The Tao of Gaming

How to solve a math problem, the engineering way:

1. Dimly recall that you studied it somewhere.


2. Dig up the likeliest book.


3. Confirm that it describes the solution.


4. Balk at the math involved.


5. Download a tool that solves the problem in a second.

To recap: while playing Lord of the Rings: the Confrontation, we have a battle we are winning 3 to 2 (before cards) and both players have the cards 1-5.

In running the calculation, I had to assign a payout matrix.

	1	2	3	4	5
1	10,-10	-2,2	-9,9	-7,7	-5,5
2	8,-8	10,-10	-2,2	-9,9	-7,7
3	6,-6	8,-8	10,-10	-2,2	-9,9
4	4,-4	6,-6	8,-8	10,-10	-2,2
5	2,-2	4,-4	6,-6	8,-8	10,-10

How to read that: Left side are our cards choices, top side is our opponents choice. If we both play the same number, then the result is (10,-10). That's the best possible result for us, since we win the battle and don't lose any ground. As a frame of reference, I valued winning a piece at 10 points. Then I called each pip two points. If we tie (my opponent plays a card one higher) then both pieces die. I'm going to call that a marginal victory for my opponent (reasonable if playing Sauron), but mitigated because he used a better card. Similarly, if my opponent wins by playing a 3 to my 1, I did't call that the full ten points, but the loss is more than six points (which I should do if using 10 points for winning and 2 point per pip difference).

Optimal play according to Gambit:

• Play the 5 and guarantee the win 62.4% of the time.


• Play the 1 (and try to win cheaply) 14.3% of the time.


• Play the 3 (and prevent the countershot) 11.0% of the time.


• Play the 4 (risking the draw, crushing a middle card) 7.2% of the time.


• Play the 2 (why not?) 5.1% of the time.

I'm glad that I put the first three choices in the correct order, but I expected a more even distribution. Discounting the 4 & 2 (as I did), I took the five 54% of the time (rounded to half). But even with two more choices, theory says take the win nearly 2/3rds of the time. This is much sharper than a power distribution.

You can quibble with the assumptions. Is one card pip really worth 2 points (given that a piece is worth 10)? I tried it with 1 point and playing '5' shot up to 80% (with my opponent conceding with a '1' 80% of the time). Intuitively that makes sense: as pieces become more valuable you take the win more often. Conversely, if I increase the value of the cards in relation to the pieces, playing the '5' becomes a weaker choice.

One ignored effect is that by playing 5:1, I may lose the ability to force a win in the next battle. Let's model that by changing the (5,1) cell to a payout of (1,-1), while leaving the rest of the model untouched. The result leaves the 5 at 60%. Amazingly, the second choice shoots from 14.3% to 18.6%, absorbing a few points from the worst result (the '2', which falls to 2%). But the rank ordering doesn't change.

Don't read too much into the 'optimal' result. [Optimal doesn't mean best, which is to know the card my opponent will play and react accordingly. Optimal means you can't improve on this against a perfect opponent. Opponents are rarely perfect.] In addition, this payout matrix is artificial. The value of a piece changes according to the game situation.

The dizzying insight: don't overthink simple situations. Even rating cards highly, pick the obvious play more than half the time, closer to 2/3rds. But remember the important caveat: this assumes a sure win. Real games rarely cooperate...