Problem

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

Randomly selects and looks at one number.
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?

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.

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!

\[ \boxed{ \text{Yes — by comparing the revealed number to a random threshold, Bob can win more than 50% of the time.} } \]

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