7 ms·
Ava always guesses the same color as Bobs forehead. Bob always guesses the opposite color as Ava's forehead.
by IIAOPSW 10y ago
Ava always guesses the same color as Bobs forehead. Bob always guesses the opposite color as Ava's forehead.
- sopooneo 10y agoI haven't reasoned through your solution fully yet, but is that a guarantee that at least one of them is correct? How? Sorry if your answer turns out to be obviously correct.
- mertd 10y agoBob is correct when they both have the same card, Ava is correct when they have opposite cards. That covers all 4 color combos.
- deleted 10y ago[deleted]
- atdt 10y agoThere are four possible scenarios: 1. Ava's card is red, and Bruce's is black. 2. Ava's card is red, and Bruce's is red. 3. Ava's card is black, and Bruce's is red. 4. Ava's card is black, and Bruce's is black. To win, Ava should pick the same color as the card she sees on Bruce's forehead, and Bruce should pick the color that is different from the card he sees on Ava's forehead. (Which player chooses which strategy doesn't actually matter.) Here is how the strategy plays out for each of the four scenarios above: 1. Ava chooses black, and Bruce chooses black. Bruce guessed correctly. 2. Ava chooses red, and Bruce chooses black. Ava guessed correctly. 3. Ava chooses red, and Bruce chooses red. Bruce guessed correctly. 4. Ava chooses black, and Bruce chooses red. Ava guessed correctly.
- sopooneo 10y agoThank you. There it is listed out. I'm unsettled that a strategy involving using information that must have 0 correlation to each person's own correct answer could yield a collectively correct answer. I have heard problems similar to this before where I have been able to say "oohhhhhh" and find out why the answer does not seem to violate other overarching truths, but I'm not there with this one yet. This is a like a person spelling out, step by step, how to build a perpetual motion machine, with no clear missteps, and the machine works, and I'm trying to understand how it jives with the second law of thermodynamics.
- BoppreH 10y agoAva is betting they have the same color, while Bruce is betting they have different colors. One of them is bound to be correct.
- shmageggy 10y agoAgreed that the independence of each card makes this seem paradoxical, in a kind of Monty Hall way. I'm reconciling this by noticing that neither person can guarantee that they will themselves be correct, individually. This satifies the independence problem, since they really do have no way of knowing their own card. Then how do they arrive at a correctively correct answer? I'm reconciling this by noticing that there is no possibility of them both being correct. That is, their strategy effectively divides up the answer space such that they are covering all two possible correct answers.
- im4w1l 10y agoNo matter the strategy Ava will get the right answer 50% of the time, and Bob will get the right answer 50% of the time. What the strategy can influence is the correlation between Ava being right and Bob being right.
- im4w1l 10y agoFor the four people puzzle. Assign every suit a number 0-3. All four people guess differently what the sum of all the cards is modulo 4, meaning exactly one person will get it right. Using the sum and the view of the other cards the one who guessed right can work out their own card.
- ulrikrasmussen 10y agoI don't think the players can see the other players' answers. The puzzle reads like an instance of the 100 hats puzzle, but instead of maximizing the number of correct answers, you just have to guarantee at least one. Edit: I should've read your answer more carefully.
- josefdlange 10y agoGist of this, since formatting got screwy: https://gist.github.com/josefdlange/04c9eb14a9ddbdfae8eb2e34b91d56df https://gist.github.com/josefdlange/04c9eb14a9ddbdfae8eb2e34... Seeing the others' answers I don't think matters in this solution. The strategy -- given that it's allowable to discuss strategy before drawing cards -- is each person has a different modulo result (x) they have to compute the number they must add [to the sum of the other three], (m) in the following algebra. Solve for m to get the suit your should guess. x = ([sum of others] + m) % 4 Suits: 0: Diamond 1: Heart 2: Club 3: Spade Example: [PLAYER]: (SUIT THEY HAVE), (CHOSEN MODULO RESULT) A: (0) (0) B: (1) (1) C: (3) (2) D: (3) (3) For A's m: ((1 + 3 + 3) + m) % 4 == 0, therefore m = 1 For B's m: ((0 + 3 + 3) + m) % 4 == 1, therefore m = 3 For C's m: ((0 + 1 + 3) + m) % 4 == 2, therefore m = 2 For D's m: ((0 + 1 + 3) + m) % 4 == 3, therefore m = 3 D guesses correct. There are many permutations of the drawn cards, but I think this one does solve it. All props to the person up this thread who initially answered the 4-player problem; I'm just explaining their methodology as I understood it.
- ulrikrasmussen 10y agoThanks! I misread the answer and took it for the solution to a different but related puzzle
- 10y ago
- deleted 10y ago[deleted]
- intended 10y agoA variant of the puzzle of identical twins? One brother always lies and the other always tells the truth.