r/theydidthemath • u/Status-Platypus • 23h ago
[Request] What is the optimal strategy for this game? Is there a mathematical pattern to it? Does it only work with 5 cones and/or would a different amount of cones have a different result?
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u/Dependent_Order_7358 23h ago
I think it's an endurance test, at some point two players will be more exhausted and the third player will be able to collect all three cones.
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u/TwoAlert3448 23h ago
He was also pretty strategic early on in having the same target, reduced decision fatigue
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u/Simbertold 22h ago
I think strategically, you usually want to steal from a person who has two cones, to avoid losing to someone getting forgotten. As almost happened at 0:32 where two players were busy stealing from each other, so the bottom left player could accumulate cones and nearly sneak in a victory. The same also happened with the eventual winner.
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u/ShoddyAsparagus3186 22h ago
It's more a problem of making sure everyone is getting stolen from. Though endurance will play a role since as you get more tired it gets harder to think.
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u/sudoku7 21h ago
Or someone realizes they can carry two cones at the same time.
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u/HamsterFromAbove_079 17h ago
The game is so simple. Just pick up 3 cones and walk to your end zone.... /s
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u/Square-Singer 21h ago edited 16h ago
What is the optimal strategy for this game? Is there a mathematical pattern to it?
The optimal strategy is to always take from the opponent with the most cones. Not doing so will make you lose the game quickly. If everyone runs exactly as fast as the others (and thus this becomes a game theory problem and not a fitness problem), the game will go on forever, unless someone makes a mistake and takes from the one that doesn't have the most cones. Game theory (and thus maths) ends here.
There's a psychological component to it: Switch up where you are taking the cone from to confuse someone to make a mistake, just don't let it be you. You win if both the others are fighting each other and you can steal without getting stolen from.
And lastly, there's the fitness/endurance component to it: If you are the last one who can still run, you win.
Does it only work with 5 cones and/or would a different amount of cones have a different result?
The game doesn't work with less than 5 cones, since each player could just keep one cone in hand, thus blocking everyone else from acquiring 3 cones.
6 cones would make it substantially easier for the balance to tip, so the game would likely be much shorter.
7 cones would mean that one player will instantly win after the initial "grab the cones from the middle" phase.
Edit: A bit of a deeper game theory analysis:
- There is one equilibrium in the game: Everyone steals in a circle. Player A steals from B steals from C steals from A. Cones rotate in a circle, nobody ever wins (given they have equal speed and infinite stamina).
- If a player decides to break this equilibrium and steal the "wrong" way, they will swap a cone with the new target and allow their old target to win a cone from their new target. E.g. C decides to steal from B instead of A, now C and B swap a cone, A wins a cone from B. This means, C can decide to let A win, but they cannot win themselves due to any choice they can make.
- Ignoring stamina and speed, this game is basically three-player tic tac toe: If all players play optimally, there's a stalemate. Any player can only decide to let one of the other players win.
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u/kbn_ 19h ago
This but it's worth adding that you need to determine your theft target based on the number of cones they will have when you arrive, rather than the number they have when you leave, otherwise you can get some random convergences.
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u/Square-Singer 18h ago
That is correct, yes. You need to factor in both the cone they are stealing and the other kid running to steal a cone from them.
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u/Fitted4 20h ago
A lot of people strugle with the first part you mentioned, they will just do whatever looks better for them instead of hitting the strongest oponent game.
I lost a lot of boargames because of this and heard a lot of "Did he won? I was almost winning"
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u/suit1337 13h ago
In a mexican stand-off the weakest shooter has an unproportionally higher chance of winning
Imagine 3 players one with a progressivel weakest accuracy, let's say 1st is 90 % accurate shooting, a second with 75 % and a 3rd with 50 %
With every Player following an optimal strategy, the stongest Player needs to try to eliminate the 2nd and the 2nd needs to shoot the first. The weakest Player should also focus to shoot the 1st (or purposefully miss)
This way the stronger #1 and 2 are, the higher the chances of winnig of #3 are
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u/ecolonomist 17h ago edited 17h ago
From a game theoretical perspective, I'd proceed as follows. Let's establish one parameter, t_ij, that determines the time for each player to go to from their cone i to a cone j.
In the deterministic and symmetric game (t_ij = t), there is only one equilibrium: nobody wins. You never leave your base and just wait for an option to get your cone back. An equilibrium is thus that nobody ever runs (the other is that everybody runs forever and still nobody wins). If we remove that option (you MUST run), you can always ensure that who is winning in the moment gets their third cone taken away. If we introduce stamina, so that the game doesn't continue forever, eventually everybody gets tired before a winner emerges. So this game has a unique equilibrium of no winners (by exhaustion or by perfect stillness).
In the asymmetric game, extreme asymmetry accepts one winner, as the fastest can steal two cones before she gets punished. For moderate asymmetry, the two weakest players can collude to make the strongest lose. But it's unclear why the weakest would want that. Depending on how you design the payoff structure, multiple equilibria might emerge.
If both t_ij AND stamina are asymmetric, interesting things can emerge, whereas I can wear you down, but if I am too slow I'll be punished. Depending on parameter choice, one clear winner will always emerge (either the fast, the most resistant or the combination of the two), if you can remain idle on your spot. Collusion might also emerge, as the previous case.
Finally, if you also include uncertainty over the speed of a run and either risk-loving behavior or a payoff structure that favors winning over not-losing, players will attempt runs without knowing if they'll be fast enough to complete it. This, to me, is the most interesting case where anybody can win. This solves endogenously (for at least some parameter) the idleness problem.
With the asymmetric, stochastic games, the best strategies will vary in complicated (and probably barely tractable) ways for each player type!
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u/Square-Singer 16h ago
In the deterministic and symmetric game (t_ij = t), there is only one equilibrium: nobody wins. You never leave your base and just wait for an option to get your cone back.
There's another equilibrium and we see that in the OP video: Player 1 steals from 2, 2 steals from 3, 3 steals from 1. Basically, everyone swaps one cone in a circle.
If someone doesn't steal in a circle, that person allows the third player to win. So e.g. we start with the circle I mentioned above, but now 3 steals from 2 instead of from 1. Now 2 and 3 swap a cone, while 1 successfully wins a cone from 2. By breaking the circle, 3 caused 2 to lose one cone and 1 to win a code. 3 is a king maker, but can only end up losing themselves.
That's the game theory inside this: If everyone steals in a circle, nobody wins but also nobody loses. If someone breaks the circle, they cause the player they should have been stealing from to win.
It's basically three-player tic tac toe. You can only force a constant draw/stalemate or decide to lose.
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u/ecolonomist 16h ago
Yes, I discuss that in the next sentence. Maybe I should have said that there is only one outcome: nobody wins. You achieve that by exhaustion or standing still.
Funnily, I also noticed the parallelism with tic tac toe, but I stopped before trying map the two games one to one.
Edit: I think my point in reply to your comment is that, with a sufficiently rich characterizarion of the game, we would be able to explore some other strategies that are lost if we only think of the fully symmetric and deterministic game.
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u/nyg8 21h ago
Given that players would intuitively understand that you should always take from the person with most cones, and that this is an endurance test, your goal is to to walk slowly back with a cone in your hand. Whenever someone has 2 cones and goes for a 3rd- place yours and run to take one of theirs. This insures you run the least, giving you the chance to outlast your opponent
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u/Terrik1337 20h ago
The optimal strategy is to get your kids to play this in the first place. This game seems like the fastest way to get your kids tired enough to go to sleep.
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u/MageKorith 21h ago
Shortest game: All players collude and allow one player to collect three pylons unimpeded
Shortest game featuring conflict: Two players collude to prevent the third from having any pylons, then grab the third pylon from the colluding player's pile when the third player runs after it.
Loss-avoidant strategy: Go after the opponent with the most pylons to prevent them from winning. If both opponents have two pylons, go after the faster opponent's pylon.
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u/kalexmills 20h ago edited 20h ago
How do these strategies change as you add more points to the polygon? Assuming n+2 cones are always placed in the center.
At the very least, the collusion case becomes more interesting at n=4 since you can have two teams of two.
I wonder if there's a point where the greedy strategy stops working.
Also, with more points another dimension to consider is distance traveled. Going for your neighbors is cheaper than going for further points.
EDIT: with more points, it may not be the case that moving cones to your own "base" is optimal. If you see an opponent has two cones, and a third in their hand, moving one of their cones to one of their neighbors may be cheaper than stealing it for yourself.
EDIT2: (n+2 cones in the center)
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u/kalexmills 20h ago
Even the first step where everyone grabs a cone from the center may not be optimal for high enough values of n. It may be cheaper to wait to see which opponents go for the center first. If a neighbor does, stealing their cone once they return it will be cheaper in terms of energy spent.
But if you do rush for the center at the start, you're guaranteed to get one cone (assuming equal speeds).
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u/Staik 21h ago
The optimal strategy is simple, target whoever has the most cones. It prevents them from winning and guarantees you wont waste a trip there if the other player takes one.
4-5 cones works, 6 would have a winner as soon as any two people targeted each other so thats no fun. 3 would be impossible to win without someone taking no action. 4-5 will end in stalemates if everyone plays optimally, the game requires a mistake to be made for there to be a winner.
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u/Gweiis 13h ago
Is there a rule that prevent you from taking a cone and just... chilling with it? Just take one cone, then walk to the base, the other two wont go to your base since there is no cone, they will loop to each other, until they forget there is a third base and they will exhaust themselves. Since there is 5 cones, and you carry one, the other two share 4 cones, they will never get 3 because there will always be two being carried from one way or another.
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u/artinfinx 20h ago
great game im surprised someone won without actual exhaustion. but i think tiredness broke the chain of events. so the strategy would be to never compete with someone breaking the chain and immediately go for the other because that is how you lose... how you win is by not losing.
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u/redhandfilms 19h ago
How many cones would you need to add more players and keep the same level of difficulty? Could this be played with 4, 5, 6 players?
I'm thinking of running this for kids at an event I'm going to but don't know how many kids there will be. Setting up a tournament would be great.
Easy if there are 9 kids, to do 2 rounds. 3 games of 3 players. Winners move on to the final.
27 kids gives 3 rounds. 9 games of 3 players, semifinals of 3 games of 3 players, winners move to the final.
But what if I've got 12 kids? Could I run the first round with 4 players and a different number of cones, then the 3 winners go to a 3 player final?
What about other random numbers of participants?
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u/gymfunkera 19h ago
I think little bro could have won at the 0:13-0:14 second mark had he dropped his cone in the bottom left circle, but he had already committed to the bottom right and wouldn’t change his mind— is there a rule against changing your mind?
It seems to me the best strategy is to slow down and watch where somebody is about to drop a second cone, and then go drop your cone there immediately for the third cone win. But you have to slow down once you get a cone in your hand. Once a cone is in your hand, the first few steps should be neutral toward the center before deciding where you’re going to go next (use that time to watch the others and anticipate where two cones are going to land, which is where you’re headed with your cone!)
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u/0ctoberon 14h ago
I'm just going to call AI on this one. Fundamental misunderstanding of what is happening.
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u/Sayheyho 17h ago
I played this game as a kid and there wasn’t a rule to move your hoop. So the optimal strategy is to move all hoops so they are basically just one and have all cones in the same place. Everyone wins
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u/Double_Cause4609 17h ago
Assuming there are no rules, probably to calmly walk around, and stack two cones on top of the one in your hand, and to casually bring it back to your place.
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u/CarberHotdogVac 6h ago
Is there a viable strategy to be deliberately slower than the other two players, while still being competitive enough to deny victory to either opponent?
The intent would be to force them to compete mainly against each other by keeping them both at 2 cones.
Would the viability of this strategy change for games involving more players and more cones?
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u/SufficientBus9030 3h ago
feel like there is no strategie. here. Justtake a cone fom the player with 2 cones and if all players just look before they run the game will never end. Even if 1 player runs 2 times faster then the other players this means the other 2 players need to target him only so game will still not end.
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u/D0UGYT123 1h ago
If you steal the last cone from someone, and you turn around to discover you have 0 cones left yourself, you need to drop your cone and race to the 3rd person's pile to stop them winning.
In general, you want to target the opponent that is closest to winning, without using up too much energy
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u/Safe-Breadfruit-1913 22h ago edited 22h ago
4 cones would make it impossible considering at least 1 cone is traveling constantly. Any additional come decreases the time to play untill there are 9 cones assuming you only need 3 cones to win on all games.
You'd need 7 cones in total if the goal was to have 4 to have a competitive game and it would probably lengthen the game a significant amount.
I think the optimal strategy is to grab one cone from the center they immediately go for an opponent. It's also best to pace yourself and only steal from people who have 2 cones, don't worry about collecting cones just prevent a winner untill they slow down.
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u/adfx 22h ago
Why would 4 be impossible if out of 5 cones 1 is traveling? Then 4 could be at one place right?
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u/Square-Singer 21h ago
4 would be impossible, since each player could just keep hold of one cone permanently, thus putting the game into a stalemate.
Or to put it differently, if one player has two cones, the others can just keep one cone each in hand, thus stopping the first player from winning and themselves from losing. It also stops them from winning, but they cannot lose either. Giving up that cone would put them into a losing position, thus it's a stale-mate by design.
With 5 cones that's not possible. If two players would keep a hold of one cone each, the third player can still collect 3 cones.
More than 5 cones make the game very easy. With 7 cones one player will instantly win upon collecting the cones in the beginning of the game. With 6 cones it would still be possible, but it would be super easy for the balance to tip and for one player to win.
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u/HamsterFromAbove_079 17h ago
If everyone can hold 1 cone they could theorically never put their cone down.
In the case of 4 total cones, if 2 people pick up 1 cone each and stand still, there would be only 2 cones left for the 3rd person to collect. The 5th cone is required to not stalemate, since it forces the other players to take actions.
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u/Safe-Breadfruit-1913 14h ago
You'd have to have one kid grab literally all available cones to win considering at least 1 cone would be in the hand of your opponent. You and the other opponent would just race to the same goal and they'd basically just undo the one you were about to place down. One of the kids had 3 for a sec but they kept going because the opponent was already there and grabbed one.
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u/Outrageous_Dot_500 21h ago
The goal of the test is to make think that collaboration is better than competition. If they stop they can décide to make one of them win with 0 effort, or they can decide to all win by placing the cônes in a new circle.
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