I have been unable to Google this game, but I am sure it is well known.
Two players get 100 warriors each. For three rounds you get to choose how many warriors to use in a fight. Those warriors cannot be used for subsequent rounds. Whoever wins 2 rounds wins the game.
What is the optimal strategy?
How do you win a round? Just by having more warriors in that fight?
That’s correct. I suspect you cannot do better than just randomize. If you try to throw one round by using 1 warrior, then a counter strategy solid be to use 2 warriors one round.
It sounds like there isn’t much strategy involved, it’ll come down to luck.
Every choice you make each round will depend on what choice your opponent makes in that round, but you have no way of knowing that before you made your decision.
It’s like asking what the optimal strategy is for paper-rock-scissors.
The only good strategy I can suggest is don’t use more than 98 in the first round. I assume there is no limit and you can even commit all 100 in the first round, and you can choose to commit none in any given round.
I’ve not heard of that–but there’s a variant I got from 538 dot com some years back that I adapted and use to teach spreadsheets to my fifth graders. In the variant, you have 100 warriors and nine kingdoms to distribute them between. You compare your distribution to another player’s, and you win every kingdom where you have more soldiers than them.
The game has three rounds:
As for the game you describe, I’m curious about whether there’s a winning strategy. It feels in a way like a variant of rock-paper-scissors.
Randomize how? I can think of multiple ways.
Hmm, this would be fun to code up.
The best way would be to figure out the possible combinations.
Assume you have to assign a minimum of one warrior to each round.
So the possible combinations would be:
1-1-98
1-2-97
1-3-96
etc.
There are a finite number of possible combinations that add up to one hundred. (4851? Did I do the math right?)
You could set up a computer to run each of these 4851 possible combinations against the same 4851 combinations.
What happens in the event that a round ties? Is that a tied game one win and one draw for each player? (Of course, if two rounds tie, then all three rounds must tie as well…)
Also, are your allocations made at the start, or do you get feedback from the first round with which to make your decision about the later rounds?
I would think that whoever wins the first round has a big advantage.
It depends on why they won the first round.
You commit 5 and they commit 4, hey you have 95 more to commit for the next two rounds. and you only need to win one of the next two. The strategy then would be to commit zero in round 2, and then as long as the other side committed at least 2 you are guaranteed a win, because they would have 94 or fewer left in the last round and you’d have 95 and they can’t win. So unless they also commit zero or only commit one in the second round, you have a clear path to winning.
But if you won the first round by committing 95 and they committed 3, well if they are smart they just have to commit 6 in the second round. You can’t beat that. And then they have 91 left in the third round.
So the statement that whoever wins the first round has a big advantage isn’t true. It’s only true if they win without committing very many warriors. (And whatever “very many” constitutes is indeterminate; I’d estimate if you won by committing less than 33 then you did well, but that’s still just speculation on my part.) And that goes back to the paper-rock-scissors random nature of the game. Your success is too dependent on unknown factors to form any kind of realistic strategy.
One of us is misunderstanding the rules. I am assuming that each player allocates the warriors to each round before the numbers are revealed. You seem to be assuming that the players can adjust the numbers after each round.
My intuition is that a fixed sequence is going to lose to a method that uses some sort of random distribution.
I’m making one assumption myself that isn’t actually stated in the OP… That you do each round one at a time. But I guess it would be possible to do them all at once. So basically you can hand each of them a piece of paper that says something like:
Round 1:
Round 2:
Round 3:
Each person fills out their sheet by putting a value for each round, and the total has to be 100 or fewer. (Nothing says you have to use all 100, though there is no advantage in using fewer than 100 that I can see.) You then reveal each sheet and compare the amount for each round and see what happens.
I assumed that wasn’t how it worked, because it feels less like a contest and is even closer to rock-paper-scissors than going round-by-round.
Also, if you can adjust the numbers after each round, what is the limit? Do you only do it once? Do you do it in secret? Do you do it one after each other? If in round one I picked 30, and you picked 25, then you change your number to 31, then I change mine to 32, and so on, the game would never end. If we just do it once… What’s the point of that? What value does it add to the game?
It only makes sense that you each pick a number for the first round, reveal it to each other, and that’s now a fixed amount. Then you go again in round two doing the same thing, and then whatever you have left after round two is what you have for round three. Doing it any other way makes it sound even less like a game of strategy than it already is.
The limit is after each round. You can’t change numbers, because once the round is over, the round is over. But given that the game is described as sequential rounds, you presumably can base your decisions for later rounds on what happened on previous rounds (which is different from the version with multiple kingdoms that @Left_Hand_of_Dorkness described).
I don’t think it’s just luck, because some strategies are clearly better than others. For instance, a split of 100-0-0 is a sure loser, or at best a tie, because you’ll win one round and lose two. On the other hand, evenly splitting your troops also can’t be the optimal strategy, because if your opponent knew you were doing that, they could win with 50-50-0. The true best strategy would presumably involve a randomized mix of strategies, to some degree. But not just “enumerate all possible allocations, and pick one at random”: In the optimal strategy, some splits would be more likely than others.
Yeah, that was the same conclusion I drew. That’s why the idea of the numbers being able to be changed each round didn’t make sense to me from what little the OP described.
It’s not just luck, but it’s going to be mostly luck. There are definitely some automatic losing strategies, as you pointed out the 100-0-0 can at best be a tie (which I also alluded to in my first post in this thread).
Really, I think the only truly valid strategy would be this…
Don’t do anything extremely stupid to guarantee a loss. (Easy to avoid.)
Play against someone predictable.
I mean, anything outside of that is going to be pretty random I’d imagine.
It is interesting, because it appears at first glance to be a game of strategy, but really it is a game of psychology, with just a little bit of strategy thrown in.
If you end the first round with fewer than half of the troops that your opponent has, you are, absent really stupid play on their part, guaranteed a loss: they could put half their troops in each of the remaining kingdoms and guarantee victory.
So the question is: do you want to win the first kingdom and then limp along to win one of the remaining two, or abandon the first kingdom and try to dominate the remaining two? Which one makes sense, and by how much, depends on your read of the opponent.
The winner of the first kingdom necessarily has fewer troops remaining, and they are guaranteed to lose one of the two remaining kingdoms. They should throw all their remaining troops into kingdom two or three. Which one they throw the troops into is a psychological call.
Of course, they could count on their opponent expecting that play, and protecting only one of the two remaining kingdoms, in which case an even split of troops would be appropriate.
And naturally the poison cannot be in front of me…
LOL, it does have that feel to it. 
I only heard about the game second hand. It was used recently in a competiton-based reality series. I don’t know all the details, but from what I understood:
You could say it’s a game of psychology, but there should be one mathematically optimal strategy. There can be strategy in rock paper scissors, but between computers it’s just random. One player can always choose at random and win 50% of the time.
In rock-paper-scissors, there are only three options for each round, and it’s easy to show by symmetry that the optimum is equal chances for all three. In this case, though, there are thousands of options, and each round is not like the previous one, and some choices are clearly worse than others.
I mean, sure, once you do find the optimal mix, if both players are using that optimal mix, then it’ll be 50-50 who wins. But finding the optimal mix is nontrivial.
It’s probably easier to find the solution if you have to assign all the rounds at the start instead of reacting after each round. You could simulate it, but it would still be hard to know that you had found the best random distribution.
Assuming that each player secretly chooses their warrior number for the first round; then the results is published and each player then chooses the number for the next round … then I don’t think there will be an optimal mix.
My reason for saying that is that if one player has to reveal their numbers each time, it would be easy to beat them.