I am not sure if there is a name associated with this game but I do not know of one.
There was this 2 player game I had once played, and I am wondering if there is a winning strategy. The game is as follows:
Two players both get 6 rocks, numbered 1-6.
Each rock number corresponds to its weight (6 is heavier than 5, is heavier than 4, etc)
Each number on the rock also corresponds to its points. (rock 6 = 6 points).
The game is played in rounds, 6 in total.
Each round,A random rock from the referee's (or computers) pile [similar to player rocks 1-6] is placed into the middle of the lever by the ref.
Each player then chooses one of their rocks to place onto the lever. (The players do not show their choice until they are revealed on the lever.)
The heavier rock chosen between the player wins the round, and the winning player receives the points for all the rocks on the lever. (Their number + random Ref number + opponents number).
The rocks used in a round are no longer able to be used in the next rounds. This includes the referee's rocks that go into the middle. (so if the ref uses rock 6, then the next round the rock in the middle must be random from 1-5)
If there is a tie, the rocks remain on the lever for the next round.
The player with the most points at the end of the game (6 rounds) wins!
My question is:
Is there a possible winning strategy to this game? [Ties will be counted as wins]
Alternatively, is there a strategy which can give you the highest chance of winning?
Assume it can be used in either situation:
- Both players are using the optimal strategy.
- Your opponent is not following the optimal strategy.
If you were both given more or less rocks, would this change the situation? (i.e 1-10, or 1-4)?