First off, let me say that I'm not familiar with Otherkind and not entirely sure I understand your system (though I think I do), and besides that I could just be making a mathematical blunder, but it seems to me that your assessment of the bonus die is not quite right, that the bonus die is more powerful. I have a d6 right here (actually, I have a d4, d8, a whole bunch of d10s, a d12, and a d20, but I assume you meant a d6 for your system), so what I'm going to do is roll it and actually see how a bonus to one die and a bonus die change things. The bonus die will be enclosed in [ ].
I'm making some assumptions here. First, that the target numbers are unknown, and therefore increasing the average of all the dice increases the average number of successes for all three dice. Second, I'm assuming there is no reason not to swap the lowest die with the bonus die if the bonus die is higher.
First roll: 2, 1, 4, [4]
Now, with no bonus, the dice average is 2.33. If we apply the +1 then the average is 2.67. If we swap the 1 with the bonus die, the average is 3.33. Hence, since the average of all three dice is higher the average number of successes should be much higher. Also, the 4 is much more likely to be a success than the 1.
Second roll: 4, 5, 4, [3]
This is an interesting case. The bonus die is no good here, since it is lower than all the other numbers. With no bonus, the average is 4.33, with the +1 to one die the average is 4.67. A 5+1=6 should beat or tie with anything, but a 4+1=5 should give better chances of all around success.
Third roll: 5, 6, 2, [4]
As we can see here, the best deal is to swap the lowest die with the bonus die, rather than get a +1 bonus. The only exception was if you specifically wanted to max out that 5, rather than applying it to the 2. Consequently, the average is also 4.33 with no bonus, 4.67 with the +1, and 5 with the bonus die.
Alright, now I'm starting to think you may be right. The question is, what is the probability that one of the three dice will end up lower than the bonus die? I'm not sure I have a firm grasp of probability, but as I read on Wikipedia, the chance of A or B happening is the probability of A plus the probability of B. Therefore, the chance of any of the three dice A, B, and C coming up as a given number is 0.167+0.167+0.167=0.5. The chance of the bonus die coming up with a specific number is 0.167 (or 1/6). So, there is a 50% chance one of the three dice will be a 1, and a 5/6 chance that the bonus die will then be higher. I think this will have to be on a case by case basis. Here's the equation we need, the chance that A and B or A and C or B and C will be greater than or equal to x, times the chance of x coming up on the remaining die. So 3[(6-x+1)(6-x+1)/36](1/6).
In the table below, I have the lowest of the three dice, the chance that that will be the lowest, and the chance that the bonus die will be higher.
| Chance | 50% | 34.72% | 22.22% | 12.5% | 5.56% | 1.39% |
| Bonus Die | 83.33% | 66.67% | 50% | 33.33% | 16.67% | 0% |
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I make no claim as to the accuracy of this table, so someone please correct me if I'm wrong. And yes, I suddenly realize how beneficial a probability course could be, I'll consider taking one if I can fit it into my last year here at college (though honestly I'll bet modern geometries is way more fun).
To get the chance of that bonus die actually being useful, I have no idea where to go from here, sorry.
In any case, you're probably right, the bonus die will be more powerful. So it sounds like a good idea to go with your gut here.