38)*(36/39)*(37/40) = 777/988. So the chances that you will get the feature (get at least another scatter on the remaining reels) when you get 2 scatters on the first 2 reels is 1-(777/988) = 21,35%.
But this is only a minority of all feature triggers of course, provided that reel1 and reel2 has a scatter. To get the probability of a feature trigger you have to sum up the probabilities of getting exactly 3, 4 and 5 scatters as written in my prevous post (Or sum up the probability of getting 0,1 and 2 scatters, and deduct the sum from 1).

 

[Rusty]

Let's see the above formulas in action for the Moonshine problem. We have fixed the first two reels, so we have 3 reels remaining. As it was written: The last 3 reels have 38,39 and 40 stops with 1 bottle on each reel. Therefore s3=s4=s5=1, while l3=38, l4=39, l5=40. We want to know what is the chance that there is no scatter on reel3, reel4 and reel5. That is (37/38)*(38/39)*(39/40) = 37/40. So the chances that you will get the feature (get at least another scatter on the remaining reels) when you get 2 scatters on the first 2 reels is 1-(37/40) = 3/40 = 7,5%.
But this is only a minority of all feature triggers of course, provided that reel1 and reel2 has a scatter. To get the probability of a feature trigger you have to sum up the probabilities of getting exactly 3, 4 and 5 scatters as written in my prevous post (Or sum up the probability of getting 0,1 and 2 scatters, and deduct the sum from 1).

Correct me if I am wrong but have you missed that a scatter can be placed in any of 3 reel positions and the figures should be (35/38)*(36/39)*37/40)=92.5%?
That is odd I get the 7.5% using these figures?
Maybe I am missing something.

Even so I can not see how this can be correct even though I see nothing wrong with the formula as a rough working tells us that we have Three 1 in 13 chances to hit another scatter which is clearly more than 7.5% chance?

 

[gerilege]

Correct me if I am wrong but have you missed that a scatter can be placed in any of 3 reel positions and the figures should be (35/38)*(36/39)*37/40)=92.5%?
That is odd I get the 7.5% using these figures?
Maybe I am missing something.

Even so I can not see how this can be correct even though I see nothing wrong with the formula as a rough working tells us that we have Three 1 in 13 chances to hit another scatter which is clearly more than 7.5% chance?

You are right. I will first edit all the above posts then I will have a look on your last question.

 

[Zoozie]

I am too lazy for the math now. But you can just do this easy with the slotanalyzer. (remember to remove the "," between the symbols on each reel)

Kimss supplied the following reels for moonshine:
0,9,3,8,5,7,6,4,7,P,9,2,8,6,3,S,7,6,5,8,6,0,9,7,4,2,P,5,8,6,S,9,1,8,7
0,6,9,7,3,8,S,7,2,8,5,9,P,7,3,6,4,2,6,S,8,7,4,P,8,5,9,0,6,1,7,5,8,6,9
0,9,7,4,2,5,8,7,6,S,9,3,8,5,7,1,6,0,9,4,8,6,7,P,9,2,8,4,7,6,P,9,1,8,3,6,5,8
0,6,9,3,7,9,4,8,7,2,0,1,4,7,6,S,9,5,8,7,6,0,9,6,8,5,4,1,P,7,3,8,P,5,2,8,4,9,5
0,7,6,8,4,7,5,9,S,3,1,4,7,5,P,8,1,9,0,7,8,2,5,6,P,1,3,9,P,6,7,4,8,5,9,2,6,8,4,9

S=scatter
(0 to 9 and P are just other symbols)
Paytable is not important if you only want the analyzer to calculate probability for feature. Just remember that S is the scatter. I did that and the result is:

Probability for getting a feature each spin: 826119/72618000~0.011376229034123771~1/88

Furthermore if you set #scatters to start feature to first 5,4 and then 3. Then by substraction you can find the exact probability for each of them.

 

[Rusty]

I am too lazy for the math now. But you can just do this easy with the slotanalyzer. (remember to remove the "," between the symbols on each reel)

Kimss supplied the following reels for moonshine:
0,9,3,8,5,7,6,4,7,P,9,2,8,6,3,S,7,6,5,8,6,0,9,7,4,2,P,5,8,6,S,9,1,8,7
0,6,9,7,3,8,S,7,2,8,5,9,P,7,3,6,4,2,6,S,8,7,4,P,8,5,9,0,6,1,7,5,8,6,9
0,9,7,4,2,5,8,7,6,S,9,3,8,5,7,1,6,0,9,4,8,6,7,P,9,2,8,4,7,6,P,9,1,8,3,6,5,8
0,6,9,3,7,9,4,8,7,2,0,1,4,7,6,S,9,5,8,7,6,0,9,6,8,5,4,1,P,7,3,8,P,5,2,8,4,9,5
0,7,6,8,4,7,5,9,S,3,1,4,7,5,P,8,1,9,0,7,8,2,5,6,P,1,3,9,P,6,7,4,8,5,9,2,6,8,4,9

S=scatter
(0 to 9 and P are just other symbols)
Paytable is not important if you only want the analyzer to calculate probability for feature. Just remember that S is the scatter. I did that and the result is:

Probability for getting a feature each spin: 826119/72618000~0.011376229034123771~1/88

Furthermore if you set #scatters to start feature to first 5,4 and then 3. Then by substraction you can find the exact probability for each of them.

:notworthy
Thanks Zoozie, talk about not seeing the Wood for the trees

So 1 in 88 chance of feature trigger? That seems rather low but is in keeping with my rough calculations.
I am unfamiliar with your slot analyzer and don't have a copy, is there a way to discover the chance of a feature trigger when the first Two scatters have appeared?
Can you send me it, I lost the original you sent me when my PC died?

 

[gerilege]

Correct me if I am wrong but have you missed that a scatter can be placed in any of 3 reel positions and the figures should be (35/38)*(36/39)*37/40)=92.5%?
That is odd I get the 7.5% using these figures?
Maybe I am missing something.

Even so I can not see how this can be correct even though I see nothing wrong with the formula as a rough working tells us that we have Three 1 in 13 chances to hit another scatter which is clearly more than 7.5% chance?

The posts are corrected now. However in your calculations (35/38)*(36/39)*(37/40) does not equal to 92.5%, but to ~78,65%. Therefore the chance of hitting at least one more scatter is 21,35%.

 

[Zoozie]

is there a way to discover the chance of a feature trigger when the first Two scatters have appeared

Can you send me it, I lost the original you sent me when my PC died?

Yes. But it depends on how early on the reels you hit them. Ie. how many reels are left to hit the scatter. (and of course take into probability for a scatter on each of these reels.). This is much easier than your original problem, because now you can just calculate chance of not hitting it=p and then it is (1-p)

For the slotanalyzer you can just look in my signature. It is an applet now.

 

[Rhyzz]

I'm a big fan of both MG and Playtech, although I have my gripes with both.

I'm 100% with what people are saying in the thread, in that MG don't seem to be realistic with expectations on their features.

 

[zap987]

I took my information about reel layout from

You do not have permission to view link Log in or register now.

which also has a bunch of interesting probabilities calculated.

 

[Rusty]

The posts are corrected now. However in your calculations (35/38)*(36/39)*(37/40) does not equal to 92.5%, but to ~78,65%. Therefore the chance of hitting at least one more scatter is 21,35%.

Yeah, not sure what happened there

Anyway a little better than1 in 5 which was my rough estimation yet they seem to trigger less than this from this situation.
I guess we would need an awful lot of data to verify this.

Does anyone have the reel layouts for Halloweenies as this is the worse offender as far as I am concerned.
Perhaps there are multiple scatters on the first Two reels here?
The thing is though the last Two scatters love to appear on this slot as well just as long as there is no other scatter in view it would really good to know the reels on this one.

Also does anyone know what percentage of the RTP is made up from the feature on these slots?
Moonshine triggers more than 1 in 80 spins on average so I assume the expected RTP on this feature is lower than most other slots?
I guess Zoozie could run the free spin reels through his analyzer and see what turns up.
Incidentally do we know why different reels are used for free spins, to me it suggest the slot is more likely to be random because there would be no need for different reels if you wanted to manipulate outcome.
Did I just say that?

I took my information about reel layout from

You do not have permission to view link Log in or register now.

which also has a bunch of interesting probabilities calculated.

Great little link that, thanks.

 

[KasinoKing]

This would mean the feature triggering at an average of less than 90 spins?

Correct - it's actually 1 in 75.
Don't forget you can view the reel layouts for this slot

You do not have permission to view link Log in or register now.

.

Is the reel layout for Halloweenies similar to Moonshine?

No.
I haven't plotted this one yet, but there are obviously 3 or 4 scatters on reel 5, and no more than 2 on any other reel.
Might be the next MG slot I look at...

 

[winbig]

I haven't plotted this one yet, but there are obviously 3 or 4 scatters on reel 5, and no more than 2 on any other reel.
Might be the next MG slot I look at...

Maybe that's why there's few to no winning SS's posted for this slot.....with that many 'dead' spots on reel 5, you'd almost have to have a horseshoe stuck up your a$$ to get 5oak...

 

[KasinoKing]

Maybe that's why there's few to no winning SS's posted for this slot.....with that many 'dead' spots on reel 5, you'd almost have to have a horseshoe stuck up your a$$ to get 5oak...

Neigh laddie!

 

[KasinoKing]

I haven't plotted this one yet, but there are obviously 3 or 4 scatters on reel 5, and no more than 2 on any other reel.
Might be the next MG slot I look at...

Just for you Rusty, I took a couple of hours out & plotted DogFather (a Halloweenies clone) using my new (top secret) method, and got a very surprising result!

Note: Not verified yet - if anyone would like to double-check - please do!

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?

DogFather reel layouts (provisional):-
Reel 1... Reel 2... Reel 3... Reel 4... Reel 5
DOGFATHER DOGFATHER DOGFATHER DOGFATHER DOGFATHER
GLOVE.... CIGAR.... BULLDOG.. REVOLVER. BULLDOG..
WHITE DOG LADY DOG. CIGAR.... GLOVE.... TWIN DOGS
KNUCKLE-D KNUCKLE-D TWIN DOGS CIGAR.... KNUCKLE-D
RABBIT... RABBIT... RABBIT... TWIN DOGS CIGAR....
HYDRANT.. GLOVE.... LADY DOG. RABBIT... REVOLVER.
GLOVE.... TWIN DOGS WHITE DOG GLOVE.... WHITE DOG
CIGAR.... KNUCKLE-D REVOLVER. CIGAR.... RABBIT...
RABBIT... WHITE DOG KNUCKLE-D GLOVE.... HYDRANT..
ItchyFlea RABBIT... CIGAR.... RABBIT... TWIN DOGS
KNUCKLE-D REVOLVER. RABBIT... CIGAR.... GLOVE....
GLOVE.... CIGAR.... KNUCKLE-D WHITE DOG REVOLVER.
CIGAR.... DOGFATHER HYDRANT.. TWIN DOGS CIGAR....
KNUCKLE-D RABBIT... REVOLVER. RABBIT... TWIN DOGS
DOGFATHER REVOLVER. TWIN DOGS HYDRANT.. REVOLVER.
CIGAR.... BULLDOG.. GLOVE.... CIGAR.... ItchyFlea
TWIN DOGS GLOVE.... REVOLVER. REVOLVER. CIGAR....
GLOVE.... HYDRANT.. WHITE DOG KNUCKLE-D LADY DOG.
RABBIT... CIGAR.... RABBIT... GLOVE.... RABBIT...
REVOLVER. KNUCKLE-D CIGAR.... RABBIT... CIGAR....
LADY DOG. RABBIT... REVOLVER. ItchyFlea TWIN DOGS
WHITE DOG WHITE DOG KNUCKLE-D LADY DOG. GLOVE....
REVOLVER. ItchyFlea BULLDOG.. BULLDOG.. ItchyFlea
RABBIT... REVOLVER. GLOVE.... KNUCKLE-D RABBIT...
KNUCKLE-D GLOVE.... RABBIT... REVOLVER. CIGAR....
BULLDOG.. RABBIT... CIGAR.... GLOVE.... REVOLVER.
REVOLVER. TWIN DOGS GLOVE.... RABBIT... KNUCKLE-D
CIGAR.... GLOVE.... ItchyFlea KNUCKLE-D RABBIT...
RABBIT... REVOLVER. KNUCKLE-D TWIN DOGS GLOVE....
KNUCKLE-D KNUCKLE-D REVOLVER. HYDRANT.. ItchyFlea
REVOLVER. CIGAR.... GLOVE.... REVOLVER. KNUCKLE-D
GLOVE.... GLOVE.... CIGAR.... WHITE DOG WHITE DOG
TWIN DOGS KNUCKLE-D RABBIT... KNUCKLE-D GLOVE....
REVOLVER. CIGAR.... TWIN DOGS TWIN DOGS HYDRANT..
CIGAR.... REVOLVER. KNUCKLE-D CIGAR.... KNUCKLE-D
. . . . . . . . . . . . . . . KNUCKLE-D REVOLVER.
. . . . . . . . . . . . . . . . . . . . ItchyFlea
. . . . . . . . . . . . . . . . . . . . RABBIT...
. . . . . . . . . . . . . . . . . . . . GLOVE....
. . . . . . . . . . . . . . . . . . . . KNUCKLE-D

(Itchy-Da-Flea is the scatter symbol - and I always thought it was a Frog before! )

 

[gerilege]

Correct - it's actually 1 in 75.
Don't forget you can view the reel layouts for this slot

You do not have permission to view link Log in or register now.

.


No.
I haven't plotted this one yet, but there are obviously 3 or 4 scatters on reel 5, and no more than 2 on any other reel.
Might be the next MG slot I look at...

I believe Zoozie's simulation is accurate @ 1:88 and not yours @ 1:75. To prove this mathematically let me make some simplifications, so the calculations become easier. Assume that the reel 3, 4 and 5 have only 38 symbol each, with one scatter on each reel. That would only increase the chance of the feature trigger, so if I show that even with this small gain the chance of feature is still less than 1:75, we are still ok. Reel 1 and two has 35 symbols each with 2 scatter. Now the cases for getting 3 scatters can be grouped as the following due to simmetry reasons:
a) Scatter on both reel 1 and 2 and exactly one other on any of the other 3 reels (3 combos).
b) Exactly one scatter on reel 1 and two, and exactly two scatters on reels 3,4 and 5 (6 combos).
c) No scatter on reel 1 and two and 3 scatters on the other reels (1 combo).
a) we have two scatters on reel 1 and 2 = 6/35*6/35. We have exactly on scatter on the remaining 3 reels, for 1 remaining reel to have scatter and the other two do not have: (3/38)*(35/38)*(35/38), and we have 3 combinations here, so altogether:
(6/35)*(6/35)*(3/38)*(35/38)(*35/38)*3
b) Example: we only have one on reel 1, do not have on reel 2 and have two on the remaining 3 reels.
The first two reels is (6/35)*(29/35), the remaining three (35/38)*(3/38)*(3/38). We have 3 combos for the last 3 reel, and the same 3 for reel 1 does not have a scatter and reel 2 does, so a total of six combos, altogether:
(6/35)*(29/35)*(35/38)*(3/38)*(3/38)*6.
c) We do not have on the first 2 = (29/35)*(29/35), we have on the remaining three: (3/38)*(3/38)*(3/38), and we only have one combination = (29/35)*(29/35)*(3/38)*(3/38)*(3/38)

4 scatters:
There are two cases due to simmetry reasons again:
a) We have scatters on both first two reels (3 combos)
b) We do not have a scatter on one of the first two reels (2 combos)
a) We have on the first two: (6/35)*(6/35), we do not have on one of the other 3 : (35/38)*(3/38)*(3/38). This is 3 times, so (6/35)*(6/35)*(35/38)*(3/38)*(3/38)*3
b) (6/35)*(29/35) for the first two reels, and *(3/38)*(3/38)*(3/38) for the remaining three, and 2 combinations, so (6/35)*(29/35)*(3/38)*(3/38)*(3/38)*2
5 scatters is 1:74710 (6*6*3*3*3/35/35/38/39/40), this is exact.
Now we have to sum these up together (5 scatter could come at the end, but it is so small that can be safely ignored):
(6/35)*(6/35)*(3/38)*(35/38)*(35/38)*3 +
(6/35)*(29/35)*(35/38)*(3/38)*(3/38)*6 +
(29/35)*(29/35)*(3/38)*(3/38)*(3/38) +
(6/35)*(6/35)*(35/38)*(3/38)*(3/38)*3 +
(6/35)*(29/35)*(3/38)*(3/38)*(3/38)*2 =

(3/38)*
{0,1438}= 0,011352 ~= 1:88

So even with the less non-scatter symbols on the last 3 reels we got 1:88, so it is absolutely sure that the chance for the trigger is less or equal than 1:88. Another intertesting fact is that the majority (52%) of the features will be triggered by scatters on the first two reel and a scatter on one of the last 3 reels.

I will have a look on the other slot you mentioned.