Outreach module (grades 9-12)
Part 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?
In Part 7 you tested a random threshold rule by simulation. This page asks whether that strategy really can do better than \(50\)-\(50\), and why.
Try the questions first. Then open the explanations to see how the pieces fit together.
Suppose that in one round the hidden amounts are \(x\) and \(2x\), where \(x\) is a fixed positive integer. Draw a threshold \(T\) independently of the envelope handoff. The rule is:
\[ \text{switch if } A<T, \qquad \text{stay if } A\ge T. \]
If you were handed the smaller envelope, what relationship between \(x\) and \(T\) leads to a win?
In that case \(A=x\). You win exactly when you switch, so you win when \(x<T\).
If you were handed the larger envelope, what relationship between \(x\) and \(T\) leads to a win?
In that case \(A=2x\). You win exactly when you stay, so you win when \(T\le 2x\).
Let \(W\) be the event that your final envelope is the larger one. The notation \(\mathbb{P}(W\mid X=x)\) means “the probability of winning, given that the hidden pair is \(x\) and \(2x\).”
Because the envelopes are shuffled,
\[ \mathbb{P}(S\mid X=x)=\mathbb{P}(L\mid X=x)=\frac12. \]
What formula do you get for \(\mathbb{P}(W\mid X=x)\) by splitting into the cases \(S\) and \(L\)?
Using the law of total probability,
\[ \mathbb{P}(W\mid X=x) = \frac12\mathbb{P}(T>x)+\frac12\mathbb{P}(T\le 2x). \]
How can you simplify that expression?
Hint: for any events \(E\) and \(F\), \[ \mathbb{P}(E)+\mathbb{P}(F)=\mathbb{P}(E\cup F)+\mathbb{P}(E\cap F). \]
The events \(\{T>x\}\) and \(\{T\le 2x\}\) together cover the whole sample space, and they overlap exactly on \(\{x<T\le 2x\}\).
So
\[ \mathbb{P}(T>x)+\mathbb{P}(T\le 2x)=1+\mathbb{P}(x<T\le 2x), \]
which gives
\[ \mathbb{P}(W\mid X=x) = \frac12+\frac12\mathbb{P}(x<T\le 2x). \]
The formula from Question 4 isolates the extra term beyond the baseline \(\frac12\).
For this fixed value of \(x\), when is the random rule exactly \(50\)-\(50\), and when is it better?
It is exactly \(50\)-\(50\) when \(\mathbb{P}(x<T\le 2x)=0\). It is better than \(50\)-\(50\) when
\[ \mathbb{P}(x<T\le 2x)>0. \]
What kind of threshold distribution makes \(\mathbb{P}(x<T\le 2x)\) positive for every positive integer \(x\)?
Any threshold distribution that gives positive probability to every positive integer will do. The geometric distribution from the lab is the cleanest example.
Since \(x\) is a positive integer, the interval \((x,2x]\) always contains at least one positive integer, namely \(x+1\). So under a geometric threshold,
\[ \mathbb{P}(x<T\le 2x)>0 \qquad\text{for every positive integer }x. \]
Now let \(X\) vary again. The overall win probability is found by averaging the fixed-\(x\) formula over the possible values of \(X\).
What do you get if you average \(\mathbb{P}(W\mid X=x)\) over all possible values of \(X\)?
Hint: if \(Y\) is a discrete random variable and \(E\) is an event, then \[ \mathbb{P}(E)=\sum_y \mathbb{P}(E\mid Y=y)\,\mathbb{P}(Y=y). \]
By the law of total probability,
\[ \mathbb{P}(W)=\sum_x \mathbb{P}(X=x)\,\mathbb{P}(W\mid X=x). \]
Substituting the formula from Question 4 gives
\[ \mathbb{P}(W) = \frac12+\frac12\sum_x \mathbb{P}(X=x)\,\mathbb{P}(x<T\le 2x). \]
Because \(X\) and \(T\) are independent, for each fixed \(x\),
\[ \mathbb{P}(x<T\le 2x)=\mathbb{P}(X<T\le 2X\mid X=x). \]
Applying the law of total probability again gives
\[ \mathbb{P}(W) = \frac12+\frac12\,\mathbb{P}(X<T\le 2X). \]
What condition guarantees that the random rule does better than \(50\)-\(50\)?
The formula from Question 7 shows that
\[ \mathbb{P}(W) = \frac12+\frac12\,\mathbb{P}(X<T\le 2X). \]
So the random rule beats \(50\)-\(50\) exactly when
\[ \mathbb{P}(X<T\le 2X)>0. \]
A geometric threshold is one clean way to make this happen, but it is not the only way.