----

Three poasters are competing for a girl's attention. She's mildly receptive to their jestermaxxing and decides to give them a few minutes of attention. She has them all face each other and tells them: "I'm going to place a hat on each of your heads so that you can only see each other's, not your own. Each hat is either red, blue, or yellow (but there may be multiple or none of a given color). You will all guess at the same time. If any of you gets your hat color right, I'll make out with each of you for like 5 seconds. You may agree on a strategy beforehand and let me know when you're ready."

What strategy can they use to ensure that at least one of them guesses their own color correctly?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845604)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845619)

edit: oh, I see.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845623)

We're assuming they all want to win the kisses despite being POASTERS.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845630)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845626)

just seeing the other two men's hat colors gives you no information about your own hat color since there are no rules governing color distribution (your hat color could still be Y, R, B regardless of the other men's colors).

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845658)

if the rule is "All or none of your hats could be Y/R/B in ANY combination regardless of the color of any other man's hat" then there can be no strategy because there is no rule. seeing the other hats provides no information as to the distribution of colors (no rule), other than the information about what the other 2 men's hat colors are, but this cannot be signaled as part of the strategy.

there is nothing we can all agree to say that guarantees one of us is right.

edit: I see it now

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845686)

edit: "other than the information about what the other 2 men's hat colors are, but this cannot be signaled as part of the strategy." – They can't "signal" that information bc they're guessing simultaneously, but they can still incorporate it into their guesses

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845689)

As for the hysterical "self-interested route" and "blood on my hands", if everyone picks red, everyone survives, and so the only people with "blood on their hands" (lol) are dumbs with their own blood on their hands.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845706)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845758)

once we show him the running tally and it is clear hitting the blue button won't save the blues from their well deserved genocide i assume he will press red.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845771)

I do think it matters if you stipulate whether kids are involved or if you limit it to people who are generally competent to consider the issue. If there’s lots of kids etc involved then you may figure that there’s enough blameless non rational actors we want to save that the ethical choice is to switch to blue and try to save them. If you’re assuming a game theory sterile abstraction where “everyone” is just rational actors then it’s absolutely red.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846369)

empirically we can see many people choose blue and cannot be dissuaded. so we know in reality many people will choose blue.

the hypo requires universal participation, so "opting out" by choosing red is bullshit. you're still participating, team red interprets it as a moral choice. but we know from observation that many people will choose blue and that humans are perfect rational actors.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846428)

(Although if you want to express it in game theoretic terms, "red" is a strictly dominant strategy for everyone, therefore you should absolutely pick "red" in all circumstances, and "all pick red" is a dominant strategy equilibrium)

Life is a big IQ test. Those that pick "blue" fail.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846645)

The funny thing is that these people are probably thinking along Kantian lines when they hit blue but 1) red is fully consistent with categorical imperative and 2) Kant’s political writings are explicit that you have to go with the world you want to see. If you’re a lib sitting and thinking “well maybe nobody should pick blue in the first instance but somebody will so I’ll save them and pick blue,” then it’s a self fulfilling prophecy.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846384)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845744)

SOLUTION:

(Before you go on - if you have a CS background and haven't figured it out, try again with the hint that it involves modular division.)

You assign each color a number 0, 1, or 2.

You reason that when the three hats' numbers are added up, the total 0-6 falls into one of these categories:

- { 0, 3, 6 }: have a remainder of 0 when divided by 3

- { 1, 4 }: have a remainder of 1 when divided by 3

- { 2, 5 }: have a remainder of 2 when divided by 3

So if you each cover one of those 3 possibilities, it's guaranteed that exactly one of you will be correct.

You assign one person to remainder 0, one to remainder 1, and one to remainder 2. Everyone adds up the other people's hats and guesses their own hat so that that the total divided by 3 has their assigned remainder.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845751)

Red = 0, Yellow = 1, Blue = 2

I am assigned remainder 0.

- If I see a red hat and a yellow hat, the total is 1, so I need to guess Blue for a total of 3 in order to get a number in my assigned category.

- If I see two red hats, the total is 0, so I need to guess Red for a total of 0 to get a number in my assigned category.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847410)

I capture you and lock you in my dungeon as my slave. I treat you well and feed you and stuff and you enjoy spending time with me BUT you'd much rather be free to go back to your life. After a couple weeks I tell you that I'm going to give you a chance to escape. There are three switches in your cell that have always been a mystery to you. They're each marked "on" and "off" and you've kept them off after some initial experimentation.

I tell you that each switch connects to a lightbulb in my room directly above the dungeon. Each lightbulb is controlled by one switch. I tell you to do whatever you want with the switches until tomorrow morning, at which point I will take you upstairs to where the lightbulbs are. If you can figure out which switch corresponds to each lightbulb, I'll set you free.

What is your strategy to escape?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845801)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845813)

look at me! i'm like Sherlock Holmes!

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845899)

sat here for 10 min trying to figure out if it could he solved in one night. this is BULLSHIT

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846392)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845785)

"A bus full of ugly people unexpectedly crashes and kills everyone on board

Everyone shows up at the Gates of Heaven where God comes to meet them all Himself.

"Gee guys, I didn't intend for that to happen, I'm really sorry. I can't just resurrect you all, but to make up for it I'll grant you all one wish before I let you in."

The first person steps up and thinks for a moment. He says "You know God, I've been ugly all my life. For once I would like to know what it is like to be beautiful. Make me beautiful." So God snaps his fingers and it is so.

Everyone else starts chattering amongst themselves at such a brilliant idea. They all start wishing for the same thing.

God steps up to the last guy in line who is laying on his side laughing so hard he is crying. After several minutes of patiently waiting, God finally says "Now what in Heaven could be so funny boy?" The man stands up, wipes a tear from his eye and says "Make them all ugly again.""

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845786)

We can break it down into:

- { 0, 3, 6 }: have a remainder of 0 when divided by 3

- { 1, 4 }: have a remainder of 1 when divided by 3

- { 2, 5 }: have a remainder of 2 when divided by 3

Since we have 3 guesses, we have each person construct his guess so that the total meets his assigned condition (remainder 0, 1, or 2). One of them will be right and the other two will be wrong.

For example:

Red = 0, Yellow = 1, Blue = 2

I am assigned remainder 0.

- If I see a red hat and a yellow hat, the total is 1, so I need to guess Blue for a total of 3 in order to get a number in my assigned category.

- If I see two red hats, the total is 0, so I need to guess Red for a total of 0 to get a number in my assigned category.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847364)

One day, the Queen tells everyone that she is observing male infidelity in the kingdom and wants to eradicate it. She says that women with cheating husbands are now legally required to shoot and kill their husbands at midnight after discovering the infidelity, starting that night.

How does each married woman know when/if to kill her husband?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49845889)

Wait it might be simpler than I was thinking. They might just count down the nights, and since they are all perfect reasoners they literally shoot at the same time on the last night.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846584)

let's use your numbers and say that there are 10 women total and 6 have unfaithful husbands. the 4 with faithful husbands will say there are 6 cheaters. the 6 with unfaithful husbands will say that there are 5 because they only can't see their own. this doesnt break the taboo because none of the wives are telling the other ones their husband is cheating. the numbers could all just be written anonymously too. the women with unfaithful husbands know the number the wrote down and can deduce their husband is the cheater.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846599)

if this was on the LSAT and both mine and chokes answers were available, id select his because it's "best", i.e. what the drafter would think is best.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846681)

Solution is generalizable to greater populations and numbers of cucks.

Edit: nm I fucked up and am retarded. On day one, cuckwife would deduce that she is cuckwife and murder her husband on day one. Oomox explains below.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846595)

in my solution with 10 women with 6 unfaithful husbands, the women who could see 5 cheaters would realize on the 6th night when the queen announces infidelity still remains that their husbands are cheaters and will shoot them after midnight. i was wondering where the perfect hearing component worked here. it's necessary for all the women to know whether murders occurred.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846636)

There are 100 prisoners in a for-profit prison in a brand new SmartCity. Their sadistic AI warden decides to play a game. The prisoners will stand in a line on a ramp, each of them able to see all of the prisoners in front of them. A Red or Blue hat will be placed on each person's head. As in the last case, they can't see their own, just those of the people in front of them. Starting at the back of the line, each person will have to guess whether he has a Red or Blue hat. If he's wrong, he's executed; if he's right, he gets an upgrade from jail to the VR data harvesting labs where at least you get to play games. Everyone can hear the guesses of everyone before him.

They can strategize about it beforehand. How can they ensure the maximum survival possible of the group?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846880)

but the last in the line cannot do this and so may croak

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846984)

So what info should they agree on beforehand for the first guy encode in his guess?

Each guess = just "red" or "blue," no numbers or additional info.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846985)

edit: The result doesn't matter *if* we assume they're perfectly rational actors, which we are. If we stipulate that they may make mistakes, it does matter.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846932)

(If there's the possibility that the prisoners make mistakes deducing from the answers before them, we do need the results to be audible to everyone.)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846977)

if the last prisioner sees 40 red hats he would announce even. if the next prisoner sees 40 red hats he knows his is blue. then each prisoner follows suit.

i guess knowing the result of the guess isn't necessary if we assume the prisoners execute this perfectly. it would be necessary to confirm the accuracy of the count if someone makes a mistake. if they execute it perfectly, 99 prisoners are sure to live. the last guy is 50/50 regardless. i assume for the purposes of the hypo he will cooperate with the strategy even though he won't benefit personally?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49846976)

this might be the best answer in a real world scenario considering human fallibility and how unlikely it would be for the last prisoner to make an accurate count.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847010)

At any point when summoned to go to The Playroom, a poaster can tell CSLG: "Every poaster has been to The Playroom." If they're correct, he'll set everyone free. If they're wrong, everyone is executed and CSLG sells the UNITS to SpacePorn to make into real playrooms. Sick stuff.

CSLG gives the poasters a chance to coordinate a plan before bringing them to their separate units, warning them that they'll never see each other again unless someone correctly tells him that they've all been to The Playroom. He tells them that the first poaster will be SUMMONED that night. What's their plan to get out alive?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847283)

im stumped. the only way i can think of to come to a certain answer is to brute force it, but that could take decades, and there's no way all 50 balding middle adged poasters all last that long.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847376)

if they can only do it once, tell the residents to toggle the switch only if this is their first time to the playroom. then you can keep track of how many first timers have gone until everyone has paid a visit

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847348)

Expand on your second idea. How would you keep track?

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847380)

for every other guy, they will turn it on if they have never flipped the switch. if it's already on, they do nothing.

whenever first guy comes around again, he turns it off. at some point he will have done this N-1 times

will take a very long time but uh guarantees everyone should get to go home within 10000k days or so

it's something like this + minus the requirement of the "first guy" (since they dont know who goes first) and instead getting everyone to agree on the special man

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847404)

unless it's something goofy like "just put ur hand on the light bulb lol"

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847459)

if you want to be certain by brute forcing with the light switch it'll take a long time. no chance everyone survives that long. and even if the did, i think everyone would choose to spent ~1 year in the unit and have a really good chance of surviving than potentially decades there and waiting for absolute certainty.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847472)

SOLUTION: The group designates a Counter. Everyone except for the Counter interacts with The Playroom as follows: If it's your first time in there and the light is off, turn the light on. Otherwise, do nothing.

The Counter interacts with The Playroom as follows: If the light is on, turn it off and add a tally to the POASTERS count. When you've recorded 49 fellow poasters, the next time you're summoned, you can tell CSLG you're ready to bounce. And yeah it will probably take a while.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847446)

if the counter goes first he turns it off. he's the only one that can switch the switch off. everyone else only switches it on if thats the first time they've been able to do so.

counter just has to count to 49 times he turned the switch off.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847468)

best bet is wait a year and yolo that everyone has gone, which is near certain by then.

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49847466)

SpacePorn says, "Tomorrow I will give you a chance to leave. I'll put a Black or White collar on each of you so that you can see each other's but not your own. You will all guess your collar colors at the same time; anyone who gets it right will be set free. You may prepare a strategy."

As soon as SP leaves, the children whisper excitedly. They already have a plan for how to escape as a group as long as they have 10 or more kids on the outside. So this could be their chance!

But can they GUARANTEE that 10 or more kids get their colors right? What's the most children they can ensure answer correctly? How do they achieve that result? How would they do the same for other numbers of children? (He might capture 1 or 2 new ones tonight.)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49848383)

If you have a total space of exactly 20 bits, the breakdown will be:

5 Parity Bits (for error checking)

15 Information Bits (your actual 1/0 data)

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49848799)

If Spaceporn gags them to prevent talking he still didn't restrict their movement. When Spaceporn steps back and says "guess" Kid 1 simply walks around the room and physically pushes everyone with a Black collar to the left wall, and everyone with a White collar to the right wall. Since Kid 1 can see everyone else's collars this is easy. Then Kid 2 grabs Kid 1 and guides them to their respective wall. Now everyone knows which group they belong to, and they all guess correctly at the exact same time

(http://www.autoadmit.com/thread.php?thread_id=5860893&forum_id=2Vannesa#49848947)