This die defines your luck in life, and is rolled by your god of choice at every important event: roll a (0) you fail miserably, (6) you succeed briliantly, (4) you pass, barely.
Would you rather have:
A: a 000666 die?
B: a 111555 die?
C: a 222444 die?
D: a 333333 die?
Do note that I made a mistake in the original post, but the conclusion was still the same. I forgot to divide the Expected Value (EV) for all dice by 6 (the number of faces).
I’m not sure how to prove this empirically, but playing with it on my whiteboard I get a sense that the die 444222 is going to have the best EV, under the given constraints and my value assignments. The real kicker is “average face value of 3”. Given that constraint, you will never be able to create a die with a positive or even zero EV using my values. Consider die 333333 and each face’s value:
This die has an average face value of 3 ( (3 * 6) / 3) and we can consider changing any face up or down. But, in order to keep the average a 3, moving one face up one number requires we move a different face down one number and vice-versa. For example, if we push one face from a 3 to a 4, we must also pull one face from a 3 to a 2 to balance out the average. And because the value for positive value numbers (4, 5, 6) starts off one doubling behind the values for the negative value numbers (3, 2, 1, 0), going any further than 4 in the positive direction on a face means that another face will be pushed down far enough to cancel out the benefit of going to a 5 or beyond.
To look at it another way (the way I did on my whiteboard), let’s just consider a two sided die (a coin flip). Using the same values for each number, we can consider a 33 coin. This has an EV of -1 ( (-1 * 2) / 2) and an average of 3 ( (3 * 2) / 2 ). Now, move the numbers, but keep the same average of 3. Moving to a 42 coin changes the EV to -1/2 ( (+1 + (-2)) / 2 ) and the average is still 3 ( (4 + 2) / 2 ). The EV got better. So, let’s take another step in each direction. We get a 51 coin with an EV of -1 ( (+2 + (-4)) / 2) and the average is unchanged at 3 ( (5 + 1) / 2 ). And going to a 60 coin takes us to an EV of -2 ( (+4 + (-8)) /2 ) with a average of 3 ( (6 + 0) / 2 ). This means that the best coin for this scenario is a 42 coin. Taking that coin idea back to the die, you can think of the die as a bunch of linked coins. If you want one face to be a 5 the one face must be a 2, which would be worse than having the pair of faces be a 4 and a 2. So, to maximize the EV, you want to create a bunch of 42 pairs.
Of course, we could fiddle with multiple faces at once. What about a 622233 die. Well, it gets worse. EV is -2/3 ( +4 + (-2) + (-2) + (-2) + (-1) + (-1))/6).
Maybe a 522333, EV is -5/6 ( (+2 + (-2) + (-2) + (-1) + (-1) + (-1)) / 6). Again, since lower numbers get a more negative valuation faster than higher numbers get a positive valuation, you just really don’t want to let numbers get any lower than necessary. The 42 paring just happens to hit a sweet spot where that effect isn’t yet pronounced enough to cause the EV to drop off.
So ya, while I don’t know the maths to prove it. I’m gonna say that the 444222 probably maximizes the EV under the given model.