Outreach module (grades 9-12)
Part 6: Bayes' theorem
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?
The previous page explained why the first argument fails: it uses a probability model that does not exist. This page does the opposite.
We will start with an actual model for how \(X\) is chosen, and then use Bayes' theorem to update our probabilities after seeing \(A\).
Suppose \(X\) is equally likely to be \(1\), \(2\), or \(4\). Then the envelope pair is \(\{1,2\}\), \(\{2,4\}\), or \(\{4,8\}\), each with probability \(\frac13\).
| Observed value \(A\) | What it means | Best action |
|---|---|---|
| \(1\) | You must have the smaller amount. | Switch |
| \(2\) | You could be holding the smaller amount or the larger amount, and the two stories are equally likely. | Tie |
| \(4\) | You could be holding the smaller amount or the larger amount, and the two stories are equally likely. | Tie |
| \(8\) | You must have the larger amount. | Stay |
Once a model for \(X\) is fixed, the observed value can matter.
Now compare two deterministic models.
\(X=1\) every time, so the envelopes are \(\{1,2\}\).
If you observe \(A=2\), then you must be holding the larger amount.
Best action: stay.
\(X=2\) every time, so the envelopes are \(\{2,4\}\).
If you observe \(A=2\), then you must be holding the smaller amount.
Best action: switch.
The value \(A\) by itself does not settle the question. What matters is how likely the different hidden stories are under the model.
Let \(S\) be the event that you were handed the smaller envelope, and let \(L\) be the event that you were handed the larger one. Before you open anything,
\[ \mathbb{P}(S)=\mathbb{P}(L)=\frac12. \]
After you observe \(A=a\), those probabilities may change. Bayes' theorem is the rule that updates them:
\[ \mathbb{P}(S\mid A=a)= \frac{\mathbb{P}(A=a\mid S)\,\mathbb{P}(S)} {\mathbb{P}(A=a\mid S)\,\mathbb{P}(S)+\mathbb{P}(A=a\mid L)\,\mathbb{P}(L)}. \]
The denominator comes from the law of total probability: it adds the contributions from the two hidden stories \(S\) and \(L\).
After you observe \(A=a\), there are again two stories to compare.
You were handed the smaller envelope.
Then \(X=a\), and the other envelope contains \(2a\).
You were handed the larger envelope.
Then \(X=\frac a2\), and the other envelope contains \(\frac a2\).
These stories tell us that
\[ \mathbb{P}(A=a\mid S)=\mathbb{P}(X=a), \qquad \mathbb{P}(A=a\mid L)=\mathbb{P}\!\left(X=\frac a2\right). \]
Using \(\mathbb{P}(S)=\mathbb{P}(L)=\frac12\) and the law of total probability,
\[ \mathbb{P}(A=a) = \mathbb{P}(A=a\mid S)\mathbb{P}(S) + \mathbb{P}(A=a\mid L)\mathbb{P}(L) \]
becomes
\[ \mathbb{P}(A=a) = \frac12\mathbb{P}(X=a) + \frac12\mathbb{P}\!\left(X=\frac a2\right). \]
Substituting into Bayes' theorem gives
\[ \mathbb{P}(S\mid A=a) = \frac{\mathbb{P}(X=a)}{\mathbb{P}(X=a)+\mathbb{P}\!\left(X=\frac a2\right)}, \]
\[ \mathbb{P}(L\mid A=a) = \frac{\mathbb{P}\!\left(X=\frac a2\right)}{\mathbb{P}(X=a)+\mathbb{P}\!\left(X=\frac a2\right)}. \]
If \(\frac a2\) is not a positive integer, or if it is excluded by the model, then \(\mathbb{P}(X=\frac a2)=0\).
Switching wins exactly when event \(S\) occurs, so the decision comes from the size of \(\mathbb{P}(S\mid A=a)\).
| Condition | Best action |
|---|---|
| \(\mathbb{P}(S\mid A=a)>\frac12\) | Switch |
| \(\mathbb{P}(S\mid A=a)<\frac12\) | Stay |
| \(\mathbb{P}(S\mid A=a)=\frac12\) | Either choice is fine |
Because of the formula above, this can be checked by comparing two probabilities in the model:
This now explains several earlier examples in one framework. Odd observations force switching because \(\mathbb{P}(X=\frac a2)=0\). If the model rules out \(X=a\), then staying is forced. The “more than half the money in the world” example is the same kind of idea: one story is impossible, so it gets probability \(0\).
This page answers what to do when a model for \(X\) is known. The remaining question is what is still possible when that model is unknown.
We have answered most of the problems we started with. The last basic question is whether it is possible to do better than \(50\)-\(50\), no matter how \(X\) is generated.
What do you think? Before moving on, pause and make a prediction.