Austin and Joshua play a game. Austin chooses a random number equal to 1, 2, 3, 4, or 5 . Joshua then chooses randomly from the remaining four numbers. Joshua's first round score is equal to the product of his number and Austin's number. Austin then chooses randomly from the remaining three numbers, and his first round score is the product of his second number and Joshua's first number. The process is repeated until each of the five numbers has been chosen. The sum of each player's two scores is their final score and the player with the highest final score wins. If Austin chooses 2 to start and Joshua then chooses 3 (making Joshua's first round score 6), what is the probability that Austin will win?
Since Austin chooses 2 to start and Joshua chooses 3 next, we can list all of the possible orders of choices of numbers:

| Order of  Numbers | Joshua's  1st Rd Score | Austin's  1st Rd Score | Joshua's  2nd Rd Score | Austin's  2nd Rd Score | Joshua's  Total | Austin's  Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2,3,1,4,5$ | 6 | 3 | 4 | 20 | 10 | 23 |
| $2,3,1,5,4$ | 6 | 3 | 5 | 20 | 11 | 23 |
| $2,3,4,1,5$ | 6 | 12 | 4 | 5 | 10 | 17 |
| $2,3,4,5,1$ | 6 | 12 | 20 | 5 | 26 | 17 |
| $2,3,5,1,4$ | 6 | 15 | 5 | 4 | 11 | 19 |
| $2,3,5,4,1$ | 6 | 15 | 20 | 4 | 26 | 19 |

Since Austin's total is larger than Joshua's total in 4 of the 6 games, then the probability that Austin wins is $\frac{4}{6}$ or $\frac{2}{3}$.

ANSWER: $\frac{2}{3}$