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

When selecting a box you only have two choices. But by taking out one note at a time each choices is split again into to options. Resulting is having four choices:

1. first $1, second $100,
2. first $100, second $1,
3. first $100, second $100,
4. first $100, second $100 (reverse of 3)

After selecting a box and before taking a note, all four options are equal. But after selecting $100 note first, only the three options (2, 3, 4) are possible. Therefore 2 out of the remaining 3, options means you have the box with $200. Resulting in 2/3 (not 1/2)

If the reverse happened, and you selected the $1 first. You have 100% chance of the box with $101, no one would argue that you still have a 50/50 (or 1/2) chance because that is the chance you started with.