interpretation of pwa() with multiple advantaged players

[RobinHankin]

In pwa(), what happens when more than one competitor has an advantage? In particular, what happens if all the players have an advantage? It is not clear to me either what pwa() returns under these circumstances, or what it should return.

As it says in pwa.Rd, "If there is more than one competitor, the resulting likelihood function does not seem to instantiate any simple situation" (this is not particularly clear, it means if there is more than one player with advantage).

In particular, what happens if all the players have an advantage? Symmetry would suggest that the advantage disappears. Even with two players something odd happens:

> ab <- letters[1:2]
> pwa(race(c("a","b")),ab)
log( (S + a) * (S + a + b)^-1)
> 

But what happens if we give each player an advantage, and then symmetrise by summing over all possible race orders?

> ab <- letters[1:2]
> pwa(race(c("a","b")),ab) + pwa(race(c("b","a")),ab)
log( (S + a) * (S + a + b)^-2 * (S + b))
> 

(the probabilites don't add to one, which is a problem). But at least it is symmetric between the real competitors. What is the maximum likelihood estimate?

> maxp(pwa(race(c("a","b")),ab) + pwa(race(c("b","a")),ab))
       a        b        S 
0.000001 0.000001 0.999998 
> 

So the advantage (as quantified by chameleon S) is infinitely strong compared to the real competitors.

[RobinHankin]

With a two-player race and both players having an advantage the proper likelihood function would be:

a beats b: (a+S)/(a+b+2S)

b beats a: (b+S)/(a+b+2S)

so then the probabilities add to one. Of course, I can't do this because the unit sum constraint is a+b+2S=1. But the likelihood function for both observations together, viz (a+S)(b+S) [the denominator =1] works out to be 1-(a-b)^2 which of course maxes out whenever a=b. Which is nice.