5 ms·

What's super surprising if that I'm "dumb" and I play against two "smart" players then I can't have a bad strategy. Assuming that the "smart" players play the N

What's super surprising if that I'm "dumb" and I play against two "smart" players then I can't have a bad strategy. Assuming that the "smart" players play the Nash-equilibrium mixed strategy then whatever strategy I execute, I have the same chance of winning.

alexmolas 4y ago
oh, I didn't think about it! It seems that it contradicts the principle of Nash equilibrium, since in Nash equilibrium you don't have any reason to change your strategy. However, if you change a "smart" strategy to a "dumb" strategy you're not increasing/decreasing your winning changes, so it still fulfills the Nash equilibrium principle.
Revery42 4y ago
This game seems to do weird things if you have a player able to disclose their choice in advance too. Stating that you will choose 42 is stronger than stating you will choose 1, assuming other players only maximize their own odds of winning. It's kind of like a convoluted game of chicken.
wikfwikf 4y ago
This is obviously wrong. If you choose one million billion trillion every time you will not win 28% of the time.
- planede 4y ago
  You will, your opponents will choose matching numbers 28% of the time if they use the Nash-equilibrium mixed strategy.
  - wikfwikf 4y ago
    You are right. Choosing a bigger number makes it less likely that you will win by being smaller than the other two, but more likely that you will win by the other two knocking each other out, since there are now more numbers available for them to do so with.
    - planede 4y ago
      I think most importantly your opponents become less likely to knock your number out. The goal is your number to win, not your opponents to lose.
- alexmolas 4y ago
  but still (1) there's some probability of both players choosing a number bigger than one million billion trillion, and (2) there's some probability of both other players choosing the same number. Actually, the probability that the other both players choose 1 is pretty big (~0.21). So even if you choose a super-big number, there's a big probability of you still winning.

alexmolas 4y ago

oh, I didn't think about it! It seems that it contradicts the principle of Nash equilibrium, since in Nash equilibrium you don't have any reason to change your strategy. However, if you change a "smart" strategy to a "dumb" strategy you're not increasing/decreasing your winning changes, so it still fulfills the Nash equilibrium principle.

Revery42 4y ago

This game seems to do weird things if you have a player able to disclose their choice in advance too. Stating that you will choose 42 is stronger than stating you will choose 1, assuming other players only maximize their own odds of winning. It's kind of like a convoluted game of chicken.

wikfwikf 4y ago

This is obviously wrong. If you choose one million billion trillion every time you will not win 28% of the time.

planede 4y ago
You will, your opponents will choose matching numbers 28% of the time if they use the Nash-equilibrium mixed strategy.
- wikfwikf 4y ago
  You are right. Choosing a bigger number makes it less likely that you will win by being smaller than the other two, but more likely that you will win by the other two knocking each other out, since there are now more numbers available for them to do so with.
  - planede 4y ago
    I think most importantly your opponents become less likely to knock your number out. The goal is your number to win, not your opponents to lose.
alexmolas 4y ago
but still (1) there's some probability of both players choosing a number bigger than one million billion trillion, and (2) there's some probability of both other players choosing the same number. Actually, the probability that the other both players choose 1 is pretty big (~0.21). So even if you choose a super-big number, there's a big probability of you still winning.