Game Theory XII: Randomness

We have seen that sometimes, it is to a player's advantage to remove one of their own options. Is it ever to a player's advantage to take their choice of action out of their own hands?

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Before we start on this discussion, a vaguely relevant video. Spoilers for a certain VN/anime/manga series, even though it manages to not name many names.

But back to the question. Let's return to the game we started with:

	A	B
A	(-1,1)	(1,-1)
B	(1,-1)	(-1,1)

As you can see, this is a zero-sum game in which your objective is to make a different move from your opponent. But suppose your opponent knows you very well. Suppose your opponent knows you so well that they know what you will do. For instance, we'll say that you're playing this against a computer that has a complete copy of your brainstate loaded into it. No matter what you think, the computer can predict your thoughts. And no matter what you decide, the computer can know what you will do.

So, since you lose if you pick A, and you lose if you pick B, what do you do?

You make your move based on something that cannot be predicted, something that even you cannot predict. You make your move random. Flip a coin, spin a roulette, or roll a die. You play what is called a mixed strategy.

Now, let's put together a general form of this:

	A	B
A	(-A,A)	(B,-B)
B	(C,-C)	(-D,D)

where A,B,C,D>0

So, if we play a mixed strategy, we need to figure out what odds we should give to each move (assuming that your opponent will assign the same odds):

Odds of Playing A	Outcome
0%	-D
25%	-(1/4*1/4)A+(1/4*3/4)B+(3/4*1/4)C-(3/4*3/4)D =-A/16+3B/16+3B/16-9D/16
50%	-(1/2*1/2)A+(1/2*1/2)B+(1/2*1/2)C-(1/2*1/2)D =-A/4+B/4+C/4-D/4
75%	-(3/4*3/4)A+(3/4*1/4)B+(1/4*3/4)C-(1/4*1/4)D =9A/16+3B/16+3C/16-D/16
100%	-A

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