The Likeliest Monopoly Square

Fiddler problem: Can You Race Around the Monopoly Board?

The probablity to reach number \(x\) in the \(n\)-th cast be \(p_n(x)\). Then the probability that the number \(x\) is hit is:
\(p(x)=p_1(x)+p_2(x)+p_3(x)+\ldots\). Now, the probability mass function after casting two dice can be calculated from square of the generated polynomial of the p.m.f of one dice: \(g(x)=x^6/6+x^5/6+x^4/6+x^3/6+x^2/6+x/6\).
That is, \(h(x)=g^2(x)=x^2/36 + x^3/18 + x^4/12 + x^5/9 + 5 x^6/36 + x^7/6 +\) \(5 x^8/36 + x^9/9 + x^{10}/12 + x^{11}/18 + x^{12}/36\).
So, for example, \(p_2(8)=5/36\)
And the probability after another two-dice cast is given by \(h^2=g^4\).
The least element of the \(h^{20}\) is \(x^{40}\) and we only are interested in the numbers \(x\) from 1 to 39. Therefore, we only have to calculate \(h, h^2,\ldots h^{19}\).
Running the program, we obtain the extremum values for two dice case:
(0.18223, 7) and (0.12470, 13)
So the maximum probability is p(7)~0.18223 and the minimum probability (out of the numbers from 10 to 39) is p(13)~0.12470 For the three dice case, the least likely case (out of the numbers from 10 to 39) is p(17)~0.07634

Here is the Python code for the problem:



```python
from numpy.polynomial import Polynomial as P


p= P([0,1/6,1/6,1/6,1/6,1/6,1/6])

p=p**2
    
r = []
for i in range(1, 40):
    r.append([i])

for i in range(1, 40):
    for j in range(1,20):
       q=p**j;
       if i<= q.degree():
        
                         r[i-1].append(q.coef[i])
       else:
           r[i-1].append(0)

s= [row[1:] for row in r]

sums=[]
for i in range(39):
    sums.append(0);
    for j in range(19):
         sums[i]+=s[i][j]
        
print(max((val,idx+1) for idx, val in enumerate(sums)))
s10=sums[10:]
print(min((val,idx+11) for idx, val in enumerate(s10)))


p= P([0,1/6,1/6,1/6,1/6,1/6,1/6])
p=p**3

r = []

for i in range(1, 40):
    r.append([i])

for i in range(1, 40):
    for j in range(1,14):
       q=p**j;
       if i<= q.degree():
        
                         r[i-1].append(q.coef[i])
       else:
           r[i-1].append(0)

s= [row[1:] for row in r]

sum=[]
for i in range(39):
    sum.append(0);
    for j in range(13):
       
        sum[i]+=s[i][j]
        
print(max((val,idx+1) for idx, val in enumerate(sums)))
s10=sum[10:]
print(min((val,idx+11) for idx, val in enumerate(s10)))

```