The probability of rolling dice and walking according to the number of steps in the nth grid

Like a monopoly, roll the dice and calculate the probability of going to the n-th grid according to the number of dice and steps.

For example, our function dp(n) calculates the probability of the nth grid, and then:

//The number of dice must be (1), which means 1 possibility.

P( 1)= 1 probability of grid = 1/6.

//The number of dice must be (2), (1, 1), two possibilities.

P(2)= probability of the second grid =1/6+(1/6) (1/6).

//Dice number (3), (1, 2), (2, 1), (1, 1)

P(3)= probability of the third grid =1/6+(1/6) (1/6) 2+(1/6) * (1/6).

There are too many permutations and combinations of the fourth one, so I won't continue to list them here.

Let's analyze how this p(n) is calculated. The probability of the nth lattice is actually calculated from the nth-6th lattice to the nth-1lattice.

If n

p( 1)= 1/6

p(2)=p( 1)/6+ 1/6

p(3)= p( 1)/6+p(2)/6+ 1/6

...

If n>6:

p(n)= p(n- 1)/6+p(n-2)/6+p(n-3)/6+p(n-4)/6+p(n-5)/6+p(n-6)/6

Then we just need to calculate the probability from 1

When calculating the probability of 1, the influence of the first lattice on the probability of 2-7 lattices must also be calculated.

p(n+ 1)+=p(n)/6

The specific algorithm is as follows: