How to program the dice to find the probability?

Hello, the probability of any dice getting 4 is 1/6, and the probability of not getting 4 is 5/6. Therefore, this is a binomial distribution problem.

C(20,5)x( 1/6)^5 x[ 1-( 1/6)]^(20-5])

= 15504 x( 1/6)^5 x 5/6)^ 15

= 0. 1294.

Every roll of the dice is an independent event and does not affect each other. Throwing 10 is another independent event. Five throws 10 fall between 1-50 and the other five fall between 5 1- 100. You can't use the obtained results to guess the following results, because they have the greatest probability. Just like drawing lots, if the people who draw lots later don't know that the results drawn by the people in front are fair, everyone will be1/n. If they know the results, the probability behind will change.