@js kelly Spoiler Wait, the two-square one - if the host says "1" and you've agreed with your friend that HH or TT is the signal for "1", but the host has already laid the coins down as HH or TT, what do you do? Given that Exohedron said that you have to flip one coin. ...Maybe you could do something like "If the host says '1' I will make sure the coin on the left is heads, if he says '2' I will make sure the coin on the right is heads. So the host says "1" and the coins are laid out like: HH - flip the second H to make it HT, coin on the left is still H HT - flip the T to make it HH, coin on the left is still H TH - flip the T to make it HH, coin on the left is now H TT - flip the first T to make it HT, coin on the left is now H Does that work? Edit: Wait NVM it doesn't work because the coin on the right is heads as well in half of those. Duh. Edit the second: ...But you could say "If the host says '1' then the coin on the left will be H, if the host says '2' the coin on the left will be T. Host says "1" and the coins are laid out like: HH - flip the second H to make it HT, coin on the left is still H HT - flip the T to make it HH, coin on the left is still H TH - flip the T to make it HH, coin on the left is now H TT - flip the first T to make it HT, coin on the left is now H So it has the same coin-flipping solutions as the first idea I had, just a different ... key? IDK. I'm gonna stop now, this is making my head hurt.
Spoiler You are right. How about this? If it is "one," then the coin in the first position will be H and if it is "two" then the coin in the first position will be T.
I think I've gotten sufficient entertainment out of watching people on this particular puzzle, so for those who really need a solution, here it is: Spoiler: Solution for 64 coins The Phoenixian was closest. Here's a solution: Label the squares on the board 0 through 63, and then translate into binary. So the top left square becomes 000000, then the next is 000001, and so on, down to the last square which is 111111. To translate from a bunch of coins to a number, you take the bitwise XOR of the labels of the squares that have heads on them. So if the heads are on the squares 5, 17, 38 and 56, we get 000101 010001 100110 111000 --------- 001010 = 10 The thing about XOR is that a XOR a XOR b = b, so "adding" and "subtracting" a number via XOR amount to the same thing. So the trick is that if the host gives you a number a, and the board XORs to b, then you translate both into binary, take their XOR to get the difference, and then flip over the corresponding coin. So if the host had said "41", we get that 101001 001010 --------- 100011 = 35 So you flip over coin 35. The trick here is that, via this encoding, there is an exact match between the amount of noise given by the original heads-tails configuration of the board and the amount of duplication in the translation of heads-tails configurations to numbers, as The Phoenixian was trying to set up, but unlike that setup, here flipping a coin goes in both directions, rather than just adding if flipped to heads and subtracting if flipped to tails. Of note is the fact that the coin on the 000000 square tells you absolutely nothing, but needs to be there in case the original heads-tails configuration matched the number that the host gave, because player 1 needs to flip over something. Spoiler The solution I originally came up with for this involved trying to 64-color a 63-dimensional hypercube, or at least asserting that such a coloring existed. Which it does, but the quickest way to see that is via this binary XOR method. The guy who told me this puzzle was kind of stunned that I came up with something so complicated.
Stacks of hats Again you're in a gameshow playing cooperatively with a friend. Again you can figure out strategy beforehand but can't communicate once the game starts. Here's the game: Each of you gets a stack of three (3) hats on your head which you can't see, but you can see the other player's hats. Each hat is either black or white with equal probability. When the host blows a whistle, each of you must point to a hat on your own head. You both win if you both end up pointing to a white hat, and lose if either or both of you point to a black hat. Instinct would tell you that since any hat you point to has a 50-50 chance of being white or black, and any hat your friend points to also has a 50-50 chance of being white or black, you have at best 25% chance of winning. Can you actually do better than 25%? Well, this problem would be really dull if you couldn't. So the question is: How can you do better than random chance? What's the best you can do? Again, you can't see any of the hats on your own head, only on your friend's.
Note: There are obvious extensions of this, where instead of three hats you have four hats, or five hats, and so on. One can even consider (countably) infinitely tall stacks of hats. Just, hats all the way up. And again the question is what is the best strategy to win most of the time? Two of my friends are actually writing a paper about the infinite hat case. As far as anyone knows, the infinite-hat case is still open, and the best-known-strategies seem to give a really bizarre number for the probability of winning.
Okay, so you flip over coin 35. And then, the other player just xors together all the heads, and gets your number?
Hmm. I'm not immediately seeing what options you have in the hat game. Hmm. Okay, I can see their hats. But I can't, presumably, see them pointing, because that would be "communicating". So I can't tell them anything about their hats by my pointing choice... But then I don't see a way to do anything. Also, consider the case where the middle hat is white, and the other two are black. Is it actually possible to point at the white hat without pointing also at a black hat? Or are these Abstract Hats, which do not actually reflect in any way the usual physical properties of stacked hats, such as overlapping?
Abstract hats, where you can point to any specific hat without pointing to anything else. And indeed, you can't see where they point.
Huh. So casual searching turned up a related puzzle: "Three people are given hats. Each hat is either red or blue, chosen at random. Each person can see the other 2 hats, but not their own. They each must simultaneously either guess their own hat's color, or pass. No communication is allowed, although they can agree on a strategy ahead of time. What strategy will give them the best chances of at least one person guessing right, and nobody guessing wrong?"
Hmm. I am not able to see a way to affect chances. The probabilities of the hats being various colors are independent, so there's no reason to think that seeing any other player's hat allows me to improve my choices in any way. You don't have a "pass" option, so you can't, say, "not point if your friend has all black hats" or something that would let you at least avoid losses. So I'm missing something, I assume.
Yeah, probability problems tend to be really unintuitive at first. But there is a way to do better than 25%. Indeed, that's the first major hurdle, to convince yourself that you can do better than 25%.
If you are going by probability, seeing the other hats tells you nothing at all unless you know there is a limited number of each color. If not, then the colors they got have no effect on the probability of your own. For example - you have a perfect coin (no physical defects that would affect which side pops up on a perfect flip). You see it land on heads 7 times, tails 3 times. What is the probability of it being heads when flipped again? .5 - the same as if it had not been flipped at all.
It is entirely true that there is nothing either of you can do to increase the probability that you, individually, point to a white hat, nor is there anything either of you can do to increase the probability that your friend, individually, points to a white hat.
Agree to point to the hat that corresponds to a white hat on your friend's head? \ Edit: won't help - there is no correlation in the rules between the hat colors on each person.
There really isn't a way to improve your odds. You cannot pass information nor do you recieve information.
There is either missing information in that puzzle, or the rules were incorrectly stated. Taken literally, there is no strategy that can change the result.
I'm not seeing any way for any strategy to affect things. In principle, a strategy has to give you Better Information. But I don't see a way to derive information.
