Outreach module (grades 9-12)
Part 5: Improper modeling
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 returns to Argument A from Part 1. Work through the questions before opening the explanations.
The goal is to inspect the reasoning one step at a time and decide which assumption needs to be checked.
After you observe \(A=a\), there are two possible stories:
The notation \(\mathbb{P}(S\mid A=a)\) means “the probability of \(S\), given that you observed \(A=a\).” Similarly, \(\mathbb{P}(L\mid A=a)\) means “the probability of \(L\), given that you observed \(A=a\).”
So the expected amount in the other envelope is
\[ \mathbb{E}[\text{other}\mid A=a] = \mathbb{P}(S\mid A=a)\cdot 2a + \mathbb{P}(L\mid A=a)\cdot \frac{a}{2}. \]
What additional assumption turns this into \(\frac{5a}{4}\)?
You would need to assume
\[ \mathbb{P}(S\mid A=a)=\mathbb{P}(L\mid A=a)=\frac12. \]
With that assumption, the formula becomes
\[ \mathbb{E}[\text{other}\mid A=a] = \frac12\cdot 2a + \frac12\cdot \frac{a}{2} = \frac{5a}{4}. \]
Start by testing the equal-split idea in cases where the possibilities are easier to track. In this page, \(X\) is assumed to be a positive integer, so the larger amount \(2X\) is always even.
Suppose you observe an odd number, so \(A=a\) is odd. Can the event \(L\) happen?
No. If \(L\) happened, then your envelope would contain \(2X\), which is always even. So an odd observation cannot come from \(L\).
What does that tell you about \(\mathbb{P}(L\mid A=a)\) and \(\mathbb{P}(S\mid A=a)\) when \(a\) is odd?
It forces
\[ \mathbb{P}(L\mid A=a)=0, \qquad \mathbb{P}(S\mid A=a)=1 \qquad \text{when } a \text{ is odd}. \]
So the equal split cannot hold for every observation.
Now turn to an even observation, say \(A=2n\). Then there are two ways to see \(2n\):
In terms of \(\mathbb{P}(X=n)\) and \(\mathbb{P}(X=2n)\), what are \(\mathbb{P}(A=2n \text{ and } L)\) and \(\mathbb{P}(A=2n \text{ and } S)\)?
These are the probabilities of the two full scenarios that can produce the observation \(A=2n\). Because the handoff itself is random, each one gets a factor of \(\frac12\):
\[ \mathbb{P}(A=2n \text{ and } L)=\frac12\mathbb{P}(X=n), \qquad \mathbb{P}(A=2n \text{ and } S)=\frac12\mathbb{P}(X=2n). \]
Suppose you try to use this equal-split idea for every even observation that can occur.
What relationship does that force the probabilities
\[ \mathbb{P}(X=1),\ \mathbb{P}(X=2),\ \mathbb{P}(X=4),\ \mathbb{P}(X=8),\dots \]
to have?
\[ \mathbb{P}(X=3),\ \mathbb{P}(X=6),\ \mathbb{P}(X=12),\ \mathbb{P}(X=24),\dots \]
What relationship would those probabilities have to satisfy?
Could one honest probability distribution on the positive integers behave that way?
Question 5 shows that whenever the even observation \(A=2n\) can occur, the equal-split idea would force
\[ \mathbb{P}(X=n)=\mathbb{P}(X=2n) \]
So along the first list of values, you would get
\[ \mathbb{P}(X=1)=\mathbb{P}(X=2)=\mathbb{P}(X=4)=\mathbb{P}(X=8)=\cdots . \]
Along the second list, you would get
\[ \mathbb{P}(X=3)=\mathbb{P}(X=6)=\mathbb{P}(X=12)=\mathbb{P}(X=24)=\cdots . \]
The same thing would happen starting from any odd number.
If some odd number had positive probability, then all of its doubles would have that same positive probability too. That would create infinitely many equal positive terms, so the total probability would be larger than \(1\).
So every odd number would have to get probability \(0\). Now suppose some positive integer had positive probability. If you keep dividing that number by \(2\) until you can no longer do so, you eventually land on an odd number, and the equalities above say the probability stays the same at each step. That would produce an odd number with positive probability, which is impossible.
So this would force \[ \mathbb{P}(X=n)=0 \qquad\text{for every positive integer } n, \] which is impossible for a probability distribution.
Therefore the equal-split assumption cannot come from one honest probability model.
The next page starts from an actual model for how \(X\) is chosen and shows how to update the probabilities correctly after observing \(A\).