boochaka
boochaka
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April 15th, 2010 at 10:38:06 AM permalink
How is the average return of a free spin combination on a slot calculated?

I've been searching all around your site, and other info I have, and until now I have no clue.
The closer I have reached, is the example and calculations you made on the Atkins Diet Slot you designed,
but I can't yet see the formulas used there.

I want to calculate the EV of a combination like the one in Atkins diet,
and how can I calculate other with less or more spins / winning multipliers.
And knowing the EV of all the other combinations.

I hope I made myself clear, and appreciate your help with this.

Thanks in advance.
Wizard
Administrator
Wizard
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April 15th, 2010 at 2:38:00 PM permalink
The simple formula is [number free spins]*[average return from each free spin].

In the Atkins Diet machine the number of free spins is not obvious. You start with 10, but free spins can earn more free spins, infinitely. In that case, the expected final number of spins is n/(1-n*p), where n is the number of free spins earned per bonus, and p is the probability of the bonus per spin. In the the Atkin's case, the bonus probability is 0.011185, so the expected number of free spins per bonus is 10/(1-0.011185*10) = 11.259335.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
boochaka
boochaka
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April 16th, 2010 at 5:06:30 AM permalink
Quote: Wizard

The simple formula is [number free spins]*[average return from each free spin].

In the Atkins Diet machine the number of free spins is not obvious. You start with 10, but free spins can earn more free spins, infinitely. In that case, the expected final number of spins is n/(1-n*p), where n is the number of free spins earned per bonus, and p is the probability of the bonus per spin. In the the Atkin's case, the bonus probability is 0.011185, so the expected number of free spins per bonus is 10/(1-0.011185*10) = 11.259335.



Thanks, I will try with that and return here if I get stuck.
boochaka
boochaka
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May 12th, 2010 at 7:33:11 AM permalink
How the [average return from each free spin] is calculated?

It is (total number of hits) * Sum (EV) / Total Slot combinations ?

or Sum ( Return of each combination * P(comb) ) / Game Total Hits ?

I'm kindda confused here, because I can't make my way to get the numbers on the Atkins Diet machine,
and I want to understand how this work.

Thanks Again!
jass
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September 18th, 2012 at 5:13:20 AM permalink
My question is about different number of free spins when you have different number of scatter symbols visible. In the Atkins Diet machine, for each number of visible scatter symbols (3, 4 or 5), the player is granted with the same number of free spins (10 free spins).
How can we calculate the expected number of free spins if for example:
for 3 scatter symbols you win 5 free spins (probability to get 3 scatter symbols is p_3)
for 4 scatter symbols you win 7 free spins (probability to get 4 scatter symbols is p_4)
for 5 scatter symbols you win 9 free spins (probability to get 5 scatter symbols is p_5)?
CrystalMath
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September 20th, 2012 at 9:22:44 PM permalink
Quote: jass

My question is about different number of free spins when you have different number of scatter symbols visible. In the Atkins Diet machine, for each number of visible scatter symbols (3, 4 or 5), the player is granted with the same number of free spins (10 free spins).
How can we calculate the expected number of free spins if for example:
for 3 scatter symbols you win 5 free spins (probability to get 3 scatter symbols is p_3)
for 4 scatter symbols you win 7 free spins (probability to get 4 scatter symbols is p_4)
for 5 scatter symbols you win 9 free spins (probability to get 5 scatter symbols is p_5)?



The expected number of free spins for 3 scatter symbols is 5/(1- (p_3*5 + p_4*7 + p_5*9)).
I heart Crystal Math.
24Bingo
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September 20th, 2012 at 9:34:54 PM permalink
Quote: boochaka

How the [average return from each free spin] is calculated?

It is (total number of hits) * Sum (EV) / Total Slot combinations ?

or Sum ( Return of each combination * P(comb) ) / Game Total Hits ?



I'm not totally sure what you mean by "game total hits," but these sound to me like they'd be the same number.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
CrystalMath
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September 20th, 2012 at 9:40:38 PM permalink
Quote: 24Bingo

Quote: boochaka

How the [average return from each free spin] is calculated?

It is (total number of hits) * Sum (EV) / Total Slot combinations ?

or Sum ( Return of each combination * P(comb) ) / Game Total Hits ?



I'm not totally sure what you mean by "game total hits," but these sound to me like they'd be the same number.



I think I know what he's getting at. I glossed over this post.

Anyhow, the return of a slot game, which is the same calculation used for the free games is: Sum (Return of each combination * P(comb)).

P(comb) = hits(comb)/total slot combinations
I heart Crystal Math.
jass
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September 21st, 2012 at 2:53:52 PM permalink
Using this formula I can calculate expected number of free spins for 3, 4 and 5 scatter symbols, separately.
But, how can I calculate the expected number of free spins for the game?
Is it (5+7+9)/(1-(p_3*5+p_4*7+p_5*9))?
MathExtremist
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September 21st, 2012 at 3:18:34 PM permalink
Quote: jass

Using this formula I can calculate expected number of free spins for 3, 4 and 5 scatter symbols, separately.
But, how can I calculate the expected number of free spins for the game?
Is it (5+7+9)/(1-(p_3*5+p_4*7+p_5*9))?


No. The prior formula gives you the expected # spins starting from the given feature trigger, so just weight each number by the odds of each trigger.

But I typically wouldn't do that aggregation without computing the individual results first. I'd recommend keeping things broken out and computing RTP based on each individual feature trigger.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
CrystalMath
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September 21st, 2012 at 3:47:59 PM permalink
Quote: MathExtremist

No. The prior formula gives you the expected # spins starting from the given feature trigger, so just weight each number by the odds of each trigger.

But I typically wouldn't do that aggregation without computing the individual results first. I'd recommend keeping things broken out and computing RTP based on each individual feature trigger.



I agree. I wouldn't aggregate them, but you can. If you do aggregate, the expected number of free games per base game is (p_3*5 + p_4*7 + p_5*9)/(1-(p_3*5 + p_4*7 + p_5*9)) .
I heart Crystal Math.
MathExtremist
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September 21st, 2012 at 5:16:30 PM permalink
Quote: CrystalMath

I agree. I wouldn't aggregate them, but you can. If you do aggregate, the expected number of free games per base game is (p_3*5 + p_4*7 + p_5*9)/(1-(p_3*5 + p_4*7 + p_5*9)) .


And if you want expected number of free games per free game trigger (regardless of which kind), divide the above result by the total probability of triggering any free game (p_3 + p_4 + p_5). That's the answer to the question "how many free spins will I get, on average, when I trigger the free spins?"
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
jass
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October 11th, 2012 at 12:37:08 PM permalink
What if instead of winning free spins, particular number of scatter symbols leads to a sub game (bonus game).
Lets say winning 3 scatter symbols starts bonus game when you can win minimum $3 and max $10
winning 4 scatter symbols starts bonus game when you can win min $8 and max $13
winning 5 scatter symbols starts bonus game when you can win min $11 and max $17?
Bonus game has particular number of levels, lets say 4 levels each.
Every player can pass first level. He can win min $ for the particular game (depending on number of scatter symbols) or more $ on this level depending on the selected field.
But, on the second level there are particular number of traps. For example, the player can choose between 5 fields on this level, but 2 of them are traps. Selecting field that is trap ends the game. Selecting other field than trap player gets particular amount of $.
On the third level there are 5 fields to choose from and 3 traps.
On the fourth level there are 4 fields and 3 traps.
On each level the player can select only 1 field.
Summing all $ that the player gets until choosing a trap or until passing all 4 levels is the amount he will get at the end of this sub game.
My question is: how to calculate average $ that the player can win playing the sub game?
Amount of $ per field is known for the slot machine. Higher levels give more $.
tringlomane
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October 14th, 2012 at 9:27:15 PM permalink
Quote: jass

What if instead of winning free spins, particular number of scatter symbols leads to a sub game (bonus game).
Details snipped...



To be able to accurately find the value of a bonus game like this, you need to know all the possible outcomes for every scenario. Unfortunately with a live machine, that means you would need to hit 5 scatters one time and make it to the final step at least. And that is assuming the values do not change between bonus games with the same amount of scatter symbols (the values remaining static is somewhat unlikely).
FleaStiff
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October 15th, 2012 at 5:48:57 AM permalink
For a non slot player such as myself there really is no Expected Value or mathematical value.
Freeplay would never affect my decision to go to a casino, it would only be an obligation once I was already there since its a Use It OR Lose It basis. Its like swiping my card at the kiosk for the car drawing... if I'm there, I'll do it but I'll never go to a casino for that purpose or with the expectation of driving home in a new car.
LordTony
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June 12th, 2016 at 8:36:17 AM permalink
Quote: Wizard

The simple formula is [number free spins]*[average return from each free spin].

You start with 10, but free spins can earn more free spins, infinitely. In that case, the expected final number of spins is n/(1-n*p), where n is the number of free spins earned per bonus, and p is the probability of the bonus per spin. In the the Atkin's case, the bonus probability is 0.011185, so the expected number of free spins per bonus is 10/(1-0.011185*10) = 11.259335.



Sorry for digging up an old thread, but Can you explain how you come up with this formular?

Thanks
kineskicovek
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May 20th, 2021 at 11:38:04 PM permalink
Hi :)
I have question about limitted number of free spin.
For example p=0,02 and n= 10 and I want to limit number of free spins up to 50.
If I have no limit the result would be 12,5.
I'm not sure how would be if maximum number of free spins must be 50.
charliepatrick
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May 21st, 2021 at 5:37:12 AM permalink
^ It's probably easier to do it by brute force.

e.g. P (trigger) = 0.02; Q (no trigger) = 0.98

So Pr(10) = Q^10; pr (20) = 10*P*Q^19; then it gets complicated, as you need one in the first ten and another in the first twenty to get to 30., etc.
kineskicovek
kineskicovek
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May 23rd, 2021 at 10:53:31 PM permalink
Quote: charliepatrick

^ It's probably easier to do it by brute force.

e.g. P (trigger) = 0.02; Q (no trigger) = 0.98

So Pr(10) = Q^10; pr (20) = 10*P*Q^19; then it gets complicated, as you need one in the first ten and another in the first twenty to get to 30., etc.



I think that this isn't good.
For Pr(10) is ok, but for Pr(20) isn't good. With this formula you can get 200 free spins. Right?
charliepatrick
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May 24th, 2021 at 12:25:31 AM permalink
I think P happens once, Q happens 19 times, and P has to happen somewhere in the first 10 spins.
kineskicovek
kineskicovek
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May 25th, 2021 at 1:08:20 AM permalink
Yes, I understand now. Tnx a lot!!!
And for 30 spins should be this formula: Pr(30)= (10 over 2) * p^2 * q^34 + 10*p * 10*p * q^34 ?
(10 over 2) is 45.
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