Certainly, if you know what your opponent will do, then it’s easy to beat them. That’s why the optimal strategy must be a mix.
Well, I carefully read the OP again and, unless I’m being obtuse, I didn’t see anything indicating which one of our interpretations is correct.
@Jasmine, they added more details in a later post.
So, if one round is a tie, and each player won one round, then the entire game is considered a tie? I ask because one of the stipulations is that you cannot reuse warriors, so neither player would have any left to play an extra round.
I’m not sure what you mean here. If any round is a tie, the game will necessarily be a tie.
Consider if round 1 is a tie. There are now two possibilities:
- In round two, one player commits more of their troops than the other. This player will win round two–and then they’ll lose round 3, because they have fewer troops for the final round.
- In round two, players tie a second time. This means they committed exactly the same number of troops in round 1 and round 2, and will have exactly the same number of troops left for round three, and will tie in round 3 as well.
A similar logic works if a player wins round 1 and ties round 2: they’re guaranteed to lose round 3, leading to a tie. And, rarely, players will tie round 3. This’ll only happen if they’ve tied 1 and 2, OR if they each won a previous round by exactly the same margin.
A tied round means a tied game.
I don’t understand. I win the first round. The second round is a tie, but then I win the third round. Despite the tie, I’ve won the prerequisite 2 rounds necessary for victory. Yes, no?
Mathematically how would you do that? If you tie 1 round and win another you’ve used more troops than your opponent. Therefore he will have more in reserve and will win the other round.
AH!! LOL
I’ve actually done post graduate studies in math all the way through 3 levels of calculus. I guess I should have thrown in a course in arithmetic. 
I haven’t done anything beyond precal, but I’m a dab hand at adding numbers :).
The rules don’t require using all warriors. A valid (and extremely sub-optimal) tactic is to send 1 warrior each round.
Such a strategy would be so pointlessly suboptimal it’s not even worthy of consideration.
I don’t think this is quite correct either. The big advantage comes if you win the first round without committing too many warriors more than your opponent. If you send 45 warriors to the first battle and your opponent sends 44, that’s a great result - the worst you can achieve now is a tie, and your opponent will have to guess very well to get away with that. However if you send 45 and they send 1, not so good - though you can turn it into a 50-50 guessing game by splitting your remaining warriors 55-5.
It looks like sending (say) 11 warriors to the first battle gives a big advantage against an opponent who sends 1, and sending 21 gives a big advantage against 11, and 31 gives a big advantage against 21 … and 51 flat-out loses against 1. So it’s a rock-paper-scissors situation, and the optimum strategy will be a mixed one of “randomly pick one of these opening bids according to this probability distribution.”
I don’t think this is quite true either. The game must have some sort of strategy to it because it’s possible to play it badly (sending 98 warriors to the first battle is a pretty terrible decision). It’s just that the best strategy will be a mixed one involving random selections that may give the best results in the long run but will not win every time.
And since the game is symmetrical, there’s nothing to stop both players playing the same optimal strategy, in which case it reduces to whose RNG is a better guesser.
At least, once the optimal strategy is figured out, which we haven’t done yet.
That’s fair, that’s what I meant anyway.
This is nitpicking to an irrelevant degree. This game is like Tic-Tac-Toe, where the game is extremely simple and it doesn’t take very long to be competent at it. The difference is that once you reach that level of competency, where you figure out how not to make simple mistakes, in Tic-Tac-Toe you end up always in a stalemate once both players are reasonably competent. In this game once both players are reasonably competent (again, it’s ridiculously easy to get to that point) then it becomes Paper-Rock-Scissors where you’re either going to win by random chance or being able to somehow predict your opponent.
“Being able to play badly” doesn’t invalidate the comparison. You can play Paper-Rock-Scissors badly too. One way is to announce what you’re going to do before you do it, which will tell your opponent how to beat you. Another is to always play the same thing every time and for your opponent to know that about you. While either of those things seems like a stupid thing to do, neither are breaking the rules of the game.
Does that mean you already have a strategy that even if your opponent knows it, he cannot beat you long term?
No, I’m pretty confident such a thing can’t exist with a game like this, or Tic-Tac-Toe, or Paper-Rock-Scissors.
“Ridiculously easy” just means learning some simple rules, like not committing almost everything in the first round, or not committing just 1 warrior in the first two rounds. Just as with Tic-Tac-Toe, “ridiculously easy” pretty much just means not leaving yourself open to someone putting 3 in a row (which isn’t hard and leads to stalemates in every game, as demonstrated in the film Wargames).
Of course such a strategy can exist. There must be a “perfect” strategy that if player A plays it, player B can at best equal.
Of course the “perfect” strategy doesn’t need to be unique, it probably has quite a large forgiving range.
OK, so what is your strategy, then? If it’s “ridiculously easy”, don’t leave us hanging. How many soldiers would you commit to the first round?
The only “perfect” strategy is to not commit only 1 warrior in two rounds. So don’t commit 98 in the first round, or don’t voluntarily commit only 1 in two of them.
Anything beyond that is either luck or predicting your opponent; basically, Paper-Rock-Scissors.
So you got nothin’
I bet if you give me any comprehensive strategy
eg.
31-35-34 33%
50-50-0 33%
52-24-24 34%
All randomly assigned in order
might be an example
I can come up with a counter strategy that will win more than it loses.
Eventually somebody could come up with an optimal unbeatable one, but I don’t think you’ve got it.