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

The probability is two thirds.

Let’s label the 100$ in the first box as A, and the two 100$ notes in the second box as B and C.

When you pick the first note, you have four equally likely cases: $1, $100A, $100B, $100C.

If you got a $100 note, you can be in one of three possible cases:

* You picked A, and the other note in the box is $1
* You picked B, and the other note in the box is C
* You picked C, and the other note in the box is B

The counter-intuitive part of this whole thing is that all the notes look the same so you can’t tell B and C apart.

If you phrase the question differently, it’s a bit easier to get the right intuition: You have a box with a 1Direction album and an Apple, and a box with a Banana and a Cabbage. If you got one of the Apple, Banana, or Cabbage from the first draw, what’s the likelihood of getting either a Banana or Cabbage from the second draw?

* You picked A(pple), and there’s a 1(Direction) in the box
* You picked B(banana), and there’s a C(cabbage) in the box
* You picked C(abbage), and there’s a B(anana) in the box.

Another way to look at it is: You have one single bag with a $1 and three $100. Two of the $100 notes have “get another $100” written on them, and the other has “get $1” written on it.