Yes, from the simplified logistic Delta = 400*log(g^N) = 400*N*log(g). For average game N=60, g=20 and Delta ~ 30,000 Elo points, the distance from the perfect player to a random mover. And another similar quantity from a random mover to a perfect loser.michiguel wrote:Exactly. In fact, we can make some extreme estimations to show it.Don wrote:I don't understand your difficulty. A random move might be the best move on the board and thus 2 random moves in a row have a chance to be two best moves in a row. 3 is even less likely but still possible and so on. Therefore it is theoretically possible for a random player to play a perfect game, i.e. every move is best. I never claimed it was likely, but the context of the argument was about playing infinitely bad and that is almost impossible to do unless you simply resign on every move.carldaman wrote:Hi Don,Random moves are much stronger than losing move because a random player will draw once in a while even against a perfect player.
the above statement doesn't make sense, as random moves are more likely to be bad, leading to a loss sooner rather than later, even if some random moves may actually be good along the way.
Interesting idea about calibrating to zero based on random moves, though.
Regards,
CL
Chances are that a random player will play horribly but due to the laws of probability it will occasionally play good enough to draw even a perfect player.
Of course I understand how rare this is, but it's not impossible.
The probability to win is
where D is the delta elo, and S a constant that is about 180 to represent the scale we know.Code: Select all
p = 1/(1+exp(-D/S))
So, given a probability we can calculate the ELO difference
But if the probability to win is veeeeery small, the above simplifies toCode: Select all
D = S * ln (p /(1-p))
If one of the players play perfect, the probability to draw for the other is to play good enough moves every single time. If an average game lasts N moves and the average probability for "good enough" is "g", thenCode: Select all
D = S * ln(p) or D = 2.3 * S * log(p) [Equation 1]
Let's ignore that g^N is the probability to score half a point rather than win, it won't make a big difference.Code: Select all
p = g^N [Equation 2]
Combining Eq 1 and 2 we get
since S = 180Code: Select all
D = 2.3 * S * log(g^N) D = 2.3 * S * N * log(g)
So, if we have to make a perfect move out of 40 (g = 1/40) for N=100 moves,Code: Select all
D = 414 * N * log(g)
we get that the maximum ELO compared to a random player is ~66,000 and that is very extreme.
A less extreme number could be g = 1/20 and N = 60 which would give a maximum ELO of ~32,000
Bigger number than those ballpark number obtained by "back in the envelope calculations" are impossible.
Miguel
Most people have a backwards model of how chess strength works. It has nothing to do with you, it's all about your opponent. You cannot "go after" the half point, you have to wait for your opponent to give it to you while not making a blunder yourself.
There is really no such thing as a "great" move. You are never in a losing position and muster so much brain power that you create a win. You hear language like that in chess books that romanticize chess sometimes, especially the older book that fawn over the old masters. But that is not how it works.
So for a random player to play a "good" game we really mean that it is "lucky" enough to avoid all the game theoretical half or full point losses.
I think a lot of positions have multiple moves that are (theoretically) the same so it's not quite so hard as having to find 1 move out of 40 for 50 or 60 moves in row. In some endings you are shuffling pieces and there are few bad moves - most of the moves are adequate. But in many position there is only 1 move that must be played to avoid throwing away the win or draw.
One theoretical truth here is that if you are losing, you cannot play a bad move - unless you define ALL moves as bad of course.
So the rule is that you don't play good chess even though we often say that. What we really mean is that you play less bad moves that other guy.
Kai
