I was curious about the math involved here, so I decided to examine it.
Dr.'s analysis doesn't seem to be right, since a) it seems to assume you will always win when you face no aces b) it seems to ignore that how much you win depends on whether someone comes in behind you c) he seems to be thinking that the other option is fold. Also, you should use your knowledge that you have no ace in calculating the odds you face one.
First, the hands that O1 (UTG+2) has are not random, and even if you flatcall the hands the blinds will call with are not random.
Second, the value of tournament chips will decrease as you get more, but this far from the big money, the amount is small. I estimate 2xStack => 1.95x$ and 3xS=> 2.85x$.
The problem asked is (approximately) coldcalling with 2xO1's allin stack. Both blinds have you covered. First, I will try the simpler problem: if you had less than or equal to O1's stack, do you want O2 to call or fold?
To simplify the math, I will start with chip EV (CEV), and use calling and losing as the baseline (0 CEV point).
The problem is amenable to some approximations: against a hand with an ace, you win about 69%(67-72%); against a hand with no ace, you win ~78-86%, let's estimate 82%.
If you face 2 Ax hands, you will win, interestingly, about the same as against one: 68% (only having 2 ace outs cancels with the greater flush, straight, trip possibilities in two hands).
For the other two situations (two hands without aces, and one with one without), the chances of each one beating you are almost independent, so we can estimate them by .82*.82=67% of winning and .82*.69=57% (these results match nicely with simulations, biggest errors come from not knowing exactly what range of hands these players would have).
So, for
<font class="small">Code:</font><hr /><pre>
O1 O2 Pvs1 Pvs1&2 CEV1 CEV2 ~$EV1 ~$EV2
Ax Ax .69 .68 .69*2=1.38 .68*3=2.04 1.35 .68*2.85=1.94
xx xx .82 .67 .82*2=1.64 .67*3=2.01 1.60 1.91
Ax xx .69 .57 1.38 .57*3=1.71 1.35 1.62
xx Ax .82 .57 1.64 1.71 1.60 1.62
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Everything is +EV, except possibly the last situation: +CEV, but neutral $EV
Obviously by CEV, you always want a call. Even if my $EV estimates are off, there is only one situation that may be questionable, and the others much more than make up for it.
Unfortunately, that only evaluated the utility to you of someone else's decision.
For the problem at hand, we have the added complication that your stack is 2xO1, so there is more action to occur. Worst case (A) is that O2 will never put in any more money when he is behind, best case (B) is that he will always call the rest of your stack. I'll just solve case B because it is easier and more likely (O2 is getting better than 4:1 to call you on the flop). The difficulty here is estimating the chance you will win the side pot. If the chances of each of your opps beating you are independent, this will be P(1wins)*P(you/O2), but since you lost to O1, the chances of you losing to O2 are greater especially in the case of Ax Ax, so I ave adjusted these numbers down. Fortunately errors propogating from here are smaller since this only gives you a 2xO1 stack, while scooping gives you 5xO2 (blinds have been ignored).
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O1 O2 Pvs1&2 P1/u/2 CEVNC CEVC ~$EV ~$EVC
Ax Ax .68 .07 2.38 3.54 2.37 3.37
xx xx .67 .15 2.64 3.65 2.62 3.48
Ax xx .57 .25 2.38 3.35 2.37 3.20
xx Ax .57 .12 2.64 3.09 2.62 2.95
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Ugh, that was tougher and less precise than I imagined. Basically, unless you are very likely to be outplayed postflop by the blinds, the blinds are very large, the blinds will almost never call or raise, the blinds are more likely to call if you raise, or you are much closer to the money, flatcalling is the better play, and it is not very close. My earlier perception of it depending strongly on amount and location of aces was incorrect. Probably would be somewhat useful to look at case A, but it isn't easy.
Craig