First find the probability of getting four twos in a row.
(1/6)^4*(5/6)^16
Then you would multiply this probability by the number of combinations of those four twos in a sequence of 20 attempts
20!/(4!(20-4)!)
And the product of the probability times the number of combinations is:
(1/1296)(152587890625/2821109907456)4845 :P which is approximately:
0.2022 or 20.22% to the nearest hundredth of a percent.
