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Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

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Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#101

Post by lmm »

npizzolato wrote: post #75
ekianjo wrote: post #73

Earlier quoted context omitted.

> what kind of useful information could be mined from a huge set of random-random games Whether there is a real advantage to start with white instead of black ?

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.

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.

Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#103

Post by lmm »

stolio wrote: post #28
afro88 wrote: post #11

Earlier quoted context omitted.

If it's impossible to checkmate then it's a stalemate. So if one side has one king and the other side one king (or even one king and a pawn, bishop, knight or rook) then it's a stalemate and the game ends. edit: forgot you can checkmate with just a king and a queen

You can easily checkmate with a king and a rook, it just has to be along an edge of the board. In rare circumstances you can checkmate with a king and a single bishop: http://www.chess.com/forum/view/more-puzzles/mate-with-one-b...

Any piece (well, except the king) can checkmate with cooperation from the opponent's pieces. The standard sets (queen, rook, two bishops, bishop and knight) are those with which you can force a checkmate against an opponent who just has a king.

Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#104

Post by jdoerrie »

ema wrote: post #88
ekianjo wrote: post #76

Earlier quoted context omitted.

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.

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.

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.

Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#106

Post by ema »

jdoerrie wrote: post #104
ema wrote: post #88

Earlier quoted context omitted.

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.

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…

Yes you're right. Quite the embarrassing mistake I made there.

Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#107

Post by Qwertious »

lmm wrote: post #101
npizzolato wrote: post #75

Earlier quoted context omitted.

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.

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.

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.

Re: Chess: Who will win in this riveting game of Math.random() vs. Math.random()?

#110

Post by bnegreve »

shultays wrote: post #108
bane wrote: post #63

Out of curiosity, anybody know the relative strength of this vs. other real engines?

Uh, zero?

Probably, but is it that obvious? Having an opponent that is not even trying to win is much harder to predict.

I've heard several poker players saying that it's hard to play against novices, because they do nonsense.

Surly there is some game theory that says something about this. Anyone?

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