Here’s Part I again:

*My 15-year-old son Adam likes game theory. He invented the following simple game, and asked me about it when I got on the phone with him while I was away at a conference last month (I’ve simplified and formalized the set-up slightly):*

*There are two players, each of whom is given a real number which is chosen randomly from a uniform distribution between 0.0 and 1.0. The players know their own number but not their opponent’s. One player moves first and has the choice of passing or challenging. If he challenges, both players reveal their number, and the player with the higher number receives a payoff of 1, while the other player receives a payoff of 0. If the first player passes, the second player has a choice of challenging or passing. If he challenges, again both players reveal their numbers and the player with the higher number receives a payoff of 1, while the other player receives a payoff of 0. If the second player also passes, both players receive a payoff of 1/2. They play the game one time, and are interested in maximizing their expected payoff.*

*What is the right strategy? For example, if you received the number 0.17, would you pass or challenge if you were the first player? What about if you were the second player? What would you do if the number you received was 0.0017?*

*I’ll tell you more in a later post, but for now why don’t you think about it….*

Here’s the promised followup:

It is clear that if any player has the advantage, it’s the second player, because he gets some information from the first player, and can use it to make his decision. Nevertheless, the first player can adopt the strategy of always challenging, and thereby guarantee that he wins half the time. So apparently he should always challenge. This is the answer that was given by “Optionalstopping” in the comments. The same answer was given to me by the evolutionary game theorist Arne Traulsen (who has recently worked on a ground-breaking theory for the emergence of punishment), after I asked him about the game at a lunch conversation.

There is another way to arrive at the same answer. Assume that each player chooses a strategy parameterized by a single value; if he receives a number above that value, he challenges, while if he receives a number below that value he passes. If you work out the Nash equilibrium (basically, that means that both you and your opponent pick the strategy that gives you the best payoff assuming that the opponent is maximizing their payoff), you’ll find (I won’t bore you with the math) that the value for both players is zero–they should always challenge. My son Adam gave an intuitive version of this argument, without the math. Of course it’s true that probabilistic strategies are also possible. I haven’t proven it, but I strongly doubt that introducing probabilistic strategies will change the result that the Nash equilibrium is to always challenge for both players.

So at first I thought the matter was settled. ** But still, there is something very weird about this result.** Would you really challenge if you were the first player and you received a .001? Would you really?

And what if you were the second player, and the first player passed, and you had a .000001? You know that the first player is not “following the rules” of Nash equilibrium. Are you really going to challenge because Nash tells you to? It’s obviously a crazy result! There must be a hole in the arguments.

So what’s wrong with the above arguments?

First let’s start with the Nash equilbrium arguments. By the way, many authors use the term “rational” for players that use strategies dictated by Nash equilibrium arguments, but I think “rational” and “irrational” are excessively loaded terms, so I prefer to instead say that a player that follows a strategy dictated by Nash equilibrium arguments is a “Nashist.”

If you are the second player, and the first player has passed, you can deduce that the first player is not a Nashist. So in order to make a correct play (maximize your expected winnings) you need to choose some probability distribution for what the first player’s strategy is, and then compute whether you will win more or less depending on whether you challenge. There doesn’t seem to be any obvious way to choose the probability distribution for the first player’s strategy, but you can definitely say that he is not certainly a Nashist! A Nashist is stuck (by definition of being a Nashist) believing that all other players are always Nashists, even in the face of clear evidence that they are not (you see why I don’t like to call Nashists “rational”) and would choose the strategy that followed from that obviously wrong belief; he would always challenge as the second player.

A non-Nashist, on the other hand, can come up with a reasonable probability distribution for the first player’s strategy, and come to the conclusion that he should pass if he is the second player and he has a .000001.

OK, so we can see why the second player might want to pass if he receives a .000001. What about the first player: is it wrong to be a Nashist? Should you pass or challenge if you get a .001, or a .000001, or a .000000001? A Nashist would be compelled, by the force of his “idealogy,” to challenge in each case. But you can make a good argument that that’s wrong. Instead, let’s say I am the first player and I have a .000001. I know that if I pass, the second player will be able to deduce that I’m not a Nashist, and will go through the argument given above for the second player. Now “all” I have to do is form my probability distribution for what his probability distribution of my strategy will be, and compute whether I will have more chance of winning depending whether I challenge or not. Again there’s no obvious way to choose these probability distributions, but it seems pretty clear to me to that reasonable probability distributions will give a result that says don’t challenge if you are the first player and you have a sufficiently low number.

Well what about the other argument, that says that the first player should always challenge, since he is at a disadvantage and if he always challenges, he’ll win half the time? It seems paradoxical to think that there is a strategy that can do better than winning half the time for the first player.

Of course, all Nashists will always win exactly half the time, whether they play first or second. If you are playing against someone who you know is a Nashist, it actually doesn’t matter what you do! But suppose instead that you are playing against an ordinary human. You should play forming the best possible probability distribution of what they will do. Many humans will challenge if they have a number above .5 and pass otherwise, whether or not they play first (a very bad strategy by the way). It is perfectly possible, indeed likely, that playing against a population of ordinary humans, there exists a strategy that wins more than half the time for the first player. I can’t prove that strategy exists by arguing in this way (one would need to run experiments to determine the probability distributions of strategies, and then it would be easy to compute), but I’m actually pretty confident that it does exist, and I’m also pretty confident that the strategy involves passing when you are given a .000001 as the first player.

So I don’t believe either argument for always challenging holds up, which is comforting, because always challenging does seem intuitively wrong. Unfortunately, I can’t tell you exactly what the optimal strategy is either, at least until you tell me what the true probability distribution is for player strategies.

By the way, Arne recommended that I pick up Game Theory Evolving, by Herbert Gintis, for my son. It’s a wonderful book, full of interesting games and solved problems in game theory. Adam and I both love it. Gintis gives other examples showing that Nashists (he calls them “self-regarding agents”) can choose bizarre strategies, including “Rosenthal’s Centipede Game:”

*The players, Mutt and Jeff, start out with $2 each, and they alternate rounds. On the first round, Mutt can defect by stealing $2 from Jeff, and the game is over. Otherwise, Mutt cooperates by not stealing, and Nature gives Mutt $1. Then Jeff can defect and steal $2 from Mutt, and the game is over, or he can cooperate and Nature gives Jeff $1. This continues until one or the other defects, or each player has $100.*

In this game, the Nash equilibrium, obtained by induction by working backwards from the end of the game, when it is clearly “correct” to defect, is that Mutt should immediately defect on his first turn. So that’s what a Nashist would do, but fortunately humans are much more “rational” than Nashists!

Tags: Game Theory, Nash equilibrium, Nashist

August 15, 2007 at 11:36 pm |

I found your website a few days ago and it’s great!

Another simple problem (not game theory) I found intriguing is

this one on Tom Cover’s website.

August 26, 2007 at 9:39 pm |

Interesting game. I think that leaving probabilistic strategies out may be the cause of the counterintuitive result you are getting. I haven’t done the game theory analysis, but I suspect that the problem with the first player passing is that the second player will interpret it as a sign of weakness and immediately challenge. Thus the first player might as well challenge immediately so as to remove any optionality from the second player.

If you allowed probabilistic strategies, then the first player would have the option of occasionally passing when he has a strong hand so that he can also pass when he has a weak hand without giving away that he has a weak hand. Basically, probabilistic strategies allow bluffing.

Another interesting variation might be to add risk aversion. That is instead of the goal being to maximize expected wealth, the goal could be to maximize expected wealth minus standard deviation of wealth.

In any case, it’s an interesting post.

August 28, 2007 at 1:37 am |

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