This game is similar to "rock, paper, scissors" in that two players independently pick one of the six things, and if one thing somehow "beats" the other, then that player wins. If both players pick the same thing, they repeat until someone wins.
Life grows on Earth.
Water douses Fire.
Air resists Cold.
Life drinks Water.
Fire consumes Air.
Cold freezes Water.
Earth smothers Fire.
Life breathes Air.
Fire and Earth both warm Cold.
Air and Water both erode Earth.
Fire and Cold both destroy Life.
Water displaces Air.
A program that plays this game has a single set of probabilities for picking each of the six things. Assuming that the program's opponent knows what these probabilities are, what probabilities will give the program the best chances of winning?
What if the rules of the game are changed so that "Water displaces Air" is replaced with "Air ripples Water"?
Compliments to Tristan for an excellent problem, Matt for the part 1
solution, Bob Smith for the part 2 solution. Nice work!
Frederico, I think you are close, but not completely correct.
If this game were played between two persons, then I think for
the first problem, both players should pick (C,F,W,E,A,L) with odds
(1/5, 1/5, 1/5, 1/5, 0, 1/5). For the second problem, both
players should use odds (1/9, 1/3, 1/3, 0, 1/9, 1/9). This leads
to a stable situation, where, the expected outcome is 0 (no one will
win) and where nobody could improve by altering their strategy.
In game 1, if either player used a strategy of (1/9, 1/3, 1/3, 0, 1/9,
1/9), then the other player could get an edge by putting a little more
Water into his/her strategy. Water beats (1/9, 1/3, 1/3, 0, 1/9,
1/9) 4/9 of the time and only loses 2/9 of the time.
Similarly, in game 2, if either player used a strategy of (1/5, 1/5,
1/5, 1/5, 0, 1/5) , then the other player could get an edge by putting
a little more Air into his/her strategy. Air beats (1/5, 1/5,
1/5, 1/5, 0, 1/5) 3/5 of the time and only loses 2/5 of the time.