Game Theory Problem with Dice

Question

I need solution to this game theory problem. It seems impossible to me.

Two players (1 and 2) play the following game. Player 1 must write the numbers from 1 to 18 on the sides of 3 dice without repeating the numbers. Then player two chooses one die and throws it. After that player 1 chooses one of the remaining 2 dice and throws it. Is there a way for player 1 to write down the numbers in a way that he has more than 50% chance of winning.

Answer 1

I believe so. What if the dice were as such?:
A{16, 13, 12, 9, 8, 5}
B{15, 14, 11, 10, 7, 6}
C{18, 17, 4, 3, 2, 1}

Dice A and B have equal chance against each other, but die C has 1/3 chance of winning.

If player 2 chooses C, then player 1 has 2/3 chances of winning, so that is a bad choice for player 2.

If player 2 chooses A and player 1 then chooses B no matter what, or the equivalent alternative B then A, then obviously the chance of winning is 50%. However, player 1 has the advantage of being able to choose die C in the case that player 2 rolls 16, 15, or 14. Therefore, he has a greater chance than 50%.

Note: Remember that 50% is the probability of winning before anyone rolls any die, not after player 2 rolls. There is no way to have a 50% chance of winning after player 2 rolls 18 for example.

Answer 2

A = {1, 2, 11, 12, 15, 16}

B = {3, 4, 7, 8, 17, 18}

C = {5, 6, 9, 10, 13, 14}

If player 2 chooses A, player 1 chooses B.

If player 2 chooses B, player 1 chooses C.

If player 2 chooses C, player 1 chooses A.

In any case, player 1 has 20/36 = 5/9 chance of winning.

This is an example of what's called nontransitive dice.