Probability and algorithms

Question

Suppose we start at the number zero, and repeatedly roll a fair 6-sided die, adding the resulting number to the total. We win if we reach the total 100, and lose if we overshoot, for example by reaching a total of 99 and then rolling a number greater than 1. Describe an algorithm for computing the probability that we win.

Answer 1

Let $P_k(n)$ be the probability that you are in position $n$ after $k$ throws. Since you always move forward you simply want $\sum\limits_{k=1}^\infty P_k(100)$.

We can calculate $P_k(n)$ recursively. It is clearly equal to $\sum\limits_{j=n-6}^{n-1} p_{k-1}(j)/6$

We only need to calculate $p_k(n)$ for $n\,k leq 100$ and $k\leq 16$, so it can be calculated in time $\mathcal O (n^2)$

Here is a c++ code:

#include <bits/stdc++.h>
using namespace std;

double P[101][101];

int main(){
    P[0][0]=1;
    for(int k=1;k<101;k++){
        for(int n=1;n<101;n++){
            for(int j=max(0,n-6); j<n;j++){
                P[n][k]+=P[j][k-1]/6;
            }
        }
    }
    double res=0;
    for(int k=0;k<101;k++){
        res+=P[100][k];
    }
    printf("%f\n",res);
}

The result given is approximately $0.285714$ which makes a lot of sense, the approximate length of each "jump" is $\frac{7}{2}$, so the approximate probability that we hit a number should be around $\frac{2}{7}=0.\overline{285714}$

Answer 2

Hint:

You can use recurrence here. $$a_n=\frac{a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}+a_{n-6}}{6}$$ where $a_n$ is the probability to get the sum $n$.

Therefore, $$a_{100}=\frac{a_{99}+a_{98}+a_{97}+a_{96}+a_{95}+a_{94}}{6}$$