> To simplify the problem, let’s start with the simple scenario where you are competing against two other persons.
> P(wining)~=0.296
Where does the other 11.2% go? Is the probability of a triple tie? Perhaps you can add it to the article.
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.
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.
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.
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.
That is what “dominant” in game theory means, which was the specific claim that’s being discussed.
> 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.
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))
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))
P_{i+1} = (t / (t-1)) * P_i
I did not try to tackle the 3-player game with tie=t.
Now I'm wondering which is the best strategy you can follow if the other players follow an unknown non-optimal strategy (like selecting 1,2 or 3 with p=1/3). Maybe it's a good scenario to try some reinforcement learning
Everyone else should not play 1 because otherwise they would not win. But if no one else plays 1, you win. So they must take turns to lose and make you lose. It's a good strategy to not meet them again ever :)
But in this case you lose no matter what strategy you use; you only get to decide which of the other players wins. That's why the article analyses the case where the other players use the same strategy.
The hard part is to establish a prior for this "unknown". Otherwise you can't say anything.
I think a triple tie means no one wins, which would explain why the probability of winning is less than 1/3. (If each of the three players is playing the same Nash equilibrium strategy, they should each win with the same frequency.)