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Probability of a Two Boxes Having The Same Number of Distinct Balls

Difficulty: Hard

Problem Description

Given 2n balls of k distinct colors, we are to distribute the balls into two boxes such that each box receives n balls. The goal is to find the probability that both boxes contain the same number of distinct colors after the distribution. The distribution of the balls is done uniformly at random.

Key Insights

  • Each box must receive exactly n balls.
  • The distinct colors in each box can vary based on the distribution of the balls.
  • The problem can be approached through combinatorial counting to calculate the total valid distributions versus the favorable ones where both boxes have the same number of distinct colors.
  • The constraints on the number of balls and colors make it feasible to use backtracking or dynamic programming approaches.

Space and Time Complexity

Time Complexity: O(2^(n+k)), where n is the number of balls in each box and k is the number of distinct colors.
Space Complexity: O(k), for storing counts of distinct colors and their distributions.

Solution

The problem can be solved using a combinatorial approach where we recursively distribute the balls into two boxes while keeping track of the distinct colors in each box. A backtracking algorithm can be employed to explore all possible distributions of the balls. By counting the total distributions and the favorable distributions where both boxes have the same number of distinct colors, we can compute the required probability.

Code Solutions

def getProbability(balls):
    from collections import Counter
    from math import comb
    
    n = sum(balls) // 2
    total_ways = 0
    valid_ways = 0
    
    def backtrack(i, box1_count, box2_count, box1_distinct, box2_distinct):
        nonlocal total_ways, valid_ways
        if box1_count + box2_count == n:
            total_ways += 1
            if box1_distinct == box2_distinct:
                valid_ways += 1
            return
        
        if i >= len(balls):
            return
        
        # Try placing balls of color i in box 1
        for x in range(balls[i] + 1):
            if box1_count + x <= n:
                new_box1_distinct = box1_distinct + (1 if x > 0 else 0)
                new_box2_count = box2_count
                new_box2_distinct = box2_distinct
                
                # Remaining balls of color i will go in box 2
                remaining = balls[i] - x
                new_box2_count += remaining
                new_box2_distinct += (1 if remaining > 0 else 0)
                
                backtrack(i + 1, box1_count + x, new_box2_count, new_box1_distinct, new_box2_distinct)
        
        # Also consider not placing any balls of this color
        backtrack(i + 1, box1_count, box2_count, box1_distinct, box2_distinct)
    
    backtrack(0, 0, 0, 0, 0)
    
    return valid_ways / total_ways

# Example usage
print(getProbability([2, 1, 1]))  # Output: 0.66667
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