Better-Than-Random Decisions in Uninformed Settings

maximios

In this article, I want to introduce you to a simple problem with an easy-to-apply, yet awfully unintuitive solution. It is one kind of envelope problem and goes like this

There are two envelopes with some different amounts x and y of money in them. The envelopes look exactly the same and are randomly shuffled before they reach your hands.

Two envelopes, filled with money. Image by the author.

Now, you can open one of them and decide whether

  • to keep the money that is inside or
  • throw away the money that is inside (sorry, that’s the rules 😅), open the other envelope and keep the money inside of this one.

Of course, you try to maximize your reward by picking the best envelope.

That’s it. It’s worth stressing out that you do not know the values of x and y in advance. If you did, the problem of picking the better envelope would be trivial. Also, let us assume that x≠y since otherwise both envelopes are the best and the chance of winning the bigger price is also 100%.

Problem Analysis

So, what to do? We don’t have any information: The letters are shuffled, look the same, and we don’t even know the amount of money that is inside. So there is not really a way to apply some strategy and boost the probability of success beyond 50%, right? Are we dealing with a lost cause?

Of course not.

Otherwise, this article would not exist, duh! There are some easy strategies everyone could come up with:

  1. Always stick with the first envelope.
  2. Always switch the envelope.

Spoiler: applying one of these two strategies leaves you off with a probability of 50% only of getting the bigger reward. That is because both envelopes were shuffled beforehand, nothing you can do about it.

Let us now cover another strategy that is better suited to increase our success probability.

Increasing the Chances

First, let us assume xx).

The case for y first: If we get y first, we have to stick with the envelope. This happens whenever our randomly sampled z is smaller than y. This means that in this case, we win with a probability of P(z