Quote:MathExtremistNo. 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)) .

Quote:CrystalMathI 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?"

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 $.

Quote:jassWhat 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).

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.

Quote:WizardThe 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

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.

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.

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?