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

2/3 would be the answer.

You start with 3 $100 and 1 $1 (a total of 4 bills)

making the initial odds 3/4 to obtain $100 (3 out of 4 bills are possible of being $100) – 1/4 to obtain $1 (1 out of 4 bills are possible of being $1)

You take away 1 bill, bringing the number on the right side (the total number of bills) down to 3.

There are 2 $100 bills left, and 1 $1 bill left

The newly calculated odds are now a 2/3 chance of obtaining a $100 (only 2 out of the 3 are now possible of being $100), and a 1/3 chance of getting a $1 (only 1 out of the 3 are possible of being a $1).

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To keep it going, if you were to take away another bill; let’s pretend you grab another $100.

The odds of grabbing a $100 or $1 bill now are 1/2 – 1/2. – there are 2 total bills left, and 1 of each bill.

Had you taken the $1. we’d know for sure, with 2 bills left and both of them being $100, the odds would be 2/2.

TLDR;

Left side: number of possible precise bills

Right side: number of overall total bills