Maths: Number of reviews needed for a stable average

When calculating the number of reviews needed to achieve a stable average rating, we use the following formula:

D=d/(n+1)D=d/(n+1)

Where DD is the difference between the new average X[n+1]X[n+1] and the old average X[n]X[n] when a new rating x[n+1]x[n+1] is left, and dd is the difference between the newest rating x[n+1]x[n+1] and the old average X[n]X[n], with d≤4d ≤ 4.

n=d/D−1n=d/D-1

We assume that an average is accurate when each new rating doesn’t change the average by more than 0.1.

With an average rating of 5/5 stars and a new rating at 1/5 (edge case, d=4d=4), we get the following result:

D<0.1=>n>39D<0.1 => n>39

In this case, we would need at least 39 reviews for the average to be stable enough.

If we choose an average of 4/5 stars (industry average) and a new rating of 1/5 stars (d=2d=2):

D<0.1=>n>29D<0.1 => n> 29

I performed a test on a Google Sheet with random values of ratings between 1 and 5. In this case, the stability seems to occur after 21 reviews.

D<0.1=>n>21D<0.1 => n> 21

If we consider stability to occur when the average changes by less than 0.05 (because a score of 4.76 dropping by 0.05 would round from 4.8 to 4.7 on Google), stability occurs after 30 reviews:

D<0.05=>n>30D<0.05 => n> 30

To go further: In reality, ratings are not random and are generally consistent with the average rating. We would need to employ probabilistic mathematics (e.g., Bayesian probabilities) to achieve a more accurate result.

The actual number of reviews needed is probably around 20.

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