Problem

Alice writes two distinct real numbers between 0 and 1 on separate sheets. Bob:

  1. Randomly selects and looks at one number.
  2. Must guess whether the number he sees is the larger or the smaller of the two.

Can Bob devise a strategy that gives him more than a 50% chance of being correct?

Naïve Strategy

If Bob has no information and randomly guesses “larger” or “smaller,” he will be correct with probability:

\[ \boxed{\frac{1}{2}} \]

But can he do better?

Clever Strategy: Use a Random Threshold

Bob chooses a random number \( r \) from a known continuous distribution on \( (0,1) \), say uniformly at random.

Then his strategy is:

If the number he sees is less than \( r \), guess it is the smaller number.
If it is greater than \( r \), guess it is the larger number.

Why This Works

Let the two hidden numbers be \( a < b \). Bob randomly picks one of them to inspect. There are two cases:

  • If he sees \( a \):
    • He guesses “smaller” if \( a < r \) — which is correct.
  • If he sees \( b \):
    • He guesses “larger” if \( b > r \) — which is also correct.

Each has probability \( \frac{1}{2} \), so:

\[ P(\text{correct}) = \frac{1}{2}P(r > b) + \frac{1}{2}P(r < a) \]

Because \( r \sim \text{Uniform}(0,1) \), these probabilities are \( 1 - b \) and \( a \), so:

\[ P(\text{correct}) = \frac{1}{2}(1 - b + a) \]

Since \( a < b \), we have:

\[ P(\text{correct}) > \frac{1}{2} \]

Thus, Bob beats 50% accuracy, regardless of what numbers Alice chooses!

Conclusion

Yes — by comparing the revealed number to a random threshold, Bob can win more than 50% of the time.

This works because the strategy introduces external randomness to exploit the structure of the problem.

Reference