Math help, and word problem with proabilities

Question

A prisoner is given 10 white balls, and 10 black balls, and 2 boxes. If it is white, the prisoner will go free, if it is black, he will remain in prison.Each box has an equally likely chance of being selected, but the prisoner can distribute the balls between the boxes to his advantage. How should he arrange the balls in the boxes to give himself the best chance of freedom"

Answer 1

Probability problems are much easier to explain with diagrams and pictures, which the answer forum doesn’t permit, but here we go!

 

First, let’s consider how the probability of success is calculated for this word problem. You have two paths to a successful outcome Box 1 or Box 2. The probability of success is the sum of the probability of choosing each box times the probability of success from each box.

 

P(Box 1 chosen) * P(# white balls/# of balls) + P(Box 2 chosen) * P((# white balls/# of balls)

 

For example, if you put 10 balls in each box, with 4 white balls in Box 1 and 6 in box 2, then your probability of success (choosing a white ball) is .5*.4 + .5*.6 = .5

 

Now, one thing I notice about this particular probability problem is that there is no rule for how many balls belong in each box. If your problem specified that 10 balls must be in each box, we could make a table of the possible situations and you would find the highest probability is the same for any of those situations. Tables are a great tool for probability problems. They help you to organize multiple scenarios and keep track of your work.

 

.5*0 + .5*1 = .5

.5*.1 + .5*.9 = .5

 

.5*.2 + .5*.8 = .5

etc.

 

However, let’s proceed with the problem as you’ve written it. We want to maximize each term (the Box 1 success and the Box 2 success). The best case would be 100% chance of success from each box, but we can’t have that unless we throw out the black balls. We can have a 100% chance of success from one of the boxes. Place 1 white ball in Box 1. Now we can place all the remaining balls in Box 2, maximizing the chance of getting a white ball in that box.

 

.5*(1/1) + .5*(9/19) = .74

 

We were able to increase the odds of success by .24 by thinking "outside of the box" :)