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
I think that if you can’t change boxes, the probability doesn’t change depending on the box you choose first.
First pick, you have 3/4 chance of getting a $100 and 1/4 chance of $1.
If you picked a 100$ on the first pick, 2 of the remaining bill are 100$ and 1 is a 1$. So there’s a 2/3 probability you’ll pick a 100$
There are 4 possible first draws.
The probability that your first draw is a $100 note is 3/4 since 3/4 notes are $100 notes.
If you can change boxes, there’s a 2/3 chance of drawing another $100 note. 2 of the 3 remaining notes are $100, thus, your chance of starting another $100 is 2/3.
If you aren’t allowed to change boxes, that second note is either a $1 or $100 note. The chance that it’s a $100 note is 1/2. However, since you can’t change your box, and you had a probability of 1/2 of choosing the box with two $100 notes, the probability is 1/2.
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