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

Let’s say before you start they have drawn a red dot on the $100 bill going in with the $1, a blue dot on one of the other $100 bills, and a green dot on the last one. So the blue and the green are together in one box. (Let’s say the dots are written in invisible ink and are only visible after the whole experiment is over, but this gives us a way to talk about each bill separately.)

Before you pick anything, there are four possibilities:
A) $1 then $100 (red)
B) $100 (red) then $1
C) $100 (blue) then $100 (green)
D) $100 (green) then $100 (blue).

But you drew a $100 bill first. You don’t know what color the invisible dot is, so you might have the red, green, or blue $100 bill in your hand. But in any case, you’re clearly not in scenario (A).

Can we narrow it down at all between (B), (C), and (D)? No. There’s no way to know. You might have the red, the blue, or the green in your hand. They are all equally likely.

But in two of those scenarios, (C) and (D), you’re going to draw another $100 out, while in one of them your second bill is going to be a $1. So there’s a 2 in 3 chance you’ll get $100 and a 1 in 3 chance you’ll get $1.