This post is, to some extent, a follow-up to my post on the two envelopes problem.

As before, you’re the subject of an experiment. Your adversaries, who I’ll call “Team E”, present you with two envelopes, each containing a slip of paper with a different number written on it. The numbers could be any (Real) numbers, including non-integers, and negative numbers. The numbers don’t have to be related in any way, other than not being equal.

You choose one of the envelopes, open it, and look at the number. Now, you have to guess which number is greater than the other.

The question is, can you win this game with probability strictly greater than 50%?

I know I’ve seen this puzzle written about as a follow-up to the two envelopes problem, but I couldn’t find a reference. One source of it is Alex Bellos’s book *So You Think You’ve Got Problems?*, as a follow-up to a different problem he names “The Three Slips of Paper”.

It’s not always stated clearly, but for this post I’m assuming your winning probability has to be some Real number greater than 50%. Until we see what was in the other envelope, it might not be possible to say what that probability actually was. But it’s not enough to have a strategy that wins ()% of the time, since isn’t a Real number.

The orthodox answer is that you *can* win this game. To do so, choose a “random Real number”, using any random distribution you want that has a nonzero value over the entire Real number line. For example, the Gaussian normal distribution will do.

Then, make the (probably false) assumption that your random number lies between the numbers in the envelopes, and make your guess accordingly. (If your number was exactly equal to the one in the envelope, just flip a coin.) When your assumption is right, you win 100% of the time. When it’s wrong, you win 50% of time. Altogether, it seems that you will win more than 50% of the time.

It’s “easy” to “prove” that this answer is “correct”.

Generating such a random number *is* possible, or at least seems to be. Actually using the normal distribution might be tricky, but here’s a different way: Construct a random decimal number from left to right, symbol by symbol, choosing randomly from the set of syntactically valid symbols (digits, “-“, “.”). Stop when your pending number is provably larger, or provably smaller, than the number on the slip of paper.

Whatever numbers Team E select, there’s a nonzero probability that your random number will lie between them, so your probability of winning is greater than 50%.

But if you’re suspicious of this solution, I’d say you have every right to be. It’s a little too amazing. And the problem not very well-defined. Consider that, if you make a computer simulation of it, whether it is solvable or not will depend on the subtle details of how you implement the simulation.

Instead of diving much deeper into this “which number is larger” problem, I want to consider a different problem that we might simply call “guess what number I’m thinking of”. In this case, there’s only one envelope, and you have to guess *exactly* what number is written on the slip of paper inside it, and you just have to succeed more than 0% of the time. You win if you can come up with a strategy for doing that. Team E wins if you can’t. At first, this seems to be a harder problem. But I think the problems are not so different.

Let’s think about it.

To be clear, both you and Team E have access to a perfect random number generator.

If there are a finite number of numbers that Team E can choose, then you win if you know what they are, and have the ability to guess an arbitrary one of them.

If you have to write your guess on a slip of paper, and Team E’s slip of paper is bigger than yours, then Team E can win by always picking a number that’s too big to fit on your slip of paper. That’s a silly observation, but it’s really what this game boils down to: Who has the biggest memory?

At some point, we should address the issue of how numbers are allowed to be represented. But why not sidestep that by allowing Team E to draw anything they want on their slip of paper paper? Your challenge will then be to draw “the same thing”, on an equal-sized slip of paper.

Now we have to consider how the referee is going to judge whether your drawing is the same. More mundane details. But since the entire observable universe contains only a finite amount of information, a drawing on a slip of paper certainly does. Given any well-defined judging criteria, the number of different drawings Team E can make is finite. If you are able draw randomly enough, and if the game is not clearly unfair, you can win.

Suppose we formalize the problem. Let the drawings be restricted to 3″×5″, printed at 600 dpi on a black and white printer. Then the number of possible drawings is precisely 2^{5400000}. You can easily choose a random one of them, and thereby guess correctly with probability 1/2^{5400000}. That’s a small number, but it’s greater than 0, so you win.

Basically, if there’s any sort of bounds on what Team E can do, and you can guess randomly from everything within those bounds, win.

If both you and Team E have *infinite* resources, and are not bound by the capacity of a slip of paper (or the universe), and are allowed to choose a number with infinite complexity, it seems like Team E should have the advantage. They could win by choosing a random Real number between 0 and 1, using a uniform distribution. Your probability of guessing it would not be larger than 0%.

The “which number is larger” problem might be a bit different, in that it seems more plausible that you can win in the case where both you and Team E have infinite resources, using e.g. the random-number algorithm given previously. For Team E to win, they’d have to choose two different Real numbers that are infinitely large, or infinitely close together. But no such numbers exist. Still, I’m highly suspicious of this type of reasoning.