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"?

Well, Oskar is right that if you are playing another player, who can
adjust his strategy to take advantage of yours, then you need to adopt
a mixed strategy just to have an expectation of breaking even.
Once you solve the linear equations, you don't even care what his
strategy is. Your strategy will be chosen to have an expectation
of breaking even against any strategy he chooses.

However, the given problem is simpler. The problem assumes our
opponent is a computer, and it randomly picks things with known
probabilities, and never adapts those probabilities based on what we
are doing. In this case, we can pick the same thing every time
(an unmixed strategy), and can expect to win (on average). Or at
worst break-even, if the computer is using an optimal mixed strategy.

For example, assume that the computer is known to pick each thing with
probability 1/6. And assume that we are betting a dollar each tme
we play. If we always pick life (or always pick
water, or always pick fire) then we will expect to win 3 times out
of 6, lose 2 times out of six, and tie one time out of
six. Total expected gain = a sixth of a dollar
per turn.

In general, let the computer's known
probabilities be represented by the letters L, E, W, F, A, and C.
(where L = probability computer picks Life, etc.).

L + E + W + F + A + C = 1

Calculate the expected gain (on a $1 bet) for each of your choices as follows:

Life gain = +E +W -F +A -C

Earth gain = -L -W +F -A +C

Water gain = -L +E +F +A -C

Fire gain = +L -E -W +A +C

Air Gain = -L +E -W -F +C

Cold Gain = +L -E +W -F -A

Example of applying these equations:

Assume the computer picks things with

L = .5, W = F = .25, E = A = C = 0

Then

Life Gain = 0.00

Earth Gain = -.50

Water gain = -.25

Fire Gain = +.25

Air Gain = -1.00

Cold Gain = +.50

So, you should always pick Cold, and collect an average of 50 cents per turn from the computer.

The rule change in part two of the problem changes the expected gain as follows:

Water gain = -L +E +F -A -C

Air Gain = -L +E +W -F +C

But you still calculate the 6 expected gains, and then pick an unmixed stategy.

*Edited on ***June 17, 2005, 1:32 am**