Dice puzzle: pips must sum to 21 A friend recently told me this little puzzle but to me it seems like there might not be a right answer. Was wondering if there is in fact a best combination in this scenario. 
Dice rolling war
You and a friend decide to have a dice rolling war consisting of 36 battles. Each player will roll 1 die. A battle is won by the player with the higher number of pips, and the winner receives a point. If both players have the same number of pips, neither player receives a point. However, before starting the battles, the players get to choose the number of pips to put on the 6 sides of the die, the constraint being that the number of pips on all sides of the die must summ to 21 and be in the range [1-6]. For example,
[4][4][4][4][4][1] would be valid choice.
Finally, the person with the most points at the end of 36 rounds wins.
Part a) What pips should you choose to give you the best chance of winning?
Part b) What is the probability that you will win?
 A: This was already solved
Nice question with an unexpected result.
Turns out the original die, $(1,2,3,4,5,6)$ is a unique solution for the optimal die when having six sided dice that sum to $21$.
Source is this paper that I've linked in the comments earlier.  It considers the optimal strategy for equal-sum dice games with $n$ sides that sum up to some number $\delta$, where all sides are a positive integer.
Your question is answered by the solution to the $(6,21)$ case (stated under "Example 9"), which boils down to that the original die is the best option to choose, which was an unexpected result to me at first.

Result explained
The reason behind that is that the regular $(1,2,3,4,5,6)$ die has at least an equal chance of winning the game of some amount of "battles" (or rolls) against any of the valid $110$ dice.
All other dice can be matched up with at least one weak spot as I call it - a die that has greater chance of winning against the given die - making your die have a greater chance of losing. 
The original die is the only die that does not have a weak spot.
Thus, The best die is the original $(1,2,3,4,5,6)$ die and both players, playing optimally, clearly have an equal chance of winning.

Some remarks
If your opponent were to randomly play a die against your regular die, you would either keep your balanced chances of wining, or have a greater chance to win. Playing the original die, you can't lose more often than your opponent.
(This is the expected result of course, since surely you can be very unlucky and roll $1$ all the time, but those cases diminish as the amount of battles or games increases)
If you were to play any other die against a random die, you could have an equal chance of winning (overall draw), have a better chance of winning (overall win), or have a greater chance of losing (overall loss).
With the original die, you can never have an expected overall loss. All other dice can end up in that situation.
The result of the optimal game boils down to flipping a fair coin to see who'll win, with some chances of the coin being stuck vertically - a draw.

User jvdhooft had a observation in the comments, where the die $(4,4,4,3,3,3)$ is the same against a regular die, like two regular dice against each other.
Also, it plays better overall against random dice by winning more rolls on average! Does this mean this is a better die?
No. Simply because if it were playing against one of its weak spots, such as $(1,1,4,5,5,5)$, it would be expected to lose the match more often. Where the regular die is never expected to lose the match more often.

I would say that the regular dice $(1,2,3,4,5,6)$ specializes to never have an expected loss more often, while the $(4,4,4,3,3,3)$ specializes at beating lesser dice with a better score on average, but by not being the optimal die, it can end up in a duel where it will lose more often.
