8. (20,8-11) Aya and Vasya have 3 coins. On different sides of one coin are scissors and paper, on the sides of another coin are rock and scissors, and on the sides of the third coin are paper and rock. Scissors beat paper, paper beats rock, and rock beats scissors. First, Aya chooses a coin for herself, then Vasya, then they flip their coins and see who won (if the same side comes up, it's a tie). They do this many times. Is there a way for Vasya to choose a coin so that his probability of winning is higher than Aya's?
# Solution.

Let's number the coins: the coin "Scissors-Paper" has number 1, the coin "Rock-Scissors" - 2, and the coin "Paper-Rock" - 3. The probabilities of winning for each coin when any pair is chosen are as follows:

|  | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: |
| 1,2 | $1 / 4$ | $1 / 2$ | - |
| 1,3 | $1 / 2$ | - | $1 / 4$ |
| 2,3 | - | $1 / 4$ | $1 / 2$ |

Notice that, no matter which coin Aseya chooses, Vasya can choose from the remaining coins such that his coin has a higher probability of winning.

If Aseya chooses the first coin, then Vasya should choose the second. If Aseya chose the second coin, then Vasya should take the third, and if Aseya took the third, then Vasya should take the first. Therefore, by acting in this described manner, Vasya has a higher chance of winning over a large number of played games.

Answer: yes.