Hilbert's Third was the first of his 23 problems to be solved. It asks whether volume is enough to determine whether one can transform one polyhedron into another via cutting into finite pieces and re-arranging with rotation and translation (such polyhedrons are called scissor congruent). The situation is true in 2 dimensions.
Dehn showed that's not the case for polyhedron. He created his eponymous invariant in order to solve the problem: two polyhedrons with the same volume but differing dehn invariants are not scissor congruent.
I love the result because it was one of the first bits of serious math I read as an undergrad and it was also the first time I got a glimpse into how tensor products could be used to package information.
A pretty clear account can be found in Harthshorne's Geometry: Euclid and Beyond.
But how does that help us? Well, after we remove a red ball the probability the next ball will be red given the first was red and there were U red balls initially in the urn is:
P(B2=red|B1=red,U) = (U-1)/99.
From this we see that if U was in the range of 51 to 100, then P(B2=red|B1=red,U) > 1/2, i.e., the second ball is more likely to be Red.
So let's compute the following where we are summing over U ∈ [51,100]:
So we find that P(U ∈ [51,100]|B1=Red) = 75.5/101 ≈ 3/4 which means that given the first ball we picked was red, then roughly 3/4 of the time we would expect the next ball being red to be more likely.
This seems like a much more straightforward way to think about it for me.
KalEl's answer[0] is more clear to me. The most natural, intuitive meaning I would attach to the question is (2), i.e., "I have two children. I have randomly selected a child by flipping a coin, and that one happens to be a boy born on a Tuesday. What is the chance I have two boys, i.e., the other child is also a boy?", so the answer would be 1/2.
Intuition is in the mind of the beholder I guess. If someone told me they had two balloons, "and one of them is green", I would probably infer that the other one is not green.
The next most natural interpretation for me would be that "at least one of my two balloons is green".
I was so confused about this question and all the answers for so long, until it finally clicked.
To me, this “puzzle” very obviously reads as your new acquaintance Dave volunteering some information about his kids. This means the probabilities are completely independent, and the answer is obviously 1/2.
But the intention is to make it sneaky and complicated, in the style of bad, poorly worded puzzles. While it is phrased as a conversation, it is actually about drawing a sample from a random cohort, filtering based on gender and birth day, then determining the probability of the result.
Which is completely different from how the puzzle is phrased.
Suppose they said, “I used to be part of this club that I liked, but they had a rule that if anyone ever missed the weekly Tuesday meeting, they were kicked out, with no exceptions. I really liked that club, but being kicked out was worth it, worth it a thousand times over, to be there as my wife gave birth to my son (which happened on a Tuesday).” and at some other point imply that the two of them have two kids?
(This contrived example is because the first example I thought of implied that their other child was probably not born on a Tuesday. This example avoids this because if the other child came second, the same reason for it being relevant if it was on a Tuesday doesn’t apply. Uh… hm, actually, not sure if this fixes the issue…)
P([they have two sons] | [they have two kids] & [they have a son born on a Tuesday]) = P([they have two kids who are both boys] & [at least one of their two kids was a boy and born on a Tuesday] | [they have two kids])/P([at least one of their two kids was a son born on a Tuesday] | [they have two kids])
If two sons, the chance that at least one was born on a Tuesday is 13/49 . If they have one son, the chance that have a son born on a Tuesday is 7/49.
So, given have two kids, chance of both boys and at least one born on a Tuesday is (1/4) * (13/49) = 13/196.
Given two kids, chance that at least one son born on a Tuesday is (2/4)(7/49) + (1/4)(13/49) = 27/196
At least for me, Monty Hall makes more intuitive sense to me when there are 10 doors than when there are 3, even though the benefit of the optimal strategy is quite small: there's a 1/10 chance you picked the car and shouldn't have switched, but a 9/10 chance that you can play the game again with 1:9 odds. Without doing the math, it makes sense to me that the optimal strategy actually does improve your odds.
Yet for some reason a 2/3 chance you can play the game with 50-50 odds is harder to accept - in particular I have to consider the 10 door case to understand why the 3 door case makes sense. I suspect it has to do with the psychology of loss aversion: a 1/3rd chance that you incorrectly switched "feels like" a reckless risk.
> [...] imagine that instead of starting with 100 balls, you start with 101 balls in a row. Pick a ball at random. Then color the balls to the left of it green and the ones to the right of it red. Throw that ball away, leaving 100 balls.
> Then pick a second ball at random. That ball corresponds to the first ball in the original problem. The problem tells you that you picked a red ball, so it was to the right of the ball you threw away. Now pick a third ball. This ball is either to the left of the first ball, between the first ball and the second, or to the right of the second. In two of the three possibilities, the third ball is red. So the probability that the ball is red is 2/3.
I think that explanation will confuse a lot of people. Let's say the first ball picked is ball 94 (numbering from 1 to 101 left to right). So 1-93 end up green and 95-101 are red. Let's say the second ball is 98.
The third ball then must be from one of these three intervals: green 1-93, red 95-97, red 99-101. Those intervals have lengths 93, 3, and 3. OK, so now we've got three intervals which sounds like the situation the explanation is talking about, and two of those three give a red ball, so the probability the second ball is red is 2/3.
But that is clearly wrong, since the probability of the ball being picked from an interval is proportional to the length of the interval. There is a 93/99 chance that ball 3 is green and only a 6/99 chance it is red for these particular intervals.
It needs to be made clear that what you have to look at is all possible ways to pick the three balls.
Maybe something like this. I'm going to call the position of the ball that divides red and green D, which is an integer in [1,101], P1 the position of the first colored ball picked, also an integer in [1,101] not equal to D, and P2 the position of the second colored ball, also an integer in [1,101] not equal to D or P1.
Imagine a list of every possible legal (D, P1, P2) 3-tuple. Note that each consists of 3 distinct integers in [1,101]. Partition that list into several sublists, where two of the 3-tuples, (D, P1, P2) and (D', P1', P2') are in the same sublist if and only if they contain the same numbers. E.g., (5, 2, 12) and (2, 5, 12) would be in the same sublist.
Each sublist will have 6 of the 3-tuples, one for each permutation of the three distinct numbers shared by all the 3-tuples in that sublist.
Each of those 6 permutations corresponds to a different ordering of D, P1, and P2. For example (5, 2, 12) corresponds to the ordering P1 < D < P2 and (2, 5, 12) corresponds to D < P1 < P2.
Here are the 6 possible orderings, and what the outcome is when the balls are drawn:
D < P1 < P2 P1 red, P2 red
D < P2 < P1 P1 red, P2 red
P1 < D < P2 P1 green, P2 red
P1 < P2 < D P1 green, P2 green
P2 < D < P1 P1 red, P2 green
P2 < P1 < D P1 green, P2 green
In the 3 cases where P1 is red, P2 is red in 2 of them and green in 1. That gives a 2/3 chance of P2 being red if P1 was red.
This is true in all of the partitions of our list of all legal(D, P1, P2), so is true in all cases.
What if there were just 4 balls instead of hundred. Second probability of 2nd ball being of same color will be higher only if the distribution of first was higher.
(Note that the article gives the answer, and discusses arriving at the answer. So go read it first.)
Article opens with a puzzle. This is commenting on the article's explanation .. avoiding some words in the comment just so this spoiler doesn't spoil at a glance, you'd have to think about the following before reading:
Curiously, this article doesn't touch on an instantaneous and no-maths intuition about what can (or what cannot) come up as the set after all draws including your first. There are a finite set of possible sets in the container. One interesting set is ruled out by the first draw: there is a set it cannot be. With that set ruled out, you're at least (sets/(sets-1)) more likely the next draw isn't from the missing and interesting set that is intuitively known to you.
Further, if one did do this in the real world, it's intuitively evident that the missing set and its inverse would be more likely chosen for the container due to their interestingness. (The article alludes to this when it points out the sets are not coin flips, they are deliberately selected by a human.) That tips the odds even more in your intuition's favor of how the probabilities must balance.
So I'd argue this one is solvable by psychology as an alternative to math.
Ypur first argument is interstellar but doesn't help solving, because the set ruled out is exactly balanced (cancelled) by the first red ball being removed from al the other sets.
The second paragraph just says that guesses can sometimes be right, which of course is true, but doesn't tell us which guesses are likely to be correct. Wha y you claim as intuitively obvious at all, and I think most people would avoid those sets. And it's not the same problem anyway, because in the problem the sets are chosen randomly according to a stated rule, not based on a personal bias.
Hilbert's Third was the first of his 23 problems to be solved. It asks whether volume is enough to determine whether one can transform one polyhedron into another via cutting into finite pieces and re-arranging with rotation and translation (such polyhedrons are called scissor congruent). The situation is true in 2 dimensions.
Dehn showed that's not the case for polyhedron. He created his eponymous invariant in order to solve the problem: two polyhedrons with the same volume but differing dehn invariants are not scissor congruent.
I love the result because it was one of the first bits of serious math I read as an undergrad and it was also the first time I got a glimpse into how tensor products could be used to package information.
A pretty clear account can be found in Harthshorne's Geometry: Euclid and Beyond.