Problem Description

You have two soups, A and B, each starting with n mL. The game proceeds in turns where you randomly select one of four serving operations (each with 0.25 probability):

1. Pour 100 mL from soup A and 0 mL from soup B
2. Pour 75 mL from soup A and 25 mL from soup B
3. Pour 50 mL from soup A and 50 mL from soup B
4. Pour 25 mL from soup A and 75 mL from soup B

Key Rules:

• There's no operation that pours 0 mL from A and 100 mL from B (the operations are biased toward depleting soup A faster)
• Both soups are poured simultaneously in each turn
• If an operation requires more soup than available, you pour all remaining soup
• The process stops immediately when at least one soup runs out

What to Calculate: Return the probability that soup A empties first, PLUS half the probability that both soups empty on the same turn.

The answer should be accurate within 10^-5 of the actual value.

Example Understanding: If soup A empties first (before B), that event contributes its full probability to the answer. If both soups empty simultaneously, that event contributes half its probability to the answer. If soup B empties first, that contributes 0 to the answer.

The solution uses dynamic programming with memoization, where states are tracked in units of 25 mL (since all operations are multiples of 25). The function dfs(i, j) calculates the probability for i and j units remaining. For very large n (> 4800), the probability approaches 1 due to the bias toward depleting soup A.

Intuition

The first observation is that the four operations are biased toward consuming more of soup A than soup B. On average, each operation consumes (100 + 75 + 50 + 25) / 4 = 62.5 mL from A but only (0 + 25 + 50 + 75) / 4 = 37.5 mL from B. This asymmetry means soup A is more likely to run out first.

Since we need to calculate probabilities, we can think of this as exploring all possible sequences of operations. Each sequence has a probability based on the operations chosen (each operation has probability 0.25). This naturally leads to a recursive approach where we track the remaining amounts of both soups.

A key optimization comes from noticing that all operation amounts are multiples of 25 mL. Instead of working with the actual mL values, we can scale everything down by 25. This transforms our operations to:

• (4, 0) units instead of (100, 0) mL
• (3, 1) units instead of (75, 25) mL
• (2, 2) units instead of (50, 50) mL
• (1, 3) units instead of (25, 75) mL

This scaling dramatically reduces the state space we need to explore.

For the recursive function dfs(i, j) representing remaining units of A and B:

• If both are empty (i <= 0 and j <= 0), both finished simultaneously, contributing 0.5 to our answer
• If only A is empty (i <= 0), A finished first, contributing 1.0
• If only B is empty (j <= 0), B finished first, contributing 0
• Otherwise, we recursively calculate the probability as the average of all four possible operations

The formula (n + 24) // 25 ensures proper ceiling division when converting mL to units.

Finally, through experimentation or mathematical analysis, we find that for large n (specifically n > 4800), the probability becomes so close to 1 that we can return 1 directly. This is because the bias toward consuming A becomes overwhelming with larger soup volumes.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution uses memoization (dynamic programming with caching) to avoid recalculating the same states multiple times.

Step 1: Scale Down the Problem

Since all operations use multiples of 25 mL, we convert the problem to work with units where 1 unit = 25 mL. The conversion uses ceiling division: (n + 24) // 25. This ensures we round up to handle any remainder properly.

Step 2: Define the Recursive Function

We define dfs(i, j) to calculate the probability when there are i units of soup A and j units of soup B remaining.

The base cases are:

• If i <= 0 and j <= 0: Both soups finished together, return 0.5
• If i <= 0: Soup A finished first, return 1.0
• If j <= 0: Soup B finished first, return 0.0

Step 3: Recursive Formula

For the general case, we have four equally likely operations (each with probability 0.25):

dfs(i, j) = 0.25 × (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3))

This formula represents:

• dfs(i - 4, j): Taking 100 mL from A, 0 mL from B
• dfs(i - 3, j - 1): Taking 75 mL from A, 25 mL from B
• dfs(i - 2, j - 2): Taking 50 mL from A, 50 mL from B
• dfs(i - 1, j - 3): Taking 25 mL from A, 75 mL from B

Step 4: Memoization

The @cache decorator stores previously computed results for dfs(i, j). This prevents redundant calculations when the same state is reached through different sequences of operations.

Step 5: Optimization for Large n

When n > 4800, the probability is so close to 1 (within the required precision of 10^-5) that we can directly return 1. This optimization avoids unnecessary computation for large inputs where the result is effectively certain.

The final solution combines all these elements:

return 1 if n > 4800 else dfs((n + 24) // 25, (n + 24) // 25)

This approach efficiently handles the probability calculation while avoiding timeout issues for large inputs through both memoization and the early termination condition.

Example Walkthrough

Let's walk through a small example with n = 50 mL to illustrate the solution approach.

Step 1: Convert to Units First, we scale down by 25 mL per unit:

• Units = (50 + 24) // 25 = 74 // 25 = 2
• So we start with dfs(2, 2) (2 units each of soup A and B)

Step 2: Calculate dfs(2, 2) From state (2, 2), we have 4 possible operations:

• Operation 1: Pour (4, 0) units → leads to dfs(-2, 2)
• Operation 2: Pour (3, 1) units → leads to dfs(-1, 1)
• Operation 3: Pour (2, 2) units → leads to dfs(0, 0)
• Operation 4: Pour (1, 3) units → leads to dfs(1, -1)

Step 3: Evaluate Base Cases

• dfs(-2, 2): Since A is empty (i ≤ 0) but B is not, A finished first → returns 1.0
• dfs(-1, 1): Since A is empty (i ≤ 0) but B is not, A finished first → returns 1.0
• dfs(0, 0): Both are empty (i ≤ 0 and j ≤ 0), finished together → returns 0.5
• dfs(1, -1): Since B is empty (j ≤ 0) but A is not, B finished first → returns 0.0

Step 4: Combine Results

dfs(2, 2) = 0.25 × (1.0 + 1.0 + 0.5 + 0.0)
         = 0.25 × 2.5
         = 0.625

The probability is 0.625, meaning there's a 62.5% chance that either soup A finishes first or both soups finish together (with the latter contributing half its probability).

This example demonstrates:

• How the scaling reduces our state space (from 50 mL to 2 units)
• How the recursive function explores all possible operation sequences
• How base cases determine the contribution to our final probability
• Why soup A has a higher probability of finishing first due to the biased operations (3 out of 4 operations consume more from A than B)

Solution Implementation

Time and Space Complexity

The code implements a dynamic programming solution with memoization for the soup serving problem. The key insight is the scaling optimization where the original value n is divided by 25 (since all serving sizes are multiples of 25).

Time Complexity: O(C²) where C = 200

The scaled value is (n + 24) // 25, which effectively reduces the problem size. The function dfs(i, j) explores all possible states where i and j represent the remaining soup amounts after scaling. Due to the early termination when n > 4800, the maximum scaled value is approximately 4800 / 25 ≈ 192, which rounds to about C = 200. Each state (i, j) is computed at most once due to memoization using @cache, and there are O(C²) possible states. Each state computation involves 4 recursive calls with constant time operations, resulting in O(1) work per state. Therefore, the total time complexity is O(C²).

Space Complexity: O(C²) where C = 200

The space is dominated by two factors:

1. The memoization cache stores all computed states (i, j), which requires O(C²) space
2. The recursion depth in the worst case is O(C) (when we repeatedly subtract 1 from the soup amounts)

Since O(C²) > O(C), the overall space complexity is O(C²).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Scaling/Rounding When Converting to Units

Pitfall: Using simple integer division n // 25 instead of ceiling division (n + 24) // 25 when converting milliliters to scaled units.

Why it's wrong: If n = 30, using n // 25 gives 1 unit (25 mL), losing 5 mL. This changes the problem since we're not accounting for all the soup available.

Solution: Always use ceiling division to ensure we don't lose any soup volume:

# Wrong
scaled_n = n // 25  # Loses remainder

# Correct
scaled_n = (n + 24) // 25  # Ceiling division

2. Forgetting the Optimization for Large n

Pitfall: Not including the early return for large values of n, causing the solution to time out or use excessive memory.

Why it happens: The recursive solution has exponential time complexity without memoization bounds. For large n, even with memoization, the number of unique states becomes massive.

Solution: Include the optimization check before processing:

if n > 4800:  # Can also use 4800-5000 range
    return 1.0

3. Incorrect Base Case Order or Logic

Pitfall: Checking individual soup depletion before checking simultaneous depletion:

# Wrong order
if soup_a <= 0:
    return 1.0
if soup_b <= 0:
    return 0.0
if soup_a <= 0 and soup_b <= 0:  # Never reached!
    return 0.5

Why it's wrong: The simultaneous depletion case would never be reached because one of the individual checks would return first.

Solution: Check the simultaneous case first:

# Correct order
if soup_a <= 0 and soup_b <= 0:
    return 0.5
if soup_a <= 0:
    return 1.0
if soup_b <= 0:
    return 0.0

4. Misunderstanding the Probability Calculation

Pitfall: Returning just the probability of A emptying first, forgetting to add half the probability of simultaneous depletion:

# Wrong interpretation
if soup_a <= 0 and soup_b <= 0:
    return 0.0  # Ignoring this case

Solution: Remember the problem asks for P(A empties first) + 0.5 × P(both empty together):

if soup_a <= 0 and soup_b <= 0:
    return 0.5  # Half probability for simultaneous

5. Not Using Memoization or Using It Incorrectly

Pitfall: Implementing without memoization or manually managing a dictionary incorrectly:

# Without memoization - exponential time complexity
def dfs(i, j):
    # ... recursive calls without caching
  
# Or incorrect manual memoization
memo = {}  # Defined outside, persists between test cases!

Solution: Use @cache decorator or properly scope the memoization:

@cache  # Automatic memoization
def calculate_probability(soup_a: int, soup_b: int) -> float:
    # ... implementation