# Probability that 3 number chosen from set is greater than other 3 numbers

Bernardo randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

This is what I got so far:

If Bernado picks 9, his number will always be greater than Silvia's. The probability that he picks 9 is $\frac{\left(\begin{array}{c} 8 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 9 \\ 3 \end{array}\right)}$ = $\frac{1}{3}$

Now what?

You've already calculated that Bernado has a $\frac13$ chance of picking 9, and hence winning. Conditioning on the event that he does not pick 9 (which thus happens with probability $\frac23$), they two players are each drawing from the same set $\{1,\dots,8\}$. Bernado wins half the time that they don't tie.
How often do they tie? They tie only if they pick the same three numbers, although the order can be different. This happens $6$ times as often as the situation where they pick the same numbers in the same order. Assume Bernado goes first; then Silvia the same three numbers in the same order with probability $\frac18 \times \frac17 \times \frac16$.
Combining everything, we have: Assuming Bernado does not choose a 9, ties occur $6 \times \frac18 \times \frac17 \times \frac16 = \frac1{56}$ of the time, and Bernado wins half of the remaining cases, or $\frac12 \times \frac{55}{56} = \frac{55}{112}$. Of course, this was conditioning on a $\frac23$ chance. So the total chance what Bernado wins is:
$$\frac13 + \frac23 \times \frac{55}{112} = \frac{111}{168}$$