How to calculate the probability of freespins

Freespins are usually launched by dropping scatter characters according to the "3 + anywhere" rule (sometimes "on reels 2-4," "2 + scatter + wild," "progress counter," etc.). Knowing the scatter frequencies by reels or having spin logs, you can estimate the probability of a trigger in one back (q) and from it get the expected waiting time (geometric distribution).

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1) Quick dictionary

(q) is the probability of running frispins in one back.

Average wait interval: (\mathbb {E} [T] = 1/q) spins.

Median interval: (\mathrm {Med} (T) =\left\lceil\dfrac {\ln 0. 5} {\ln (1-q) }\right\rceil) (approximately (0 {,} 693/q) at low (q)).

Chance not to wait for (N) spins: ((1-q) ^ N).

Chance to wait ≥1 times for (N) spins: (1- (1-q) ^ N).

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2) Accurate counting by drum strips (strip-count)

If you know the tapes (character lists) and the number of steps on each reel:

1. For each reel (i), count

[
s_i=\frac{#\text{pozitsy scatter on reel} i} {#\text {total positions on} i}.
]

2. Rule "3 + scatter on 5 reels" (one character per reel):

[
q=\sum_{k=3}^{5}\ \sum_{\substack{A\subset{1..5}\	A	=k}}\ \prod_{i\in A} s_i\ \prod_{j\notin A} (1-s_j).
]

If the drums are "equally probable" (s_i=s), then binomially:

[
q=\sum_{k=3}^{5}\binom{5}{k}s^k(1-s)^{5-k}.
]

3. Scatter on reels 2-4 only (3 reels):

[
q=\prod_{i=2}^{4} s_i.
]

4. Rule "2 scatter + wild instead of third": Mark (w_i) - the probability of wild on the drum (i). Then the chance of ≥3 "conditional" hits:

[
q=\sum_{k=3}^{5}\ \sum_{A}\prod_{i\in A}(s_i+w_i)\ \prod_{j\notin A}(1-s_j-w_j),  ]

where (A) are subsets of drums of size (k). (Often an approximation on the first three drums is enough if wilds are not counted at 4-5.)

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3) Megaways and variable row number slots

In Megaways, the number of positions on the drum changes. It is practical to consider conditionally by configuration:

1. For each reel (i), the scatter probability at position: (p_i=\frac{#\text{scatter-taylov}}{#\text{vsekh tiles}}) (usually 1/types of characters if equilibrium; some games have their own weight).

2. With the realized height (h_i), the chance of at least one scatter on the drum: (s_i (h_i) = 1- (1-p_i) ^ {h _ i}).

3. Conditional (q (h_1,...,h_6)) - according to the formulas from § 2, but with (s_i (h_i)).

4. The final (q) is the mean (\mathbb {E} _ {h} [, q (h),]) of the height distribution (better by simulation).

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4) When there are no tables: an empiricist by logs

If you have a spin log (demo or real): Score (\hat q):

[
\ hat q =\frac {#\text {triggers}} {#\text {spins}}.
]

Confidence interval (rare event): Use Bayesian estimation with Jeffries prior (\text {Beta} (0 {,} 5.0 {,} 5)) or Wilson interval - they are more stable on small samples.

How many spins do you need? With (q\approach 1/200) (0.5%), it is reasonable to collect tens of thousands of spins, otherwise the spread is large.

Transfer to "waiting": median/mean interval from § 1.

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5) "Combined" mechanics and progress triggers

Progress counter (for example, collect 3 parts): this is a negative binomial scheme. If the chance of getting "part" for spin (p), then the chance of completing for (n) spins:

[
\mathbb{P}(T\le n)=\sum_{k=3}^{n}\binom{k-1}{2} p^3 (1-p)^{k-3}.
]

Average expectation (\mathbb {E} [T] = 3/p), median - by summation/simulation.

Wheels/trails before freespins: first a chance to hit the wheel, then a chance of the freespin sector. The total probability is the product of the stages (or the sum over the branches of the outcome tree).

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6) Calculation examples

A) 5 reels, rule 3 +, equal to tapes, on each (s = 0 {,} 12).

[
q=\binom{5}{3}s^3(1-s)^2+\binom{5}{4}s^4(1-s)+s^5
]

[
=\ 10\cdot0{,}12^3\cdot0{,}88^2\ +\ 5\cdot0{,}12^4\cdot0{,}88\ +\ 0{,}12^5\ \approx 0{,}0167.
]

Waiting for: (\mathbb {E} [T ]\approx 60) spins; median (\approach 0 {,} 693/0 {,} 0167\approach 41) spin.

Chance to see ≥1 trigger per 100 spins: (1- (1-0 {,} 0167) ^ {100 }\approach 80%).

B) Reels 2-4 only: (s_2=0{,}15,\ s_3=0{,}12,\ s_4=0{,}10).

[
q=s_2 s_3 s_4=0{,}0018 \Rightarrow \mathbb{E}[T]\approx 556,\ \mathrm{Med}\approx 385.
]

C) Megaways (conditional example): each of the 6 reels receives (h_i\in{2..7}) equally likely, (p_i=p=1/12).

Then (s_i (h) = 1- (1-p) ^ h).

Next - count (q (h)) according to § 2 (3 + out of 6) and average over all (h) (better Monte Carlo per 100k configurations).

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7) From probability to practice

Session plan. Knowing the median/75th percentile of the trigger expectation, plan the session length and bank for several such intervals.

Slot comparison. Slots with the same RTP may differ (q): one gives frispins more often, but "weaker," the other less often, but "fatter." See both (q) and quantiles of the bonus win.

Communication in articles. Give the reader a "freespin passport": (q), (\mathbb {E} [T]), median, 75th percentile and "chance to see ≥1 for (N) spins."

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8) What can distort the assessment

Different RTP versions of the same game - (s_i) and (q) may differ.

Buffer/missions/cashback do not change (q), but change the economy - do not confuse frequency with value.

Short samples for rare (q) → huge intervals of uncertainty; use Bayes/Wilson and/or simulations.

Megaways without a conditional model of heights - better immediately Monte Carlo.

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9) Ready-made "freespin passport" (template)

Trigger rule: 3 + scatter (1/5 reels; or 2-4; or 2 + scatter + wild)

Score (q): ... (method: strip-count/empirical/simulation)

Waiting intervals: average (1/q =...) spins; median...; 75th percentile...

Trigger ≥1 chance for (N =...): ...%

Risk comment: frequency vs strength of bonus; typical "deserts."

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Bottom line: the probability of frispins can be calculated "from above" (by tapes and rules) or "from below" (by logs/simulations). The key is to correctly formalize the trigger rule, take into account the features of the mechanics (limited drums, wilds-replacements, Megaways), and then translate (q) into time guidelines that the player understands: the average/median interval and the chance to meet the selected session length.