Guess the Largest Number Paradox

6 minutes to read

Many statistical paradoxes boil down to an imprecise problem statement, but this is one paradox that is truly astonishing, where we seem to get information from absolutely nothing.

The Problem

Consider a simple game. There are two pieces of face-down paper on the table, and there is a number written on each. You don’t know anything about the numbers. You flip over one of the pieces of paper at random, and read the number on it. Now, you need to decide if you think this is the larger of the two numbers. What do you do?

Intuitively, there is no better solution that guessing randomly, and you cannot guess which number is bigger more than 50% of the time. You have no information about the other number, so you might as well guess randomly.

The Solution

However, this is not actually the best that you can do. There is a way to guess correctly strictly more than 50% of the time, formalized by Thomas Cover in 1987 in only a single paragraph.1

Pick a “splitting number” TT at random according to a normal distribution. When you reveal the number XX, compare TT and XX. If X>=TX >= T, guess that XX is the larger number. If X<TX < T, guess that XX is the smaller number.

Why does it work?

This solution works because you always answer correctly when TT is between the two numbers, and answer correctly 50% of the time otherwise. It’s easiest to see this visually.

Let’s call the smaller of the two numbers AA, and the larger of the two numbers BB. Alone, that looks like this:

A then B

Now, we create our random guess TT. We first investigate the easy options, where TT is smaller than both numbers or larger than both numbers.

T, then A, then B

In this case, we always guess that the number revealed is the larger of the two, based on our strategy from before. We will be correct if BB is chosen, and wrong if AA is chosen. Because we choose between AA and BB randomly, we will pick BB 50% of the time, and thus will guess correctly 50% of the time.

A, then B, then T

In this case, we always guess that the number revealed is the smaller of the two, based on our strategy from before. We will be correct if AA is chosen, and wrong if BB is chosen. Similarly to before, we will pick AA 50% of the time, and thus will guess correctly 50% of the time.

A, then T, then B

If TT is between AA and BB, our guess depends on which of AA and BB is chosen.

ATB A Transparent

If BB is chosen, we will correctly say that BB is the larger of the two.

ATB B Transparent

Similarly, If AA is chosen, we will correctly say that AA is the smaller of the two. If TT is between AA and BB, we always answer correctly.

We can summarize our results like the below.

explanation

Analysis

How likely is it that we guess correctly? We can summarize it into the formula below by splitting into three cases based on how TT compares to AA and BB.

P(correct)=P(TA)2+P(A<TB)+P(T>B)2=P(TA)+P(T>B)2+P(A<TB)=1P(A<TB)2+P(A<TB)=12+P(A<TB)2\begin{align} P(\text{correct}) &= \frac{P(T \leq A)}{2} + P(A < T \leq B) + \frac{P(T > B)}{2} \\ &= \frac{P(T \leq A) + P(T > B)}{2} + P(A < T \leq B) \\ &= \frac{1 - P(A < T \leq B)}{2} + P(A < T \leq B) \\ &= \frac{1}{2} + \frac{P(A < T \leq B)}{2} \end{align}

Now, all we need to do is figure out how likely we are to have TT between AA and BB. Because we picked out TT from a normal distribution, we can directly solve for this probability. With a little inspection, we can see that this probability is strictly greater than 0 for all positive numbers, and as such the odds that TT is between AA and BB is also strictly positive.

P(A<TB)=AB12πex22dxP(A < T \leq B) = \int_A^B \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx

Thus, our final equation for the probability we guess correctly is

P(correct)=12+AB18πex22dxP(\text{correct}) = \frac{1}{2} + \int_A^B \frac{1}{\sqrt{8\pi}} e^{-\frac{x^2}{2}} dx

We can evaluate this for some different values of AA and BB to investigate how this behaves.

A: 0  B: 1
   Probability: 0.6707
A: 0  B: 10
   Probability: 0.75
A: 5  B: 10
   Probability: 0.50000014
A: 7  B: 10
   Probability: 0.500000000001

It’s clear that the probability goes down when AA and BB are closer to each other, and also goes down when AA and BB are farther from 1. In fact, by the time AA and BB are both greater than 10, the difference from 0.5 is below the precision of most software. However, it always positive.

This is the real caveat to the strong statement of “strictly greater than .5”. Although the probability of guessing correctly is better than one half, it is not uniformly greater than one half. Unless we assume some prior distribution of AA and BB, this strategy is basically exactly as effective as randomly guessing.

More precisely, as long as you know nothing about AA and BB, you cannot ever get measurably far from a probability of one half. For any ϵ>0\epsilon > 0, you can never achieve a probability of guessing correctly that is at least 12+ϵ\frac{1}{2} + \epsilon.

Even further, if we do know something about the distribution of AA and BB, this is still not the best strategy. If we know that AA and BB come independently from some distribution FF, we should still use a “splitting number” strategy, but our splitting point TT should be exactly the median of the distribution FF.

This lines up with intuition about this type of problem. If you know that AA and BB are drawn uniformly from the range [0,10][0, 10], we guess that numbers less than 5 are the smaller of the two, and numbers greater than 5 are the greater of the two.

Conclusion

In summary, although this paradox is incredibly surprising, it does not ultimately subvert our understanding of the wordl. It seems like we get information, out of nothing, but it turns out that the advantage gained is so small as to be negligible.

For further reading on this topic, you can consider looking at this Quanta article on the same topic.