An infinite debate?
March 23, 2011 10:54 PM   Subscribe

Got into a debate with a friend regarding a math question - I think our debate reflects his misapplication of finite math reasoning to infinities, but I'm having trouble dredging up what I learned in set theory 10 years ago. Imagine you have a bag that has all of the real numbers from 0 to 2, let's say. What are the odds that if you reach into the bag you pull out a number between 0 and 1.5 ? 75%? 50%? Why? Depending on the answer to this I may have followup questions, but I'll leave it here for now.

Screw it - followup question now. Same question for 0 to 180, with 0 to 90 being what you are trying to pull?

Now imagine the bag contains all triangles with two sides of 1 unit length and an angle between those two sides of 0 to 180. Thus, the bag contains an infinite number of triangles representing all possible acute, right, and obtuse triangles. What are the odds that if you reach into the bag you pull out an acute triangle? If you consider the angle between the two sides as the determining factor, it is between 0 and 180 and anything less than 90 is acute. But if you consider the length of the far side to be the determining factor, it is between 0 and 2, and anything less than sqrt(2) or ~1.4 is acute. So for the former, it is equivalent to the 0 to 180 question; for the latter, the 0 to 2 question. Yet the bag of triangles is the same, so the probability has to be the same no matter how you look at the problem, right? Or am I missing something?

My contention is that if you have two disjoint infinite sets of the same cardinality, then if you combine those sets, the probability of selecting an element of either set at random will always be the same. That seems odd in the context of the 0 to 2 problem above, so I thought I'd appeal to a higher cardinality of intellects.
posted by slide to Education (9 answers total) 5 users marked this as a favorite
"The misapplication of finite math reasoning to infinities" is indeed what's going on here.

When we say we choose an element "randomly" from a finite set, we are using the fact that there's a unique probability distribution, the uniform distribution, which assigns equal probability to each element.

There are lots of probability distributions on [0,2] which assign equal probability (namely, zero!) to each point on the interval. In your question you are implicitly thinking about two different choices of distribution. You say

"the bag of triangles is the same, so the probability has to be the same no matter how you look at the problem, right?"

and the answer is no, this is not right -- this is where you are mistakenly using your intuition about bags with finitely many things inside.
posted by escabeche at 11:09 PM on March 23, 2011 [2 favorites]

Agree with escabeche. The second probability is conditional on the first. Suppose x ranges uniformly from 0 to 1, and x^2 ranges from 0 to 1. P(x<0.5) = 50%. Thus, P(x^2 < 0.25) = 50%. Both sets, x, and x^2, have the same span from 0 to 1 but have different probability distributions. In essence you're not selecting an element from x^2 "at random"; its value is determined by your random selection of x.
posted by PercussivePaul at 11:34 PM on March 23, 2011

You can actually apply this to the linear one as well. If you have numbers 0-2 as you describe with an even probability, and then you square all the numbers in the bag, now you have 0-4 - but your probability distribution isn't flat anymore because half your numbers are still smaller than 1, just as they were before.
posted by Lady Li at 12:25 AM on March 24, 2011 [2 favorites]

Yep, previous answers have it.

You're assuming a uniform distribution in all these examples, but with the triangles you need to specify which parameter is uniformly distributed: is it angle or it is the far side? As you've already figured out, the two options are mutually exclusive.
posted by d. z. wang at 12:30 AM on March 24, 2011 [2 favorites]

If 'a' is the angle of the triangle then the length L of the side is 2sin(a/2). If 'a' is uniformly distributed over 0..Pi then the average value of L is 1/Pi * ∫2sin(x/2), x=0..Pi = 4/Pi = 1.273. If L was uniformly distributed its average would be 1.
posted by Rhomboid at 1:12 AM on March 24, 2011

Part of the confusion arises from the fact that you can't actually have a bag containing all the real numbers from 0 to 2 (see Cantor's diagonal argument). Therefore, any intuition you may have about pulling things out of bags likely won't work.

A better mental model for playing with real numbers is pieces of string. Here's the rough analog of your first bag example: Get a two foot piece of string, dip one end in red dye, and cut it at some random point. You now have two bits of string. If one of those is entirely red, keep it and discard the other one; otherwise keep only the bit with a red end. What are the odds that the piece you still have is shorter than 1.5 feet?
posted by flabdablet at 3:24 AM on March 24, 2011

Yet the bag of triangles is the same, so the probability has to be the same no matter how you look at the problem, right? Or am I missing something?

No it's not, or rather, the probabilities do not depend on just the bag: they depend on the bag as well as the selection process. Your argument wouldn't hold even for a finite bag of physical triangles, unless you took great pains in setting up the selection process to produce each triangle with an equal probability.

The difficulties from applying this kind of thinking to geometrical problems were one of the key points in the development of modern probability theory. See, for example Bertrand's Paradox.
posted by Dr Dracator at 5:06 AM on March 24, 2011 [1 favorite]

People above have the right answers. Elements of sets do not have probabilities themselves, you need to assign some probability measure to the space. You then compute probabilities by integrating (or summing, if things are discrete) over the measure of the elements of whatever subset you are considering, e.g. The real numbers between 0 and 1.5. The cardinality doesn't explicitly come into it at all. Google measure theory if you want to read more.
posted by Schismatic at 5:17 AM on March 24, 2011

Wikipedia has a great writeup of the similar Bertrand Paradox. It's pretty easy to understand, and the ultimate resolution to the paradox is that the different methods employed to solve the problem (which give different probabilities) are all correct: what is incorrect is thinking that you can meaningfully "randomly" choose a chord in a circle (in this case. For you, a triangle.) without specifying the method of random selection. In fact, you can't, and the way you choose the random object determines the probability, which can be calculated only once the method of selection is known.
posted by milestogo at 7:43 AM on March 24, 2011

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