Probability that two equally likely values are ordered

Key concept: This problem combines equal-likelihood outcomes with symmetry. Once the probability of equality is known, the remaining probability splits evenly between the cases Y1 > Y2 and Y1 < Y2.

Step by step

Step 1: Understand the sample space

The random variable $Y$ has $n$ equally likely values. Two values $Y_1$ and $Y_2$ are chosen at random, and we are told

$P(Y_1=Y_2)=0.02.$

We want $P(Y_1>Y_2)$.

Because the two values are chosen in a symmetric way, the events $Y_1>Y_2$ and $Y_1<Y_2$ have the same probability.

Step 2: Use symmetry

All possible outcomes fall into three categories:

1. $Y_1=Y_2$
2. $Y_1>Y_2$
3. $Y_1<Y_2$

These probabilities add to 1:

$P(Y_1=Y_2)+P(Y_1>Y_2)+P(Y_1<Y_2)=1.$

By symmetry,

$P(Y_1>Y_2)=P(Y_1<Y_2).$

So the remaining probability after equality is split equally:

$P(Y_1>Y_2)=\frac{1-P(Y_1=Y_2)}{2}.$

Substitute the given value:

$P(Y_1>Y_2)=\frac{1-0.02}{2}=\frac{0.98}{2}=0.49.$

Step 3: Why the symmetry works

This works because there is no built-in preference for one order over the other. If a pair of values is not equal, then exactly one of the two orderings must be true: either the first is larger or the second is larger. Since the setup treats the two draws evenly, those two cases must have the same probability.

So the required probability is

$\boxed{0.49}.$

This is a classic symmetry shortcut that avoids listing every possible pair.

Pitfall alert

A common mistake is to subtract 0.02 from 1 and stop there, giving 0.98. That ignores the fact that the non-equal outcomes are split between two opposite orderings. Another error is to assume $P(Y_1>Y_2)$ depends on the actual number of values $n$. In this question, the exact value of $n$ is not needed because the equality probability already tells you everything required. The symmetry argument is the key.

Try different conditions

If the question changed to $P(Y_1\ge Y_2)$, then you would include both the greater-than case and the equality case. Using the same numbers, $P(Y_1\ge Y_2)=0.49+0.02=0.51$. If the given equality probability were 0.10 instead of 0.02, then $P(Y_1>Y_2)=\frac{1-0.10}{2}=0.45$. The method stays the same: remove the equal case, then split the remainder evenly.

Further reading

symmetry in probability, equally likely outcomes, ordered pairs