It is unlikely, but possible... I recently judged a
similar thing at CMU (granted, it was Magic).
Not even close. It's WAY easier for that to happen on Magic. 3x3 solves are all different
Interesting stats question then: given a Poisson distribution of solves, how much more likely is it to happen with a 1.51 average than a 21.05 average?
I'll give it a try, but I'm afraid I'll regret this
Let's assume µ=1,51 and 21,05 respectively (I don't know if this matters)
Let's also assume the intervals of outcomes are µ+-50%
We need a probability of getting 1,51 and 21,05, wich is 1/the number of outcomes. The number of outcomes are 151 and 2105.
λ for magic= 1/151= 0,0066
λ for 3x3= 1/2105= 0,00048
Lets say you get your most likely outcome, 1,51 on the first solve, now you need 2 identical times, gives n=2.
For magic: P(Y=2)= (0,0066^2*e^-0,0066)/2= 0,000022
For 3x3: P(Y=2)= (0,00048^2*e^-0,00048)/2= 0,00000012
Divide and you get about 183 times bigger chance to get two 1,51 in a row than two 21,05 in a row.
EDIT: Two in a row actually means three in a row, the first solve is x, and then you calculate the probability to get 2 more x.