If there are two boxes. The first has a 100$ bill and a 1$ bill, and the second has two 100$ bills. If I puck a random box and take out a 100$ bill, whaat is the chance of me taking out another 100$ bill?

Question

I’m honestly stuck. I’ve seen people say 1/2, others 2/3. Something Monty Hall Problem, Bayes Theorem but I’m still confused so here I am.

Edit: I believe you are not allowed to change your box choice on the 2nd “turn” as that would make having two boxes pointless, wouldn’t it?

In: 10

Answer 1

Omgggg I get to finally use my actuarial test skills for something other than work!!!

Ok so first we need to find the probability of getting a 100 dollar bill. This is Pr(N1=100) = (1/2)*1+(1/2)*(1/2) = 3/4

Now we simply need to calculate the Pr of the second bill being $100 given the first bill was $100.

Pr(N2=100|N1=100) = ((1)*(1/2)+(0)*1/2)/(3/4) = 2/3

This is Bayes Theorem.

To help with other problems like this, think about probabilities like a shape, where 100% of all possible probabilities are within the shape. When you are given information that has already happened, you can take away areas of the shape because you have more information, meaning you can be more sure of your answer. You must readjust the total scope of the total probabilities, so you divide by the amount of remaining area of the shape.

In the first step, we found what the probability of the first outcome was. The second outcome is dependent on the first outcome, so we have to adjust the scope of the probabilities to account for this.

To make this problem more simple, we knew a $100 bill was taken from the randomly chosen box. We know that there is 3 bills left, 2 of which are $100. Since the PR of each box is equal, we can jump straight to 2 $100 bills / 3 total bills