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.