Outreach module (grades 9-12)

The Envelope Problem

Part 4: The 50-50 baseline

Module Sequence

1. Problem setup 2. Investigation plan 3. Envelope Problem Lab 4. The 50-50 baseline 5. Improper modeling 6. Bayes' theorem 7. Random strategy lab 8. Doing better than 50-50?

Problem Setup

Two envelopes contain X dollars and 2X dollars, where X is a positive integer. The envelopes are shuffled, and one is handed to you at random. You open it and observe A dollars. Before payment, you may decide to keep your envelope or switch to the other one. You are paid whatever is in your final envelope. In this module, we call it a win if your final envelope has the larger amount. Should you switch if your goal is to end with the larger amount?

This page answers one question from the earlier parts: what can be proved before we say anything at all about how X is generated?

The answer is that the two fixed rules, always stay and always switch, are tied. The reason is the shuffle.

1. What the shuffle tells us

Let \(S\) be the event that you were handed the smaller envelope, and let \(L\) be the event that you were handed the larger envelope. The envelopes are shuffled before one is handed to you, so there is no bias toward either one.

What can happen when the envelopes are handed over
Event	Amount in your envelope	Probability
\(S\)	The smaller amount \(X\)	\(\frac12\)
\(L\)	The larger amount \(2X\)	\(\frac12\)

So, before you use the observed value \(A\),

\[ \mathbb{P}(S)=\mathbb{P}(L)=\frac12. \]

This statement does not depend on any model for \(X\). It comes only from the random handoff.

2. Two fixed rules

Now compare the two simplest decision rules.

What the fixed rules do in each case
Hidden event	Always stay	Always switch
\(S\)	lose	win
\(L\)	win	lose

Always stay wins exactly when \(L\) happens, so

\[ \mathbb{P}(\text{win by staying})=\mathbb{P}(L)=\frac12. \]

Always switch wins exactly when \(S\) happens, so

\[ \mathbb{P}(\text{win by switching})=\mathbb{P}(S)=\frac12. \]

This is true for every probability model on \(X\).

3. What remains open

Part 4 settles only the two fixed rules. It does not say that every possible strategy must also be \(50\)-\(50\).

It does not explain why Argument A sounds persuasive.
It does not yet tell us how to use the observed value \(A\) when a model for \(X\) is known.
It does show that if we want to do better than \(50\)-\(50\), we need to be more clever than always switching or staying.

The next page explains the flaw in Argument A. After that, we will return to the question of how a model for \(X\) changes the decision, before thinking about whether doing better than \(50\)-\(50\) is possible.

3: Envelope Problem Lab

Return to the first simulation lab.

5: Improper modeling

Track down the flaw in the always-switch argument.