I don’t feel 100% comfortable with this wording, but it seems to consistently match the results that I get using Bayes theorem so I’ll go with it:
You need to treat each $100 bill separately. So you could have chosen the box with 2x100 and chosen the first 100 bill, or you could have chosen the box with 2x100 and chosen the second 100 bill
May I ask what strand of maths this inquiry, is in? if you do not mind me asking.
Probabilities, statistics
So you are saying that if you got a $100 - there are two ways that could have happened -
So you conclude – ?
I meant what theorem/s specifically, to calculate this/a/any probability?
Yes, that seems like a fair way to phrase it.
1/6 that you got 100 from boxA with 1 and 100,
1/6 that you got 100 from boxB with 1 and 100,
1/6 that you got 100#1 from boxC with 100 and 100,
1/6 that you got 100#2 from boxC with 100 and 100,
1/6 x2 for each of the possibilities of choosing the $1 from a and b, but we know that didn’t happen so we can throw those possibilities away
Leaving us with 1/2 chance of having chose the box with 2x100
I use Bayes Theorem personally, but that’s not the only way to calculate it.
The question was about the probabilities concerning what is left to choose from - not the probabilities concerning the prior choice that got us here. You seem to be explaining the prior choice.
Initially we had 1/6 chance of choosing any one of the bills. After choosing one - we are left with 5 bills (two $1 and three $100).
Knowing nothing else - 3:5 probability of choosing another $100 – except that we do know something else – we chose a box - not merely a bill. We want to know what is in that SAME box.
We know that the 5 remaining bills have been prearranged such that two of the 3 boxes would have only $1 left if either was the one chosen. That leaves us with a 1:3 probability that we chose a box that had two $100’s - and thus a $100 still remaining in that box.
And your answer to the first question, then, is 1/2 by the same reasoning?
“The first question”?
Yes, the very first question in the thread.
There are 2 questions, one involving 2 boxes, one involving 3 boxes. You’ve answered about the one involving 3 boxes, I didn’t see that you’ve answered the question about 2 boxes though.
The way I solve such problems is decision-by-decision – then multiply the probabilities.
When given 2 boxes (regardless of their content) - I have a decision point - 1:2.
I have been told that only 1 of the 2 had two $100’s (so 1:2 that I chose that one)
After having revealed a $100 in my box (instead of a $1) -
I now know that of the two boxes - only one still has a $100.
But my choice has already been made.
It is stipulated that to “win the other $100” I have to choose the right box - there are only two boxes and only 1 opportunity.
That leaves me a 50% chance from the beginning to the end. 
The question isn’t really about winning anything. The question is only about the probability that the other bill in the box you selected is also a $100.
I suppose I could rephrase it to make it about winning something, but that would make it a bit messy and wouldn’t clarify anything.
Right, so in your opinion, the fact that you selected a $100 bill doesn’t change the likelihood at all that the initial box choice left you with the box that has 2x $100. The first bill being $100, in your opinion, gives you no information at all about which box you chose, is that correct?
Whether winning or losing - has nothing to do with the probabilities other than to claim the preferred outcome.
Yes - finding the $100 gave you no new information. It only told you that there are still two boxes with only one having $1 - no change – EXCEPT that your decision making is over.
It didn’t matter what was in either box - you only got to make one blind choice and then stuck with the results (nothing else to calculate).
So you need a motivation to continue calculating, okay, here’s your motivation:
you only get to keep the money if you choose the box with both $100 bills, BUT you get the option to switch boxes after pulling out the first bill – once you pull out the first bill, you can say “I want to switch”, then you put the bill you selected back into the box and I will exhcange boxes with you.
And to make the deal a little bit sweeter, I’ll also throw in a snickers (or a treat of your desire) if you switch boxes and your switch was the right choice. Just to make it a little more enticing to switch.
Now that there’s an incentive to continue calculating, will you calculate the right choice?
The boxes are not what is being calculated, it’s the bills that are being calculated.
If there were 3 $100 bills and 2 $1 bills on a table, and you were blindfolded, what is the probability of picking a $100 bill? What is the probability of picking a $100 bill if they were in 1 box? How about 2 boxes? How about 5 boxes?
The boxes are NOT what is being calculated. The probability of picking a specific bill is calculated by how many of each bill there are. 3 of 5 means a probability of 60% of the 3, and 40% of the other 2, for a total of 100%.
It’s the other bill in the box that’s being calculated, though.
Would you be persuaded if I could show you experimentally that the results are not what you think they are?
You said the probability is 50% of choosing another $100 bill. What is the probability of choosing a $1 bill? Show your work for both.
Seems like you either ground the probability of first choice in # of bills, number of boxes, or number of bills in initial box, and second choice in number of bill in remaining box versus either number of bills in remaining boxes, or number of remaining boxes (and now you might just switch for the snickers, because you only keep the money if you choose the box with both 100s…boxes def not distraction).
When we threw in how drawing a 1 first would emphasize the importance of the boxes, things got interesting. Did we end up concluding the boxes are never a distraction using Bayes decision-by-decision? Motor Daddy says no…
Here’s the bayesian statistics formulation of the math:
boxA 1 100
boxB 1 100
boxC 100 100
probabilty i picked box C given i chose a 100
A = I picked the box with 2 100$
B = I chose a $100
P(B|A) = probability of B given A, which is 1
P(A) is 1/3
P(B) is 4/6
Bayes formula:
P(A|B) = P(B|A) * P(A) / P(B)
(1 * (1/3)) / (4/6)
0.5
The probably of the second bill being a $1 is also 0.5, 50%.