Understanding the two-envelope paradox

I’m a fan of maths problems, and I’m particularly keen on those seemingly impossible puzzles that seem to see the maths go against logical thinking. I’ve written about some of them before, including the Chinese boat captain riddle and the Monty Hall problem, a maths challenge that baffled even the most gifted mathematical minds of our generation. But I happened upon another fascinating and paradoxical puzzle that seems to pit everyday thinking against mathematical reasoning, and all it takes its two envelopes and some hypothetical money. 

   The problem is a very simple one to understand. There are two sealed identical envelopes, both containing money – you know that the amount of money in one of the envelopes is twice as much as the other, and that is the only information you have. You’re shown the two envelopes, and told to pick one – you can keep however much money is within that envelope. But, after you’ve chosen, you’re them given the option to switch envelopes – and the question is whether you should or not. Is there a statistically better play in order to maximise the amount of money you win? 

   Now, on its own, the answer seems to be a logical one – no, and how could there be? If the two envelopes are identical, and you know nothing about either, then it doesn’t seem like there’s any advantage to switching. However, when you bring maths into the situation, it seems like switching is always the most advantageous play, almost counter to the initial logic, because the expected value of the other envelope is always higher. 

   Let me give you an example. Say you open one of the envelopes, and it contains £60. That means the other envelope must contain either £30 or £120, meaning the expected value of that envelope is £(30+120)/2, or £75 – as you can see, it’s higher, and so it makes mathematical sense to swap. For any number x, the expected value within the other envelope is 1.25x, and so this solution implies you should always switch. And that’s the heart of the paradox – the composition of the problem is distinctly symmetrical, but the maths indicates that there is an optimal play, and that’s to keep switching between the envelopes ad infinitum. That obviously can’t be right, so what has gone wrong with the reasoning? 

   The challenge here is that there’s not a single solution – lots have been proposed, and then are lots of debates about whose solution is the right one. As a result, you can find a lot of mathematical writing on the puzzle, and on the many alterations to the puzzle. A seemingly simple problem has produced a wealth of maths and thinking, despite the answer appearing an intuitive one. I’m going to lay out one of the ways to think about the problem, but note that it’s far from the only one. 

   In many respects, the issue is the same one as in the famous puzzle with the waiter, a refund and an extra pound that comes out of nowhere – it is the framing that complicates matters. We start with a faulty premise, and then use it to derive a faulty conclusion, leading to the paradox. In case you haven’t spotted the issue, it’s this – we’ve set up a situation where the total amount is necessarily fixed, but we’re imagining that the amount in the first envelope we choose is fixed (x), but the second can be one of two amounts – 2x or 1/2x. This is not the case, and it relies on using two different meanings of a term at the same time. 

   I personally think that it’s better to think of the problem in this way. Imagine that the total amount in the two envelopes is some fixed quantity c, which equals 3x – there’s x money in one envelope, and 2x in the other. If you pick the x envelope and swap, you gain x – the other way round, you lose x by swapping. As a result, the expected average gain if you swap is 1/2(x) + 1/2(-x), or zero – i.e. there is no intrinsic benefit to swapping or keeping, the intuitive answer. Using this supposition, we can determine that the expected value for each of the envelopes is 3/2x, the same, and so there is no contradiction. 

   If you’re a maths fan, it’s worth digging into many of the technical explorations of this problem, because there’s a lot to work with. But I love it as a problem, if only as proof that the assumptions you make can produce radically different endpoints when you reason them out.