# The Two Envelopes Problem: a ‘Back of the Envelope’ Solution – by Raam Gokhale

## Abstract

We give a simple solution of the two envelopes problem by considering an alternate situation to which the problem’s expectation formula more properly applies. The flaw in reasoning in the original problem is apparent once the difference in the two problems is examined.

## The Problem

You’re presented with two envelopes and told that each contains a check. One contains a check for \$100 and the other contains a check for \$200. You do not know which is which but are asked to pick one. After having picked one, but before peeking inside, you’re asked if you want to switch envelopes. You happen to reason as follows: ‘if x is the amount in my envelope, the amount in the other envelope would be 2x if I’ve chosen the smaller amount or x/2 if I’ve chosen the larger amount. Since the probability that I’ve chosen the smaller amount equals the probability that I’ve chosen the larger amount equals 0.5, the expected value of switching is 0.5(2x) + 0.5(x/2) or 1.25x. But this is greater than the x which by supposition I presently have so I should switch.’ But of course once you’ve taken this first step you’re well on the way to switching ad-infinitum because before peeking inside the new envelope you could go through the same reasoning to switch again…and again.

## The Generalization

The problem is usually stated with the larger amount being 2 times the smaller, but analogous reasoning yields the paradox as long as the two amounts are different and non-zero. This can be demonstrated by letting the smaller amount be y and the larger amount be cy for some c>1. If we let the amount in the first envelope be x then the amount in the other envelope is cx if I’ve chosen the smaller amount and x/c if I’ve chosen the larger amount. Since the probability that I’ve chosen the smaller amount equals the probability that I’ve chosen the larger amount equals 0.5, the expected value—or so the story goes—of switching is 0.5(cx) + 0.5(x/c) or (c2+1)x/2c. This is greater than the original amount x as long as c>1, which it has to be since, without loss of generality, cy was assumed to be the larger amount.

## The Solution

So much for generalizing the fallacious reasoning. Let’s stick with the \$100 and \$200 because it’s easier to see the misstep, which is surprisingly easy to see once you see it. Consider another two envelopes situation where the reasoning is valid. Suppose after you’ve chosen the first envelope, someone offers you double or half of whatever is in the envelope you’ve chosen depending on the flip of a coin. The expected value of switching this time really is  0.5(2x) + 0.5(x/2) or 1.25x and so you should switch.

What’s the difference? Well clearly here the highest potential payoff is 2x no matter what x is and the lowest potential payoff is x/2 no matter what x is. In the original example this is not the case: if the amount in the first envelope is \$200, you’re not offered a 50/50 chance of getting \$400 and if the amount in the first envelope is \$100, you’re not offered a 50/50 chance of getting \$50. The expectation formula is wrong not because of the probabilities as some have urged (most famously, mathematician Keith Devlin, ‘The Math Guy’ on NPR’s Weekend Edition). The problem is with the formula for the expectation. The situation where you’re offered 2 times the amount in the original envelope with probability 0.5 and half the amount in the original envelope with probability 0.5 can be diagrammed as follows (“<” represents a 50/50 chance):

<(100<(200,50),200<(400,100)).

Here if we let x equal the amount in the initially chosen envelope, we can notice that the expected value of the subsequent gamble G is E(G|x) = 0.5(2x) + 0.5(x/2). This is because letting x equal the original amount, there is a 50% chance of doubling and a 50% chance of halving—and this is important—no matter what x is.

The proper diagram of the original situation is quite different. It is:

<(100–>200, 200–>100).

Here there is only one chance element: a 50/50 chance of picking any given envelope initially. The expected value of switching given x is:

E(switching|x) = x/2 if x = 200; and 2x if x = 100.

There is no second gamble; the initial choice has already been made (symbolized by conditionalizing the expectation on x). Here using the 50/50 chance in the initial choice to get an expected value of 0.5(2x) + 0.5(x/2) is mistaken because as we see from the definition of E(switching|x),  x is not an unknown, fixed for the equation as was the case in our modified example where you’re offered double or half no matter what the amount in the original envelope is; it is a variable, being 100 in the first term and 200 in the second; it is really misleading to use the same letter x in both terms. Substituting for the two different x’s, we have:

E(switching| “x”) = 0.5(200) + 0.5(100) = 150. This is the same as E(staying|x) since there the probability is 50% that we’ve chosen the \$200 envelope and 50% that we’ve chosen the \$100 envelope.

So there’s no point to switching—the math agrees with our intuitions.

The Moral

The moral is x in the expectation formula has to be a fixed unknown not a variable that can assume different values depending on which term it appears in. The misstep, if you want to pin it down to something specific, is after the initial envelope is chosen x is a fixed unknown; in the ersatz expected value formula it is a genuine variable that takes on different values depending on where it appears.

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### 6 Responses to The Two Envelopes Problem: a ‘Back of the Envelope’ Solution – by Raam Gokhale

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