No, I am not missing anything. Whatever you do, Eab is always the same as Eac and Ecb, since you do exactly the same.Daniel Shawul wrote:No you are missing inclusion of co-variance completely. In the first A vs B test you have a big covariance so that affects the variance of A - B big time. Even HGM agreed that for the example I gave two standard errors of 5 elo each , the std(A-B) = 10 which your calculation ignores..HGM explanation is correct.
A plays B and the the difference (deltaAB) has an error Eab.
Then, with the same number games, we can calculate deltaAC, and it will have error Eac = Eab (since number of games are the same).
Also, with the same number games, we can calculate deltaCB, and it will have error Ecb = Eab (since number of games are the same).
So, we can calculate indirectly
deltaAB = deltaAC + deltaCB
Here we can already see that the error of this indirect calculation is bigger than Eab, no matter what, and we are already playing twice as many games.
deltaAC and deltaCB are independent, so the error for the indirect calculation is
IndirectError_ab = sqrt(Eac^2 + Ecb^2)
IndirectError_ab = sqrt(Eac^2 + Eac^2)
IndirectError_ab = sqrt(2*Eac^2)
IndirectError_ab = sqrt(2) * Eac
If we want the IndirectError_ab to be Eab, we have to make Eac = Eab/sqrt(2). We can do that playing twice as many games, which makes the total 4x.
Miguel
Miguel
PS: Anyway, If you calculate the Eab correctly, there is no covariance, it is direct measure. But this is not relevant.