Re: Heuristic probability quickie.

On 16 Oct 2007 17:02:39 +0300, Phil Carmody
<thefatphil_demunged@xxxxxxxxxxx> wrote:

Imagine a game of skill, something like go would fit.

Imagine 3 experienced players of this game, A, B, and C say,
such that over their game histories A beats B about 60% of
the time, and B beats C about 60% of the time.

What proportion of the time would you expect A to beat C?

I know that there's no single correct answer, I'm just
wondering what kinds of values were supportable, based
on what assumptions.

The reason I'm interested is because I noticed a few things
that I consider to be anomalous in some commonly-used
ELO-like ratings schemes.

Let's model the strength of the players by using biased coins. Assume
Heads beats Tails. Thus the stronger player will have a coin more
biased towards Heads than that of a weaker player.

When 2 players play a single game, they flip their coins
simultaneously, repeating until there is a mismatch. A mismatch ends
the game, with Heads beating Tails.

Now suppose we have 3 players A, B, C such that

A's coin has P(H) = a
B's coin has P(H) = b
C's coin has P(H) = c

where a,b,c are real numbers in the interval (0,1).

Let

x = the probability that A beats B
y = the probability that B beats C
z = the probability that A beats C

It's easy to show that

z = (xy) / (1 + 2xy - x - y)

Hence, if x = 3/5 and y = 3/5, then z = 9/13

quasi
.

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