An Interesting Probability Scenario

Are we agreed, at least, that if you knew specifically that the envelope values were drawn from some finite range with some particular (possibly uniform, possibly not) probability distribution, then it would be advantageous to adopt any course of action that made you more likely to switch on lower values and less likely to switch on higher values? Is the issue only about tackling “uniform distributions on an infinite range”?

Uniform distributions on an infinite range are difficult to make sense of in all the ways we might like to, and steeped in paradox accordingly. For example:

Consider the problem of picking a point (x, y) where x and y may take on any positive values at all, uniformly. (That is, imagine picking a point at uniform random in the “first quadrant”, as people say…). If you like, x here represents the amount of money you see in the envelope you peek at, and y represents the amount of money in the other envelope.

Right off the bat, let’s ask: What’s the probability you should switch? I.e., the probability y > x; i.e., the probability the chosen point is above the diagonal line y = x.

One intuition is that this probability is 1/2: that diagonal line cuts through this infinite square with as perfect symmetry as can be, so that we might say half of its area lies on one side and half on the other. Put another way, (a, b) and (b, a) should be equally likely for each {a, b}, so that there’s a 50-50 chance of either ordering.

Another intuition (supported by your claim “the amounts are infinite and it is always preferable to switch (because any finite number will be eclipsed by the infinite possibilities above)”) is that the probability of y > x is 100% (or infinitesimally close to this): for any particular value of x, there is an infinite range of possibilities for y, out of which only a finite range is below x; thus, for any particular value of x, it is almost certain that y > x, and since this is true for each x, it must furthermore be true unconditionally that almost certainly y > x.

But, again, we can apply symmetry to this argument and say “The probability that x < y must be 0 (or infinitesimally close to it)!”. For, conditioning first on the value of y, we find that almost certainly x (with its infinite range of possibilities, only finitely many of which are below whatever particular value of y) must be larger than y, and since this is true for each possible y, so, in the same as before, we conclude that almost certainly x > y.

Three different arguments yielding three different conclusions for the probability that x > y (i.e., the proportion of our infinite square lying to the right of the line y = x through it).

Which one is right? …Well. They’re all right, in different senses. None of them are erroneous, anyway; those are all indeed arguments one can legitimately make, via the usual rules of probabilistic reasoning augmented with the idea of this infinite uniform distribution. That’s all there is to it. There’s not a single well-defined notion of how to speak about the relative proportions of area in this case. Uniform distributions on infinite ranges are paradoxical, in this way, by just this sort of argument. So it goes.

It’s like the question of the mean value of y - x, which is to say, the mean value of a random signed number. Perhaps it should be 0, in that the values above 0 are symmetric to the values below 0. But in the same way, perhaps it should be 1, since there is symmetry around 1 as well. Or -1. Or any value you like. There is not a single well-defined center of a bidirectionally infinite line. Again, there is paradox lurking here. So it goes.

[Now, non-uniform distributions on an infinite range, such as “1/2 probability of being between 0 and 1, 1/4 probability of being between 1 and 2, 1/8 probability of being between 2 and 3, etc.”, are easy enough to handle, but not often what we actually have in mind and desire to formalize. Alas, alas, alas.]

I doubt if this is true.

The only true and valid answer to that “paradox” is that it involves a theoretical infinite distribution. You can get all sorts of “paradoxes” if you involve infinity. (E.g. the number of odd numbers in the universe is equal to the number of odd plus even numbers in the universe, et al.)

If you try to limit the distribution to a non-infinite distribution, then the paradox falls apart, because the first envelope would then give you information about the second.

The over-under is predicated on what you would consider to be a trivial amount, along with your evaluation or the potential for generosity by the stuffers of the envelopes.

Part one, is if the amount of money is trivial and you are willing to forego it, you take the other envelope. But if the amount is enough to make a difference in your life, you take it and run. Your own particular circumstances is the variable.

Part two, if you are answering an easy trivia question between innings at a local ball game, and you get five bucks in the first envelope, there is little chance there is more than that in the othe and y0u’ve probably already found the jackpot… But if you’re on a nationally televised network quiz show, blow off the five and see how deep their pockets are. There WILL be a context to the offer that you can read.

I don’t think there’s enough information to improve on anything, but I’ll give my answer before looking through the thread to see if there was something.

But for me, the strategy I would follow is:

  • Is the amount in the envelope enough for me to care about losing?
    Yes –> Keep it
    No –> Try the other envelope

Hm, well looking at some of the other answers (and the XKCD link), it does seem like one can take the knowledge that 0 is a lower bound to infer that more numbers lie to the right of any value than to the left. Of course, one can also infer that the higher the number, the less probable the number is.

Over a large number of interations of the envelope test, most reasonable skewed normal distributions would probably give you a better than 50/50 return rate. But for a single iteration, my previous solution is probably more useful.