Hi,
The formula is pretty straightforward,
R=Sqrroot(Zin/Zout), or R^2=Zin/Zout
Where : R = Ratio
Zin = Impedance "seen" by the OT's primary
Zout = Load impedance
The output impedance of six 6L6s in parallel push-pull operation is in the 1.7 KOhm range, and, since your OT has two taps ( 4 and 8 Ohms ) we can calculate the transformer's ratios.
R(4) = Sqrroot(1700/4) -----> 20.6
R(8) = Sqrroot(1700/8) -----> 14.6
By removing two tubes you raised the impedance "seen" by the primary winding ( remember OT don't have an inherent impedance of their own, they have ratios ) to about 2.5 KOhms, so, by reversing the formula to solve for the output impedance we have (for the 4 Ohms tap):
Zout(4) = Zin/(R^2) -----> 2500/425 = 5.8 Ohms
While, for the 8 Ohms tap we have :
Zout(8) = Zin/(R^2) ----> 2500/213 = 11.7 Ohms
So, while it's true that the power transferred from the primary to the secondary will not be at its maximum due to the slight mismatch, it's still advisable to use the 4 Ohms tap IMHO ( connecting the 8 (now 11.7) Ohms tap to a 4 Ohms load would result in worsening the mismatch ).
Hope this helps
Best regards
Bob