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| ▲ | abetusk 6 hours ago | parent | next [-] |
| An entropy argument for optimal strategy when winning? Finite size/bounds arguments for losing? If you have 100 options available to you, the maximum information gain is if you eliminate half. So, if you can, you always want to employ that strategy. The problem comes with when you're losing, you might get maximum entropy gain by eliminating half but, because of finite effects of the game ending, that might not matter. Take the example I gave: the next move you lose and you have 4 options to choose from. Eliminating half (2 in this case) will give you maximum entropy gain but guarantee a loss, since you're not whittling down the remaining list to 1 option. Better to take the hit on entropy in order to at least have a chance at winning. I don't claim to have deep knowledge but this seems like finite size scaling effects. There's a kind of "continuum limit" of these processes but when you get to actual real-world/finite instances, there are issues "at the edges", where the continuum becomes finite. The finite size of the problems creates a kind of non-linear issue at the edges. All this is very hand-waivy, so don't take it too seriously but that's the intuition I have, at least. |
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| ▲ | thaumasiotes 5 hours ago | parent [-] | | You can see an edge effect in bidding-based card games when someone is close to victory. Say you're in a game to 500 points and you're losing 460 to 480. There are 13 tricks and a trick is worth 10 points if you bid it. The other team bids 5 tricks. Assuming they can make this (very safe) bid, they will have 530 points. You are collectively good for about 6 tricks. What should you bid? If you bid reflecting your hand, you'll score 60 points and lose the game 520 to 530. You could go higher; you can take 8 tricks without even needing to set the other team. That would convert your loss into a win. But it's extremely unlikely that you'll be able to make those 8 tricks. If you're playing duplicates and getting scored based on how good your result was compared to other teams playing the same hand, you should bid 6. If you're playing this as a one-off and getting scored based on whether you win or lose, you should bid 8 despite the fact that you can't make 8. This becomes a manners issue in some games where your bid is an important input into later players' bids. | | |
| ▲ | aidenn0 2 hours ago | parent [-] | | > This becomes a manners issue in some games where your bid is an important input into later players' bids. Yes, I learned bridge playing duplicate where preemptive bids[1] are totally fine, but at some rubber bridge tables you won't get invited back if you have a habit of bidding them. 1: https://en.wikipedia.org/wiki/Preempt |
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| ▲ | ironSkillet 6 hours ago | parent | prev | next [-] |
| Binary search minimizes the number of expected moves until you find the target. If you are already ahead, this is a natural thing to want to do. The reason why this doesn't work when you're behind is that your opponent can also do that and probabilistically maintain their lead. |
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| ▲ | gregdeon 4 hours ago | parent [-] | | I know that it minimizes the expected number of moves. But, the goal is to maximize the probability that you win in fewer moves than your opponent, not minimize the expected number of moves. Given that your opponent is playing some riskier strategy, it's not intuitively obvious to me that your optimal moves for those two objectives are the same. |
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| ▲ | SatvikBeri 3 hours ago | parent | prev | next [-] |
| In addition to other answers, one way to think about it is that the options are symmetric around the midpoint: a guess that partitions the space into (1/4 of options, 3/4 of options) is the same as one that does (3/4, 1/4). So (1/2, 1/2) is special in some way – it has to be either a local minimum or local maximum. And if the function is convex (or close enough), then (1/2, 1/2) is a global minimum/maximum. But (1/2, 1/2) is clearly a better choice than just guessing a specific individual. So it must be the best choice. |
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| ▲ | thaumasiotes 6 hours ago | parent | prev | next [-] |
| > Why not use my lead to have some chance of sealing my victory immediately, while still maintaining my lead if I'm wrong? There's no advantage to winning immediately. You aren't scored on time taken. So, it's better to use the better strategy. |
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| ▲ | IncreasePosts 6 hours ago | parent | prev [-] |
| If you're behind and you're doing the same strategy as your opponent, you'll never catch up. If you're behind doing the risky bet strategy, most times you will never catch up either because your risky bets don't pay off, but a few times they will pay off. |
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| ▲ | Supermancho 5 hours ago | parent | next [-] | | This is largely how all complex competitive games work. At some point there is a shared valuation of which player is ahead and the behind player must take steps that are outside of optimal play to attempt to leapfrog ahead. The GWENT card game was particularly well designed for this. ie How many extra cards to play/sacrifice for round control if you drew badly, based on meta-matchups. I have always asserted that some games (like Heroes of the Storm) suffer from not having catch up mechanics beyond player skill. This is problematic, when player skill can be quantized to an average value that has led to the losing state. This makes it much less likely to ever be a useful catchup mechanic, in comparison to some intrinsic gamble mechanics. The lack of catch up mechanics also means the games are less interesting because risks are only worth taking after the known state, not casually during as a chaotic factor that might be capitalized on. | |
| ▲ | gregdeon 4 hours ago | parent | prev | next [-] | | Sure, I think it makes intuitive sense to me that you should play riskier when you're behind. The surprising part to me is that when you're ahead, even if you know that your opponent will play "sub-optimally", that doesn't change your own optimal move. | |
| ▲ | sgerenser 2 hours ago | parent | prev [-] | | Hockey fans will recognize this strategy as “pulling the goalie.” |
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