# Chess can be a Game of Luck

Can chess be a game of luck?

Let us consider the following two scenarios:

A) We have a chess tournament where each of forty chess players pay 50 dollars entrance fee and the winner takes the prize which is 80% of the the total entrance fees.

B)  We have a chess tournament where each of forty chess players pay 20,000 dollars entrance fee and the winner takes the prize which is 80% of the the total entrance fees.

Before dealing with these two rather realistic scenarios let us consider the following more hypothetical situations.

C) Suppose that chess players have a quality measure that allows us to determine the probability that any one player will beat the other. Two players play and bet. The strong player bets 10 dollars  and the waek player bets according to the probability he will win. (So the expected gain of both player is zero.)

D)  Suppose again that chess players have a quality measure that allows us to determine the probability that any one players will beat the other. Two players play and bet. The strong player bets 100,000 dollars and the weak player bets according to the probability he will wins. (Again, the expected gain of both players is zero.)

When we analyze scenarios C and D the first question to ask is “What is the game?” In my opinion we need to consider the entire setting, so the “game” consists of both the chess itself and the betting around it. In cases C and D the betting aspects of the game are completely separated from the chess itself. We can suppose that the higher the stakes are, the higher the ingredient of luck of the combined game. It is reasonable to assume that version C) is mainly a game of skill and version D) is mainly a game of luck.

Now what about the following scenarios:

E) Two players play chess and bet 5 dollars.

Here the main ingredient is skill; the bet only adds a little spice to the game.

F) Two players play chess and bet 100,000 dollars.

Well, to the extent that such a game takes place at all, I would expect that the luck factor will be dominant. (Note that scenario F is not equivalent to the scenario where two players play, the winner gets 300,000 dollars and the loser gets 100,000 dollars.)

Let us go back to the original scenarios A) and B). Here too, I would consider the ingredients of luck and skill to be strongly dependant on the stakes. The setting of scenario A) can be quite compatible with a game of skill where the prizes give some extra incentives to participants (and rewards for the organizers), while in scenario B) it stands to reason that the luck/gambling factor will be dominant.

One critique against my opinion is: What about tennis tournaments where professional tennis players are playing on large amounts of prize money? Are professional tennis tournaments  games of luck? There is one major difference between this example and examples A and B above. In tennis tournaments there are very large prizes but the expected gain for a player is positive, all (or at least most) players can make a living by participating. This changes entirely the incentives. This is also the case for various high level professional chess tournaments.

For mathematicians there are a few things that sound strange in this analysis. The luck ingredient is not invariant under multiplying the stakes by a constant, and it is not invariant under giving (or taking) a fixed sum of money to the participants before the game starts. However, these aspects are crucial when we try to analyze the incentives and motives of players and, in my opinion,  it is a mistake to ignore them.

So my answer is: yes, chess can be a game of luck.

# Raigorodskii’s Theorem: Follow Up on Subsets of the Sphere without a Pair of Orthogonal Vectors

Andrei Raigorodskii

(This post follows an email by Aicke Hinrichs.)

In a previous post we discussed the following problem:

Problem: Let $A$ be a measurable subset of the $d$-dimensional sphere $S^d = \{x\in {\bf R}^{d+1}:\|x\|=1\}$. Suppose that $A$ does not contain two orthogonal vectors. How large can the $d$-dimensional volume of $A$ be?

Setting the volume of the sphere to be 1, the Frankl-Wilson theorem gives a lower bound (for large $d$) of  $1.203^{-d}$,
2) The double cap conjecture would give a lower bound (for large $d$) of $1.414^{-d}$.

A result of A. M. Raigorodskii from 1999 gives a better bound of $1.225^{-d}$. (This has led to an improvement concerning the dimensions where a counterexample for Borsuk’s conjecture exists; we will come back to that.) Raigorodskii’s method supports the hope that by considering clever configurations of points instead of just $\pm 1$-vectors and applying the polynomial method (the method of proof we described for the Frankl-Wilson theorem) we may get closer to and perhaps even prove the double-cap conjecture.

What Raigorodskii did was to prove a Frankl-Wilson type result for vectors with $0,\pm1$ coordinates with a prescribed number of zeros. Here is the paper.

Now, how can we beat the $1.225^{-d}$ record???