Two envelopes conceal different amounts. Opening one seems to leave you entirely at the mercy of chance, yet that partial information can be used in an unexpected way.
After Opening the Envelope
Riddle statement
Two sealed envelopes contain different positive amounts of money.
You do not know the amounts, how they were chosen, or any range containing them.
Choose one envelope at random and open it. After seeing its contents, you must either:
- keep the opened envelope;
- switch to the other envelope, which remains sealed.
You may prepare a randomized strategy in advance, but you may obtain no additional information about the amounts.
Is there a strategy that, for every possible pair of distinct amounts, gives you a probability strictly greater than \(1/2\) of ending with the larger amount?
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Solution
Answer: yes.
Before choosing an envelope, generate a positive random threshold \(R\) whose distribution assigns positive probability to every open interval of positive numbers. For example, take \(U\) uniform on \((0,1)\) and set
Then choose an envelope at random and open it. If the observed amount is \(x>R\), keep it; if \(x<R\), switch.
Now fix any two amounts \(a<b\).
- If \(a<R<b\), the strategy always succeeds: after seeing \(a\), you switch; after seeing \(b\), you keep it.
- If \(R<a<b\), both amounts exceed the threshold, so you always keep the first envelope.
- If \(a<b<R\), both are below the threshold, so you always switch.
In the last two cases, the success probability is \(1/2\), because the initial envelope was chosen at random. In the first case, it is \(1\). Therefore,
Since the distribution of \(R\) gives positive probability to the interval \((a,b)\), this is strictly greater than \(1/2\) for every distinct pair \(a<b\).
A fixed threshold would fail because both amounts could lie on the same side of it. The advantage need not be uniformly bounded away from \(1/2\); it may be tiny, but it is always positive.