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

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?

$$\mathbb P\{X_1
Now, $$|\{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
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.