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Comment by tromp | Hacker News Reader

Yes, the only case in which there's no winner is if all 3 pick the same number.

Which is why the author skips the case of two players. Because then picking 1 always is a dominating strategy in which neither player wins.

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I wonder what would happen to the two player game if the rewards were set up so that win=1, lose=0, tie=-1? It would be interesting to see how the strategy changed as the penalty for tying is increased.

picking 1 would still dominate over any other strategy, in a head-to-head.

However, if you are trying to maximize your H2H score against a pool of contestants, a cooperation strategy, where you both alternated taking the lower number would be close to optimal.

For a short intuitive proof of that 1st sentence, imagine both players pick 1 every time. You both lose 1 point. If you're tied for score, you're still tied after the round.

Now imagine you change your number to stop ties. Well now you've picked a number bigger than 1, and you lose every time. If you were tied, now you're losing.

This is only true if your goal is to have a score higher than your opponent. It's not true if your goal is to have the highest possible score.

If ties are with 0, you always choose 1 in either situation. If ties are worth -1, choosing 1 always achieves a greater than or equal to score but the number will be much lower than if you some percentage of the time choose another number.

> This is only true if your goal is to have a score higher than your opponent

That is what “dominant” in game theory means, which was the specific claim that’s being discussed.

No you are incorrect in what a "dominant" strategy means. It has nothing to do with dominating your opponent.

A dominant strategy in game theory means it's your best strategy, regardless of what strategy your opponents pick.

> https://en.wikipedia.org/wiki/Strategic_dominance

> In game theory, strategic dominance (commonly called simply dominance) occurs when one strategy is better than another strategy for one player, no matter how that player's opponents may play.

That is what is being described above.

All the other answers in the thread focus on maximizing head-to-head score. I'll try to answer the question for the goal of greedily maximizing your own score.

Using the same definitions from the article, we now have (for a 2 person game):

  Q_i = -P_i + (1 - sum_{j=1..i}(P_j))

Where the first term is for the case of choosing the same number as your opponent, and the second when it's larger than your opponent.

Now solve the same set of equations, but with our new Q_i. Solving Q_1 = Q_2 analytically is easy, then Q_2 = Q_3 and so on... You get P = (1/2, 1/4, 1/8, ...)

So that's the result for this specific 2-player game. You could also ask about the 2-player game with tie=t for any negative t (the above is for t=-1). Now we get

  Q_i = t*P_i + (1 - sum(P_j))

Again, solving Q_i=Q_{i+1} is easy and gives

  P_{i+1} = (t / (t-1)) * P_i

For example, if t=-2 then (using the fact that all P_i's sum to 1): P = (1/3, 2/9, 3/27, ...)

I did not try to tackle the 3-player game with tie=t.

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