How to resolve the algorithm Dice game probabilities step by step in the Wren programming language

Published on 12 May 2024 09:40 PM
#Wren

How to resolve the algorithm Dice game probabilities step by step in the Wren programming language

Table of Contents

Problem Statement

Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six. They roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player? Later the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player? This task was adapted from the Project Euler Problem n.205: https://projecteuler.net/problem=205

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Dice game probabilities step by step in the Wren programming language

Source code in the wren programming language

var throwDie // recursive
throwDie = Fn.new { |nSides, nDice, s, counts|
    if (nDice == 0) {
        counts[s] = counts[s] + 1
        return
    }
    for (i in 1..nSides) throwDie.call(nSides, nDice-1, s + i, counts)
}

var beatingProbability = Fn.new { |nSides1, nDice1, nSides2, nDice2|
    var len1 = (nSides1 + 1) * nDice1
    var c1 = List.filled(len1, 0)
    throwDie.call(nSides1, nDice1, 0, c1)

    var len2 = (nSides2 + 1) * nDice2
    var c2 = List.filled(len2, 0)
    throwDie.call(nSides2, nDice2, 0, c2)

    var p12 = nSides1.pow(nDice1) * nSides2.pow(nDice2)
    var tot = 0
    for (i in 0...len1) {
        for (j in 0...i.min(len2)) {
            tot = tot + c1[i] * c2[j] / p12
        }
    }
    return tot
}

System.print(beatingProbability.call(4, 9, 6, 6))
System.print(beatingProbability.call(10, 5, 7, 6))

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