Two dice: $\Pr(X_1 \lt X_2 \mid X_1 \leq X_2)$

Question

This question comes from a completed, marked, and returned exam. It will not likely be reused.

Problem

The problem statement gives a fair die labeled 1-6. Let $X_1,X_2$ denote the two observed labels on two rolls. True or False: $\Pr(X_1 \lt X_2 \mid X_1 \leq X_2)=\frac{5}{7}$?

Work

I erroneously concluded during the exam that the conditional probability was certain, having flipped the two expressions in my work. I marked false.

The given answer is true, yet I am unable to convince myself of it, even with simple counting arguments. I was able to conclude from definitions that the probability given is equivalent to $$\frac{\Pr(X_1 \lt X_2 \cap X_1 \le X_2)}{\Pr(X_1 \le X_2)}$$

I have been unable to precede: my other argument goes something like

• if $X_2$ is 2, the conditional probability is $\frac{1}{2}$;
• if $X_2$ is 3, the conditional probability is $\frac{2}{3}$;

&c. But the sum will be greater than 1. Of course, the probability for $X_2$ is $\frac{1}{6}$, so we try $$\frac{1}{6}\sum_{1 \le i \le 5}{\frac{i}{i+1}} \neq \frac{5}{7}$$

Question

At any rate, I am unable to reach $\frac{5}{7}$ via these arguments. What reasoning leads to the conclusion?

Answer 1

$$\mathbb P\{X_1<X_2\mid X_1\leq X_2\}=1-\mathbb P\{X_1=X_2\mid X_1\leq X_2\}.$$

Now, $$|\{X_1\leq X_2\}|=1+2+3+4+5+6=21$$ and $$|\{X_1=X_2\mid X_1\leq X_2\}|=6.$$ Therefore, $$\mathbb P\{X_1=X_2\mid X_1\leq X_2\}=\frac{6}{21},$$ and thus $$\mathbb P\{X_1<X_2\mid X_1\leq X_2\}=1-\frac{6}{21}=\frac{15}{21}=\frac{5}{7}.$$

Answer 2

There are $21$ die rolls where $X_1 \leq X_2$, each equally likely. Of those, in $6$ of them, $X_1 = X_2$, so in the remaining $15, X_1 \lt X_2.$ Therefore, your answer is $15/21 = 5/7.

The problem with your reasoning is that given that $X_1 \leq X_2$, the possible values of $X_2$ are not equally likely.