Friday, December 9, 2016

Today's Riddler Puzzle and the Prisoner's Dilemma

I've been following FiveThirtyEight's Riddler Puzzles for a little while now. Each puzzle deals with probability. Today's puzzle has a little game theory mixed in:
Consider the following war game: Two countries are eyeing each other’s gold. At the beginning of the game, the “strength” of each country’s army is drawn from a continuous uniform distribution and lies somewhere between 0 (very weak) and 1 (very strong). Each country knows its own strength but not that of its opponent. The countries observe their own strength and then simultaneously announce “peace” or “war.”

If both announce “peace,” then they each stay quietly in their own territory, with their own gold, which is worth $1 trillion (so each “wins” $1 trillion).

If at least one announces “war,” then they go to war, and the country with the stronger army wins the other’s gold. (That is, the stronger country wins $2 trillion, and the other wins $0.)

What is the optimal strategy of each country (declaring “peace” or “war”) given its strength?
So we have a continuous uniform distribution to determine country's strength. This means that any value of strength between 0 and 1 is equally likely. The problem discusses only one distribution, and doesn't give any information on what would happen if two country's tied in terms of strength, suggesting that the situation is set up so that each country will have a different strength rating - that is, once one value is selected from the distribution for one country, it cannot be selected again for the other country (random sampling without replacement). The only information a country has, then, is its own strength rating, which it can then use to determine the probability that the other country is stronger or weaker.

If we use the self-interested approach: If a country has a high strength rating (more likely that the other country is weaker instead of stronger), it should select "war." If a country has a lower strength rating (more likely that the other country is stronger instead of weaker), it should select "peace." Of course, it is still possible to win with a low strength rating or lose with a high strength rating; such is the nature of probability - unlikely is not the same thing as impossible.

This particular situation involves a zero-sum game. At least, that's how I conceptualize it, even though the problem talks about the possibility that each country could "win" $1 trillion. Each country already has $1 trillion of gold. If both countries choose peace, they get to keep their money (0 gain for either). If a country chooses war, the winner gets the other's gold ($1 trillion gain), leaving the other country with nothing ($1 trillion loss). So a player in a zero-sum game can only benefit due to the loss of the other. So the situation above is not a true prisoner's dilemma, which is a non-zero-sum game:

Frankly, in a zero-sum game, the best approach would be select "peace" every time; everyone gets to keep their money and no one gets hurt. But I'm a pacifist, so what do I know? :) And in fact, the work of Amos Tversky and Daniel Kahneman on loss aversion suggests that people will work harder to avoid a loss than obtain a gain of the same magnitude, suggesting that each country would be more hesitant to lose $1 trillion than to gain $1 trillion. In that case, they would probably only select "war" if they perceive it to be a sure thing - that is, they have a very high strength rating and the probability that the other country has a higher rating is very low. But this problem is purely probabilistic, and I'm sure they don't want social psychology mixing in.

The way to solve it, then, is to figure out the combinations that maximize a country's winnings - which in some cases would be to do nothing and accept a gain of 0. (I'm not actually going to solve the problem here; mostly just using it as an excuse to discuss some of my favorite topics. Yep, I'm using it as a teachable moment - I'm evil.)

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