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 Elemental game (Posted on 2005-06-16)
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"?

 See The Solution Submitted by Tristan Rating: 3.6667 (6 votes)

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 Spoiler: Computers can be beat! | Comment 3 of 17 |
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
 Posted by Steve Herman on 2005-06-17 01:03:13

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