5 ms·

Does it matter if there is an advantage to playing white vs black if both players are playing completely randomly? It should only really matter if there is an a

Does it matter if there is an advantage to playing white vs black if both players are playing completely randomly? It should only really matter if there is an advantage when both players are playing optimally.

ekianjo 12y ago
if completely random, you may prove that there is an empirical advantage to starting the game without taking in account the level of any player, because two players playing randomly should be equivalent. I'm not entirely sure, though.
- mehrdada 12y ago
  This is technically correct, but does not result in a very useful information, as P(A|B) can be a very different beast than P(A).
- ema 12y ago
  Imagine a really simple game where there are only 2 moves: saying 1 and saying 0. The player who first says 1 wins. So of course the player who has the first turn will in practice always win, but when both players move randomly it's 50/50.
  - jdoerrie 12y ago
    Not true. Assuming both players move completely random (both choose either 1 or 0 with 50% probability each) and independently (past actions don't influence the current event) player 1 has a 2:1 advantage, i.e. he will win with probability 2/3. This is because player 1 can win after an even number of rounds (including 0) with 50% probability. So his total chance of winning is 1/2 * (1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3.
    - ema 12y ago
      Yes you're right. Quite the embarrassing mistake I made there.
lmm 12y ago
One of the biggest innovations in computer Go has been the idea that you can evaluate a position in "monte carlo" fashion, by just playing a few random games from that position and seeing who wins. I don't know whether the same idea extends to Chess.
- Qwertious 12y ago
  Monte carlo chooses some possible moves at random, then evaluates which is most successful and has a bias towards the more successful ones. This has the randomness but without any bias or evaluation of effectiveness. But really, Monte Carlo in Go can basically be thought of as a lazy, intelligent brute-forcing method.

ekianjo 12y ago

if completely random, you may prove that there is an empirical advantage to starting the game without taking in account the level of any player, because two players playing randomly should be equivalent. I'm not entirely sure, though.

mehrdada 12y ago
This is technically correct, but does not result in a very useful information, as P(A|B) can be a very different beast than P(A).
ema 12y ago
Imagine a really simple game where there are only 2 moves: saying 1 and saying 0. The player who first says 1 wins. So of course the player who has the first turn will in practice always win, but when both players move randomly it's 50/50.
- jdoerrie 12y ago
  Not true. Assuming both players move completely random (both choose either 1 or 0 with 50% probability each) and independently (past actions don't influence the current event) player 1 has a 2:1 advantage, i.e. he will win with probability 2/3. This is because player 1 can win after an even number of rounds (including 0) with 50% probability. So his total chance of winning is 1/2 * (1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3.
  - ema 12y ago
    Yes you're right. Quite the embarrassing mistake I made there.

lmm 12y ago

One of the biggest innovations in computer Go has been the idea that you can evaluate a position in "monte carlo" fashion, by just playing a few random games from that position and seeing who wins. I don't know whether the same idea extends to Chess.

Qwertious 12y ago
Monte carlo chooses some possible moves at random, then evaluates which is most successful and has a bias towards the more successful ones. This has the randomness but without any bias or evaluation of effectiveness. But really, Monte Carlo in Go can basically be thought of as a lazy, intelligent brute-forcing method.