About modified rock-paper-scissors game

Question

(The game that I am about to ask is not my original creation; it is from a Korean game program 'RunningMan'. Also, I am also Korean, and please understand my poor English.)
The game is about a modified rock-paper-scissors game.

1. 8 people are playing this game, and each of them have 3 cards written rock, paper or scissor.
   
2. At each round, the referee(not one of the 8 people) shows a card written either rock, or paper,or scissor.
3. 8 people then play one of rock, paper, scissor after seeing the referee's card.
   
4. The least number of used cards(excluding the referee's card) get the points.
5. If the least number of cards wins the referee's card, the person(or the people with the least number of cards) get 5 points. If they are the same as the referee's card, they get 3 points. If they lose by the referee's card, they get 1 point.
6. Game is continued until there is a person who reaches 11 points.

The number of people playing the game, the points given to the winning people, the ending point could be changed.
(For example, when the referee shows the rock card, and 4 people show rock, 3 people show paper, 1 person shows scissor, the least used card is scissor, and since rock wins scissor, the person with scissor gets 1 point.)

My question is, Is there a winning strategy to this game, or is there at least a partial winning strategy?
I am not familiar with the game theory, and to me, the game seems pretty much complicated to have a winning strategy. Any help would be very nice. Thanks a lot in advance.

Answer 1

In a Nash equilibrium, suppose the other seven all play the same mixed strategy; Win $x\%$ of the time, Draw $y\%$ of the time and Lose $z\%$ of the time. Note $x+y+z=100\%$.
Suppose, when player 8 plays Win, he is in the smallest group $p\%$ of the time; when he plays Draw, he is in the smallest group $q\%$ of the time; and when he plays Lose he is in the smallest group $r\%$ of the time. Note $p+q+r\gt100\%$ because there are sometimes two smallest groups.
You can calculate $p,q,r$ as functions of $x,y,z$. It's going to be messy.
For example, the probability the other seven play one Win and six Draws is $7xy^6$. That contributes to both $p$ and $r$ because Player 8 scores in that case if he plays either Win or Lose.
Player 8's expected winnings are $5p$, $3q$ or $r$ depending on his choice of card. It is a Nash equilibrium if these three values are equal. Finding the right $x,y,z$ is a tricky numerical problem.