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Probability of hitting a checker from a long running start

Posted By: ah_clem
Date: Sunday, 6 July 2025, at 8:53 p.m.

In Response To: Probability of hitting a checker from a long running start (Jason Lee)


It rained all day today, so I had nothing better to do than dust off some of my old programming skills and write a routine that implements the recursive approach above.

It converges to 0.39446366782006914 in a little over 300 recursive calculations of p(n), which is remarkably close to Bob's answer.

Here's the code in case anyone is interested: 

 

 

# Online Python - IDE, Editor, Compiler, Interpreter 

 

def prob(length): 

    hit = [0,11,12,14,15,15,17,6,6,5,3,2,3] #number of shots from n away 

    hit.extend([0,0,1,1,0,1,0,1,0,0,0,1]) 

    short = [0,0,0,2,3,4,5,6,5,4,2,1,1] #number of ways to roll exactly n 

    short.extend([0,0,0,1,0,0,0,1,0,0,0,1]) 

    p=[0] # p[n] is the probability of hitting from n pips away 

    for n in range(length +1): 

        if n == 0: 

            continue 

        if  n <= 24: #possibly a shot this roll 

            p.append(hit[n]/36) 

        else: 

            p.append(0) #no possibility of a shot this roll 

        for i in range(n):  

            #add the probabilities of all the rolls that are short of the blot 

            if i == 0: 

                continue 

            if n-i > 0 and n-i < len(short): 

                p[n] = p[n] + short[n-i]*p[i]/36 

    for n in range(length +1): 

      print(f'{n} {p[n]}') 

     

prob(1000)


There's probably a more elegant way to deal with the zero indexed lists that Python provides; I don't really care that I'm using magic numbers (24 & 36) here.




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