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Giant versus average open player

Posted By: Rich Munitz
Date: Wednesday, 15 September 2010, at 2:57 p.m.

In Response To: Giant versus average open player (Joe Russell)

"If skill was 60%, wouldn't the Giant win ~80% of his matches against the average open player?"

Does the better chess player win 100% of the time since there is zero luck? As I recall, the Elo rating difference between two chess players determines a probability of each player winning - just like in backgammon. It is certainly not 100% skill in backgammon, but don't forget, normally the weaker player also has skill and like in chess has a chance of outplaying his opponent and winning the match if the "luck" is equal.

Aside from the dice rolls themselves arbitrarily giving and taking winning chances, the outcome is also affected by the volatility of skill. Two players who are not perfect will play better or worse than their average in any given match. Giant and Average Player only suggest average performance, not absolute performance. This is further complicated by the fact that the luck (as measured by the change in evaluated equity due to the dice assuming perfect play of the specific roll) does not factor in the difficulty of the decision resulting from that particular roll.

In the end, I don't believe that luck and skill cannot really be expressed as a ratio at all. All we can say is: "the probability that a player with Elo Rating X will beat a player with Elo Rating Y in an 11 point match is P".

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