Approximate general solution
In the game with $1 dead money and $5 stacks, I made the following simplifying assumption:
If A (UTG) moves in at a, then anyone at the table will call the all-in with the top half of A's raising range. Due to the other players remaining to act, this isn't quite right, but it's pretty close. Also, the odds one is getting are slightly better than 2:1, which also changes things a bit, but again, it's still pretty close to accurate.
Now at the point a (where A is indifferent), A will ALWAYS lose if anyone calls according to these calling criteria. Hence, in order for A to be indifferent, the probability of everyone folding has to be 5/6. If anyone calls, A loses $5, but if everyone folds, then A has successfully stolen the blinds and wins $1.
So, for n players at the table, we simply have:
5/6 = (1/2 + a/2)^(n-1)
This gives the following solutions for up to 10 players:
2: 66.67%
3: 82.57%
4: 88.21%
5: 91.09%
6: 92.84%
7: 94.01%
8: 94.86%
9: 95.49%
10: 95.99%
The EV for all players is extremely complicated, so I'm not even going to try figuring that. The completely accurate solution for 2 players is 69.44%, which is for my taste close enough to the 66.67% given by the approximation. Anyhow, all of these solutions are going to be a little on the loose side, I'm assuming.
Finally, to translate into poker hands, we need a hand ranking system. But the top ones should stay pretty much the same. The 95% should translate roughly to 99-AA and AQ/AK--unless one wants to quibble over whether 99 is better than AJs. That's pretty re-assuring to me, as those are exactly the hands I'd move in with as shortstack even UTG.
As one gets to around 90% (5-player game), one can start thinking about hands like 55/66, A9s, ATo and KQs.
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