I mathematically proved the best "Guess Who?" strategy [video]
youtube.com34 points by surprisetalk 6 days ago
34 points by surprisetalk 6 days ago
The idea is that if you're winning you can just do a binary search, but if you're losing, it's better to take some risks by making narrower guesses.
For example, let's say it's the last turn and your opponent is about to win. Say you may have 2 options but your opponent has 4 options. Instead of whittling it down to 2 options, it's better to guess one of the four. How outrageous should your guesses be is the content of the result and paper.
Paper is on archive (and linked from the video):
I find it somewhat surprising that the optimal play when you're ahead is still just binary search. Is there an intuitive reason why it's not productive to make riskier guesses? Why not use my lead to have some chance of sealing my victory immediately, while still maintaining my lead if I'm wrong?
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.
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.
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.
> 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.
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.