Progressive euchre tourney question
I can't see anyway that this would, or even could, be a winning strategy.

The odds of going alone and taking all 5 tricks cannot be more than about 10%. If we estimate the chance of Dealer's side making all 5 tricks as about 15%, and of being euchred as say 15%, then Dealer's side's point expectation is:


4 * 0.10 = + 0.40
2 * 0.15 = + 0.30
-2 * 0.15 = - 0.30
1 * 0.60 = + 0.60
Total = 1.00


Even in Third Seat (Eldest Hand's partner) any decision on ordering up Dealer as a sacrifice risks losing a big plus score by Eldest Hand.

One could quibble over my estimates above, but no reasonable values will ever approach remotely near the near-guaranteed loss of a euchre from over-aggressive oordering up.

Yes, the success estimates above assume average play by some definition. I am well aware that skillful play, relative to one's group, will significantly increase one's own point expectations, and likewise decrease the expectations of one's opponents. That is all the more reason not to make bad sacrifices - just collect one's skill reward and be content. In Contract Bridge it's called: "Partner; please stop trying so hard for a bottom."

Exception

The one exception to the above that I can see is if, with full knowledge of the score, the situation is exactly one of:


Up by precisely three or four points over precisely one opponent whom one is currently at the table with, having exactly one hand left to play. Conceding two points eliminates the low probability of a Lone Hand overtaking.
Up by seven or eight points over precisely one opponent whom one is currently at the table with, having exactly two hands left to play. Conceding two points eliminates the very low probability of two successive Lone Hands overtaking.