Perplexing the Web, One Probability Puzzle at a Time

The guy being interviewed (Daniel Litt) has a great expository video on Hilbert's third problem at Numberphile. (Here)[https://www.youtube.com/watch?v=eYfpSAxGakI]

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.

This balls in the urn problem is relatively straightforward with Bayesian Updating.

We have a uniform prior for the number of red balls that were placed in the Urn, P(U) = 1/101 where U ∈ [0, 100].

We then have the probability the first ball was red given a value of U, P(B1=Red|U) = U/100.

Lastly we have the P(B1=Red) = ∑ P(B1=Red|U)P(U), where the sum is over all values of U.

But we don't actually need to compute this sum because from symmetry arguments we can see that P(B1=Red) = 1/2.

This follows because there are the same number of configurations with U red balls as U green balls and each configuration is equally likely.

So now we can compute P(U|B1=Red) using Bayes Rule.

P(U|B1=Red) = P(B1=Red|U)P(U)/P(B1=Red) = (U/100)*(2/101)

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]:

P(U ∈ [51,100]|B1=Red) = ∑ P(B1=Red|U)P(U)/P(B1=Red) = ∑ (U/100)*(2/101) = (2/101) ∑ U/100 = (2/101) (50/100) * 75.5

P(U ∈ [51,100]|B1=Red) = 75.5/101

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.

my favorite problem, which I originally read here on hn:

> I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

Apparently the answer is 13/27

https://math.stackexchange.com/questions/4400/boy-born-on-a-...

Quite nice. Probability can be hard, and sometimes the correct intuition has to be found and/or cultivated.

As TFA says, "The interesting thing is that many people wouldn’t accept a mathematical argument, but found a simulation persuasive."

> [...] 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.

SPOILER

(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.