Don’t look up the answer or read further, try to solve it yourself first.
I present to you 4 USD bills, 1 $1 bill and 3 $100 bills. I then tell you I’m going to put 2 of these bills in one box, and the other 2 in another box - I shut a curtain and do so out of your sight. I then present you with the two boxes.
So, in front of you now are 2 boxes, both apparently identical from the outside, but one has a $1 and a $100 in it, and the other has a $100 and another $100 in it. You don’t know which one is which.
Now I say, you may choose a box, so you do so - I put the other box away. I now say, reach inside and grab one of the bills inside the box. You do so, and you find that you’ve selected a $100.
What is the probability that the other bill remaining in the box you selected is also a $100?
That means there is only a 1 in 4 probability (25% chance) of selecting the $1 bill at the start, and a 3 in 4 probability (75% chance) of selecting a 100 dollar bill.
Since one of the $100 bills is already removed, then there is a 1 in 3 probablility of selecting the $1 bill and a 2 in 3 probablility of selecting another 100 dollar bill.
So at that point there is a 66.666…% probablilty of selecting a $100 bill, and a 33.333…% probablity of selecting a $1 bill.
You got the right answer here, Motor, numerically. So congrats on that!
This might sound silly, but I don’t know if the logic you used to get there makes sense though. I assume this bit is the meat of your reasoning:
So I’ve devised an alternative scenario, to see if you can apply the same reasoning to this alternate scenario and still come to the correct conclusion.
This scenario is just like the first, but instead of 2 boxes, there are 3 boxes.
1 box has $1 and $100
1 box has $1 and $100
1 box has $100 and $100
Just like the first scenario, in this scenario you choose a box, essentially at random - you have no idea if you chose the box with 2x $100 or if you chose one of the 1 + 100 boxes. And once again, you pick out a bill at random, and you happen to find that you picked a $100.
So, just as in the first scenario, the question is, what’ the probability that the other bill remaining in the box is also $100?
Because in the first you have 4 total bills spread over two boxes, and in the second you have 6 spread over 3… 2 per box in each case… and same ratio of 1s to 100s.
There are 6 bills total, 4 of which are $100 bills, and 2 of which are $1 bills.
At first there is a 4 in 6 chance (66.666…%) of pulling a $100 bill, and a 2 in 6 chance (33.333…%) of pulling a $1 bill.
After picking the $100 bill out there is a 3 in 5 chance (60%) of pulling a $100 bill, and a 2 in 5 chance (40%) of pulling a $1 bill.
The boxes are a distraction, they mean nothing.
4 divided by 6 = .666… (66.666…%)
2 divided by 6 = .333… (33.333…%)
3 divided by 5 = .60 (60%)
2 divided by 5 = .40 (40%)
If there were 6 bills at the start, 4 being $100 and 2 being $1, then after drawing a $1 there would be 4 $100 and 1 $1 remaining.
So the percentages would be 4 in 5 (80%) for $100 bills, and 1 in 5 (20%) for $1 bills.
Put 17 marbles in a hat, of which 16 are red and 1 is blue. There is a 1 in 17 chance (5.88%) of you picking the blue marble and a 16 in 17 chance (94.11%) of picking a red marble.
If there are 6 numbers, each between 1 and 50, then there are 50 x 50 x 50 x 50 x 50 x 50 (15,625,000,000) possible combinations. If you buy one ticket with one combination of numbers you have a 1 in 15,625,000,000 chance (.0000000064%) of winning. If you have two sets of numbers you have a 2 in 15,625,000,000 chance of winning. In order to have a 100% probability you would need to buy 15,625,000,000 tickets with every different combination of numbers, and you would be in debt after winning, from buying all those tickets, even if you won a billion dollars!. At $1 per set of numbers you would have paid 15.6 BILLION dollars to win 1 billion LOL. It’s not economically feasible.
Ever buy a lottery ticket and look at the odds of winning?
P.s. I bought two lottery tickets in my life. Once when I turned the legal age to buy a lottery ticket. Once when I was delusional. Someone always wins. Eventually.
Really? You found a box with a 100% chance of drawing $100 & you’re going to, what, take your chances with another box? Those other two boxes are a distraction, yes, but that first box… not a distraction.
The boxes are not just a distraction, they mean something. I’m glad I asked the second scenario, because my intuition was right, I see the logic you’re using and it’s not generalisable to other scenarios.
The probability of pulling another 100 out of the same box in the second scenario is 1/2
I think you’re confused how I figured the ratio. Whenever there is a 1, there is a 100. In the first scenario, that happens 1/2 boxes. In second scenario, that happens 2/3 boxes. Do it again… 3/4 boxes. And so on.
In scenario 1, one $1, three $100s. 4 bills. 2 boxes.
Box 1: $1, $100
Box 2: $100, $100
In scenario 2: two $1s, four $100s. 6 bills. 3 boxes.
Box 1: $1, $100
Box 2: $1, $100
Box 3: $100, $100
That there are 4 bills over 2 boxes, and 6 bills over 3 boxes, two in each box… I don’t know how to explain or say it to you, but the proportion of 1s and 100s is balanced between the scenarios. If you did 8 bills over 4 boxes, you’d have three 1s and five 100s. Balanced the same as the first two scenarios.
Box 1: 1, 100
Box 2: 1, 100
Box 3: 1, 100
Box 4: 100, 100
The ratio is literally not the same. Neither before you select the firt $100, nor after.
In the first scenario, before you select the $100 out of the box, you have
1x $1
3x $100
1:3
In the second scenario, before you select the $100, you have
2x $1
4x $100
2:4
So in the first scenario, you have 3 times as many 100s. In the second scenario, you have twice as many 100s.
But you’re not being very clear or specific in your wording, so I’ll give you the benefit of the doubt and assume that MAYBE you meant after you select the $100, the ratios are the same, so let’s look at that scenario:
In the first scenario, after you select the $100 out of the box, you have
1x $1
2x $100
1:2
In the second scenario, after you select the $100, you have
2x $1
3x $100
2:3
So looking at “after” also does not give you the same ratio of 1s to 100s.
You’re right, I am very confused about how you figured out the ratio.
Ichthus77, the ratio of bills to boxes is the same: there are always twice as many bills as boxes (because we put two bills in each box) but it doesn’t sound like that’s what you’re saying.
There’s is a consistent patterned relationship between the number of 1s and the number of 100s, but that consistent relationship is not “the ratio is always the same”. The relationship is, there’s always 2 more 100s than there’s are 1s
I could prove it in two ways, experimentally with code, or using Bayes theorem. Choose your poison
But in simple terms, there are two ways that you could have chosen a box with $1 and $100, and also two ways you could have chosen the box with 2x $100.
That above logic works for the first example too, there are two ways you could have chosen 100 from the 2x 100 box but only one way from the other box, hence the 2/3 probability
At $1 per set of numbers you would have paid 15.6 BILLION dollars to win 1 billion LOL. It’s not economically feasible. 
