4 ms·

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 chan

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.

ZetaZero 4y ago
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.
- StevenWaterman 4y ago
  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.
  - HWR_14 4y ago
    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.
    - theptip 4y ago
      > 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.
      - HWR_14 4y ago
        No you are incorrect in what a "dominant" strategy means. It has nothing to do with dominating your opponent.
        theptip 4y ago
        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 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.
tomerv 4y ago
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.

ZetaZero 4y ago

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.

StevenWaterman 4y ago
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.
- HWR_14 4y ago
  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.
  - theptip 4y ago
    > 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.
    - HWR_14 4y ago
      No you are incorrect in what a "dominant" strategy means. It has nothing to do with dominating your opponent.
      - theptip 4y ago
        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 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.

tomerv 4y ago

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.