When To Stop Pulling Balls From An Urn: What's The Right Way To Solve?

Introduction

The problem of pulling balls from an urn is a classic example of a probability and combinatorics problem. It's a simple yet intriguing scenario that has been debated by mathematicians and statisticians for centuries. In this article, we'll delve into the world of probability and combinatorics to explore the right way to solve this problem.

The Problem

Suppose that there are $n$ balls, of which $g$ balls are green. We can pull as many balls as we want, and for every non-green ball pulled, we get $1. If we ever pull a green ball, we stop pulling and our reward is the number of non-green balls we've pulled so far. The question is, when should we stop pulling balls from the urn to maximize our reward?

The Classic Solution

The classic solution to this problem is to use the concept of expected value. The expected value of a random variable is the sum of the product of each possible outcome and its probability. In this case, the expected value of the reward is the sum of the product of each possible number of non-green balls pulled and its probability.

Let's denote the number of non-green balls pulled as $x$. The probability of pulling $x$ non-green balls is given by the binomial distribution:

$$P(x) = \binom{n-g}{x} \left(\frac{g}{n}\right)^x \left(\frac{n-g}{n}\right)^{n-x}$$

The expected value of the reward is then given by:

$$E[R] = \sum_{x=0}^{n-g} xP(x)$$

This is a complex calculation that requires a lot of computation. However, we can simplify the problem by using the concept of linearity of expectation.

Linearity of Expectation

The linearity of expectation states that the expected value of a sum of random variables is equal to the sum of their expected values. In this case, we can write the expected value of the reward as:

$$E[R] = E[X] + E[Y]$$

where $X$ is the number of non-green balls pulled and $Y$ is the reward for each non-green ball pulled.

The expected value of $X$ is given by:

$$E[X] = \sum_{x=0}^{n-g} xP(x)$$

The expected value of $Y$ is given by:

$$E[Y] = \sum_{x=0}^{n-g} xP(x)$$

Since $Y$ is a constant, we can simplify the calculation by using the fact that the expected value of a constant is equal to the constant itself.

Simplifying the Calculation

Using the linearity of expectation, we can simplify the calculation of the expected value of the reward. We can write:

$$E[R] = E[X] + E[Y] = E[X] + n-g$$

The expected value of $X$ is given by:

$$E[X] = \sum_{x=0}^{n-g} xP(x)$$

We can simplify the calculation by using the fact that the expected value of a binomial distribution is equal to the product of the mean and the variance.

The Mean and Variance of a Binomial Distribution

The mean of a binomial distribution is given by:

$$\mu = np$$

where $n$ is the number of trials and $p$ is the probability of success.

The variance of a binomial distribution is given by:

$$\sigma^2 = np(1-p)$$

In this case, the mean of the binomial distribution is given by:

$$\mu = (n-g)\left(\frac{n-g}{n}\right)$$

The variance of the binomial distribution is given by:

$$\sigma^2 = (n-g)\left(\frac{n-g}{n}\right)\left(1-\frac{n-g}{n}\right)$$

Simplifying the Calculation

Using the mean and variance of the binomial distribution, we can simplify the calculation of the expected value of the reward. We can write:

$$E[R] = E[X] + n-g = \mu + n-g$$

Substituting the values of the mean and variance, we get:

$$E[R] = (n-g)\left(\frac{n-g}{n}\right) + n-g$$

Simplifying the expression, we get:

$$E[R] = \frac{n-g}{n}$$

The Optimal Strategy

The optimal strategy is to stop pulling balls from the urn when the expected value of the reward is equal to the number of non-green balls pulled. In other words, we should stop pulling balls when:

$$E[R] = \frac{n-g}{n}$$

This is the optimal strategy because it maximizes the expected value of the reward.

Conclusion

In this article, we explored the problem of pulling balls from an urn and found the optimal strategy to maximize the expected value of the reward. We used the concept of expected value and the linearity of expectation to simplify the calculation. We also used the mean and variance of the binomial distribution to further simplify the calculation. The optimal strategy is to stop pulling balls from the urn when the expected value of the reward is equal to the number of non-green balls pulled.

References

• [1] Feller, W. (1968). An Introduction to Probability Theory and Its Applications. John Wiley & Sons.
• [2] Ross, S. M. (2010). A First Course in Probability. Pearson Education.
• [3] Grimmett, G. R., & Stirzaker, D. R. (2001). Probability and Random Processes. Oxford University Press.

Further Reading

• [1] Probability and Statistics: A Course for Students in Mathematics, Engineering, and the Sciences. (2013). Pearson Education.
• [2] Probability, Random Processes, and Statistical Analysis. (2011). John Wiley & Sons.
• [3] Stochastic Processes: Theory for Applications. (2012). Cambridge University Press.
  When to Stop Pulling Balls from an Urn: Q&A =============================================

Q: What is the problem of pulling balls from an urn?

A: The problem of pulling balls from an urn is a classic example of a probability and combinatorics problem. It's a simple yet intriguing scenario that has been debated by mathematicians and statisticians for centuries. In this problem, we have an urn containing $n$ balls, of which $g$ balls are green. We can pull as many balls as we want, and for every non-green ball pulled, we get $1. If we ever pull a green ball, we stop pulling and our reward is the number of non-green balls we've pulled so far.

Q: What is the optimal strategy to maximize the expected value of the reward?

A: The optimal strategy is to stop pulling balls from the urn when the expected value of the reward is equal to the number of non-green balls pulled. In other words, we should stop pulling balls when:

$$E[R] = \frac{n-g}{n}$$

Q: How do we calculate the expected value of the reward?

A: We can calculate the expected value of the reward using the concept of expected value and the linearity of expectation. We can write the expected value of the reward as:

$$E[R] = E[X] + E[Y]$$

where $X$ is the number of non-green balls pulled and $Y$ is the reward for each non-green ball pulled.

Q: What is the mean and variance of a binomial distribution?

A: The mean of a binomial distribution is given by:

$$\mu = np$$

where $n$ is the number of trials and $p$ is the probability of success.

The variance of a binomial distribution is given by:

$$\sigma^2 = np(1-p)$$

Q: How do we simplify the calculation of the expected value of the reward?

A: We can simplify the calculation of the expected value of the reward by using the mean and variance of the binomial distribution. We can write:

$$E[R] = E[X] + n-g = \mu + n-g$$

Substituting the values of the mean and variance, we get:

$$E[R] = (n-g)\left(\frac{n-g}{n}\right) + n-g$$

Simplifying the expression, we get:

$$E[R] = \frac{n-g}{n}$$

Q: What is the significance of the optimal strategy?

A: The optimal strategy is significant because it maximizes the expected value of the reward. In other words, it gives us the highest possible reward for the given number of non-green balls pulled.

Q: Can we apply this strategy to other problems?

A: Yes, we can apply this strategy to other problems that involve probability and combinatorics. The key idea is to use the concept of expected value and the linearity of expectation to simplify the calculation of the expected value of the reward.

Q: What are some real-world applications of this problem?

A: Some real-world applications of this problem include:

• Insurance: Insurance companies use probability and combinatorics to calculate the expected value of claims.
• Finance: Financial institutions use probability and combinatorics to calculate the expected value of investments.
• Engineering: Engineers use probability and combinatorics to calculate the expected value of system failures.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

• Not using the concept of expected value: Failing to use the concept of expected value can lead to incorrect solutions.
• Not using the linearity of expectation: Failing to use the linearity of expectation can lead to incorrect solutions.
• Not simplifying the calculation: Failing to simplify the calculation can lead to incorrect solutions.

Q: What are some resources for further learning?

A: Some resources for further learning include:

• Books: "Probability and Statistics: A Course for Students in Mathematics, Engineering, and the Sciences" by James E. Gentle
• Online courses: "Probability and Statistics" on Coursera
• Websites: "Probability and Statistics" on Khan Academy