There are two types of soup: type A and type B. Initially, we have n
ml of each type of soup. There are four kinds of operations:
When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5
of the actual answer will be accepted.
Input: n = 50
Output: 0.62500
Explanation: If we choose the first operation, A will become empty first.
If we choose the second operation, A and B will become empty at the same time.
If we choose the third operation, B will become empty first.
If we choose the fourth operation, B will become empty first.
So the probability that A will be empty first plus half the probability that A and B become empty at the same time = 0.25 + 0.25 * 0.5 = 0.375.
Input: n = 100
Output: 0.71875
Explanation: The probability is calculated using dynamic programming, considering all possible operations and their outcomes.
To solve this problem, we need to:
Apply string manipulation concepts to solve a real-world problem.
There are two types of soup: type A and type B. Initially, we have n
ml of each type of soup. There are four kinds of operations:
When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5
of the actual answer will be accepted.
If we choose the first operation, A will become empty first. If we choose the second operation, A and B will become empty at the same time. If we choose the third operation, B will become empty first. If we choose the fourth operation, B will become empty first. So the probability that A will be empty first plus half the probability that A and B become empty at the same time = 0.25 + 0.25 * 0.5 = 0.375.
The probability is calculated using dynamic programming, considering all possible operations and their outcomes.
This problem can be solved using dynamic programming with memoization.
The key insight is to define a recursive function that calculates the probability for given amounts of soup A and soup B.
We can use memoization to avoid redundant calculations.
For large values of n, the probability approaches 1, so we can return 1 directly for n > 4800.
We can scale down the problem by dividing n by 25 and working with smaller units to simplify the calculations.
This problem has several practical applications:
Calculating probabilities in scenarios with multiple possible outcomes and equal likelihood of each action.
Modeling the consumption of two different resources with various consumption patterns.
Planning for food service operations where different serving options have different resource requirements.
There are two types of soup: type A and type B. Initially, we have n
ml of each type of soup. There are four kinds of operations:
When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5
of the actual answer will be accepted.
Input: n = 50
Output: 0.62500
Explanation: If we choose the first operation, A will become empty first.
If we choose the second operation, A and B will become empty at the same time.
If we choose the third operation, B will become empty first.
If we choose the fourth operation, B will become empty first.
So the probability that A will be empty first plus half the probability that A and B become empty at the same time = 0.25 + 0.25 * 0.5 = 0.375.
Input: n = 100
Output: 0.71875
Explanation: The probability is calculated using dynamic programming, considering all possible operations and their outcomes.
To solve this problem, we need to:
Apply string manipulation concepts to solve a real-world problem.
There are two types of soup: type A and type B. Initially, we have n
ml of each type of soup. There are four kinds of operations:
When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5
of the actual answer will be accepted.
If we choose the first operation, A will become empty first. If we choose the second operation, A and B will become empty at the same time. If we choose the third operation, B will become empty first. If we choose the fourth operation, B will become empty first. So the probability that A will be empty first plus half the probability that A and B become empty at the same time = 0.25 + 0.25 * 0.5 = 0.375.
The probability is calculated using dynamic programming, considering all possible operations and their outcomes.
This problem can be solved using dynamic programming with memoization.
The key insight is to define a recursive function that calculates the probability for given amounts of soup A and soup B.
We can use memoization to avoid redundant calculations.
For large values of n, the probability approaches 1, so we can return 1 directly for n > 4800.
We can scale down the problem by dividing n by 25 and working with smaller units to simplify the calculations.
This problem has several practical applications:
Calculating probabilities in scenarios with multiple possible outcomes and equal likelihood of each action.
Modeling the consumption of two different resources with various consumption patterns.
Planning for food service operations where different serving options have different resource requirements.