In high school, I made it to the final interview for a math competition, where I was asked only this one question! I didn't solve it until after my interview ended, but it's such an amazing problem.
For a 4 person version of this problem, it's tempting to do the same thing: only guess if all the other hats match, and guess the opposite color. And this does ensure that you only ever give an incorrect guess in 2 of the 16 possible distributions....however it also means that any situation where there are 2 red and 2 blue hats nobody will guess. This happens 6/16 of the time, presumably if no guesses are made it's a loss, which means that this strategy only wins 50% of the time. We can add an additional condition to this: If you see 3 hats that are the same color, guess yours is the opposite, OR if you see 2 red hats, guess yours is blue. (you could swap this to guess red if you see 2 blue hats, but only pick one, not both). This ensures that in the 2blue/2red situations that the 2 people wearing blue hats will correctly guess their color (and the 2 in red hats will not guess), letting us win those 6/16 situations. However, this also results in us loosing the 4 situations 1blue/3red that we originally won to be lost now. This improves our total odds to 10/16. I had assumed this was the optimal solution. In terms of counting all right/wrong guesses and trying to 'bunch up' the wrong guesses, this solution has 20 correct and 20 wrong guesses, and all of the wrong guesses are bunched up into groups of 3 or 4, and I didn't think there was a way to group up the wrong guesses anymore than that. But then I had a strange thought that ended up working, and here is my further improved solution: 1) if you see 3 matching hats: guess red 2) if you see exactly 2 red hats: guess blue This strategy wins 11/16 times. But it's weird isn't it? Why guess red when you see 3 red hats? This seems like it'll be wrong more often than it's right: because it is. However, the situations where it's wrong (where you are wearing the only blue hat), we're already going to loose because the 3 red hat people see 2 red hats and guess blue. So it doesn't matter that guessing red is wrong here. The only time it makes a difference is in the 4 red hat situation: now instead of everyone guessing wrong here, now everyone guesses right. Overall this method still results in 20 right and 20 wrong guesses, but now the wrong guesses are all grouped into sets of 4. I'm tentative to claim this to be optimal. However, any improvement over 11/16 would mean matching the odds of the 3 person problem, and it seems very unlikely that there wouldn't be at least some loss in probability as you increase the number of people. (then again, with only 2 people I don't see any way of beating a 50% chance, so the odds increase when going from 2->3, so maybe I'm wrong and the optimal strategy just keeps getting better with more people).
@riddlecolo8198 thats exactly why i edit my videos a ton. I can barely get through 1 sentence without sounding like an idiot, let alone an entire video. His speaking ability is impressive
I figured out the strategy of how to guess, but explanation of why it's optimal was both surprising and insightful for me. Never saw it from that perspective.
Same here, based on another problem involving hats and a wall I started of with the idea of one of them seeing matching would be the key and adding the Monty Hall gave me the solution. But I would not have been able to really explain exactly why, and I would not have been absolutely certain it really was THE best solution, just that it would be better than 50 %.
This is very similar to the Monty hall/deal or no deal box swap question You feel like it shouldn’t work but the probability calculation tells you otherwise
The prisoners are free to make a guess or not make a guess (how this is kept secret from the other two prisoners is not explained but is just a given part of the scenario). At least one of them must guess correctly. If any of them guess incorrectly, they lose. So if one guesses correctly and the others don’t guess, they win (likewise if two or three guess correctly). But any incorrect guess by any one person is an automatic loss for all.
Well presented. Good content. As for a different strategy for 3 hats. There is an approach we could get 100% certainty of the right answer, provided we define communication as speaking, writing or gesticulating. If they each move 1 step forward if they only see blue hats and 1 step back if they only see red hats, then they should all be able to guess correctly. (Note: if the definition of "no communication" limits these visual clues then I would argue that seeing the hat colours of the others does also constitute communication as it imparts information and therefor the problem rules would contradict itself)
One day I encountered a similar problem in which the prisoners had the right to be wrong in their answers, but at least one of them must be correct. The other small difference is that there were as many hat colors as there were prisoners. And in this case, there is a strategy that works for sure regardless of the number of hats. I propose to illustrate it in the case of two prisoners. We then have prisoners A and B, the idea is for prisoner A to state the color of prisoner B's hat while prisoner B will state the opposite color of A's hat. In this case, either the hats of A and B are identical and A is then correct, or the two colors are different and it is then B who guesses correctly. I'll let you ponder the generalization for more prisoners (it's not easy).
I think you can just assign each hat a number from 1 to N, then add them all up modulo N as a checksum, to get a number from 0 to N-1. Each person is missing the color of one hat, but they can each be assigned a particular checksum from 0 to N-1 before the game starts, and state the color of the hat that would make that checksum correct. N prisoners and N possible checksums so someone must be correct.
Hmm... I figured out your optimal solution in terms of probability pretty much intuitively before you explained it. That's surprising to me, as I did enter some math competitions in high school, but I never really knew what I was doing and never did particularly well.
i’ll try 4 hats with some similar strategies: 1. pick opposite to the majority, which means everyone will guess. This strategy only works when there are equal number of red and blue hats, which is only (4 choose 2) or 6 cases out of 16 so it’s worse than letting one person guess with 50/50. 2. only guess opposite if you see everyone else’s hat to be the same color. This strategy fails outright in 6/16 cases when there are equal numbers of both colors as stated in the previous strategy. Of the remaining, it works in 8 cases where one hat has a different color to the three others and fails in the 2 cases where all four hats are the same. Meaning this strategy has a success rate of 8/16 or the same as letting only one person guess. There might be more clever strategies that takes advantage of ordering the prisoners such that their behavior is not identical, but I have not considered them.
i didn’t consider just completely ignoring one of the players (basically they sit out the game and doesn’t guess or count towards what the others see). in this case the game simplifies to the 3 player case with winning probability 3/4 which is better than the two strategies I mentioned.
There is a general solution of this problem for a team with N members. In case N = 2^k -1 (like 3, 7, 15, ...) one can apply Hamming codes to obtain (2^k -1) / 2^k probability of win. In other cases we can just ignore some players (as you already wrote). So chances are increasing with growing N ;-)
By writing out all the RB combinations and thinking "what do I see" in each case, it quickly became apparent that 6 cases are covered when seeing 2 different hats. But I've seen this puzzle decades ago, maybe a remnant of my brain remembered this. Btw, if you are allowed to know if someone else has guessed, you can win a higher percentage of the time.
hloo sir , i saw ur vids remarkable i understood everysecond of ur videos, thanks to ur plane rigorous explaination, i have request ,can u please make video of probability and its mind map, also consider involving how its links with combinatorics
Depending on the interpretation of the question, I believe you could actually get a 7/8 chance instead of a 3/4 chance. The strategy: If anyone sees 2 red hats in the first 10 seconds, they guess they have blue. If anyone sees 1 red hat and 1 blue hat in the 11th-20th seconds and nobody has already guessed, they guess they have blue (if they had red then someone else would see 2 reds and would've already guessed so this is guaranteed) If nobody has guessed after 20 seconds, anyone guesses blue (if anyone had a red hat someone would've guessed at one of the previous steps already) This strategy only loses when all 3 have a red hat, instead of losing when they either all have red or all have blue. There's nothing in the question that says they all have to guess "at the same time" so to speak, so they can change their decisions based on whether someone else has already guessed or not.
@@swenji9113 you refuted my own solution with that explanation: if all three can see each other but are otherwise completely prevented from communicating after the game has begun, this won't work. I would quibble that the agreement could be this: "If I see two hats of the same color, I frown, otherwise I don't. If you see two frowns and two blue hats, you say blue. If you see one frown and one unfrowning face, your hat is the color of the one who is not frowning."
I agree this could work, depending on the interpretation of the game rules. If this was a real life game show then whilst they would be unaware of each other's guesses taking place they would presumably be aware if the game had ended, even if they didn't know if it was due to a failure (wrong guess) or success (correct guess) or who had made that guess. If the host of the game show had confirmed the round was over then they would then know that somebody had made some sort of guess, even though there was never any direct communication of any kind between any of the three contestants, so no rule break. This means that if they have agreed this strategy beforehand when they were allowed to discuss a strategy, and assuming they each have their own digital watch to accurately measure the time elapsed so that can each independently know when the 10th and 20th seconds happen (so no rule break to communicate that to each other either) then it does work. The problem really is that we don't know how soon after the end of the game they will be told if they won or loss (or at least that the game is over). If a win or loss (or even just 'game over') is confirmed instantly (which is not the contestants communicating with each other, and not a rule break) then yes this strategy would work. I feel like if the original problem setter originally intended for the solution in the video to be optimal then they probably also originally had an additional line that was something like "the prisoners would only find out the next day if they won or not, as they'll either be released, or not, on that next day".
I like that this gives a detailed but easy to follow derivation of an answer... right up to the point there the unsupported assertion is made that half the guesses *have* to be incorrect. (That might be true, but it's not justified here.)
It is justified here though by the fact that regardless of what color the other prisoners' hats are, the color of your hat was determined by a fair (independent) coin flip. Your guess ALWAYS 50/50. No logic or strategy can increase the odds of your guess being correct. Regardless of what you see when you look at the other prisoners' hats, there will be exactly 2 situations that match that: 1 where you have a red hat, and 1 where you have a blue hat. There's no way for you to differentiate between these situations so you're going to do the same thing in both of them, so you're either going to get 1 right and the other wrong, or you won't guess in either of them. Regardless the number or right and wrong guesses is equal.
Maybe I missed it, but the parameters are not completely clear to me. If the players can all first announce whether they want to make a guess, then you could push it to 100%. If everyone wants to make a guess, the colors are all the same.
The rules say that only 1 correct guess is needed to win, but also says that any incorrect guess is a loss for everyone. The only way to have 1 correct guess and no incorrect guesses is if the others can choose not to guess.
Another (non mathematical) strategy is to preassign each corner of the room, for each player to go to independently. If you see two red hats go to the left side of the back wall. Two blue, go to the left side of the near wall. For cases of one of each colour, you also find out in advance the other players names. If the player with the first name alphabetically has a red (and the last name alphabetically has a blue) then go right back and vice versa. Then everyone can see what colour they have from where their colleagues are standing.
You can scan over all possible strategies in the 3-person game pretty easily with a computer (there are ~500,000 strategies). Turns out there are 3 more strategies that reach the optimal 75% efficiency, but they are asymmetric: Prisoner A only guesses if they see two matching hats on the other prisoners, and their guess is the same as the colour they see. Prisoners B and C only guess if they see two different colours, and they guess the colour prisoner A is wearing. The prisoners decide who is A beforehand, and this choice gives the 3 extra strategies.
this is the same as the original strategy in a way - you can take any existing strategy, and before applying it, flip the color of A's hat, apply the strategy, and then flip A's hat color back to what it was. That way you can get this strategy from the regular, symmetric strategy.
Just by the silence of the other two seeing two different hats you know your hat is different. You get an answer without a guess being made the odds are 100% if everyone understands this before the hats are placed
Any strategy that guesses 3 red hats correctly must include a "see 2 red, guess red" rule which fails in the 3 cases of 1 blue and 2 red hats. The same is true for the 3 blue hats. Thus you get an upper limit of guessing 6/8 cases correct, namely the ones with one color different than the other two. Finding the correct strategy is then much easier when you only think about solving these cases.
the problem is this requires they coordinate this strategy but the rules state they can't communicate. It qualifies it with "during the game" but that needs to be better emphasized - i.e. by illustrating they can communicate all they want before the game - because it's not part of the puzzle to catch that.
If the number of prisoners increases to 4, the possibilities are 2×"all same colour", 8×"3 of one, and one different, and 6×"two each". It's handy to think of the possibilities laid out across a row of pascal's triangle: 1-3-3-1 for the puzzle in the video, and for 4 I read these numbers off from 1-4-6-4-1 Applying the same rule as for 3 "If I only see one colour, guess the opposite." Gives us 1/2 success, 1/8 failure, and 3/8 still waiting for anyone to make a guess. The question then is "when can I make further deductions on the basis that I'm still here?" For 4 prisoners we might designate the most patient prisoner to make a guess in any case, but only when thoroughly bored of waiting. They see 2:1, they think no one is ever going to guess at all because the true ratio is 2:2, so any player with this information can guess correctly the hat colour of which they only see 1. But it only works if the gaolers aren't unduly delaying the release of the prisoners following a true guess. It's leaving the bounds of the set problem, but not unreasonably. We were not told what would happen if no one made a choice ever, so we must presume. This can be adapted to 5 prisoners: moving down the 1-5-10 of the pascal's triangle row we get the most likely scenario of two prisoners seeing 3:1, and when the 1 hasn't done anything about it after a significant period, clearly they also see 3:1, so either of those prisoners can guess that their hat matches the 1 prisoner's hat. For 6 and more it's more involved, as there are a greater number of possibilities to differentiate from through the passage of time, so precisely how much time has elapsed will be important. So let us find some logical clock the prisoners can use- be it hours elapsed (if they can see a clock), days passed (if they can see the sun), or meals served (provided they are served regularly) while waiting for these strangely logical prisoners to make a guess, then we can win. The initial rule applied in time period 1 is the same- either we're all wearing the same colour and guessing wrong, or one person is different and guessing right. If 2 people are different, then in time period 2, they, seeing one prisoner different, guess that they match the one different. In time period 3, anyone seeing 2 different to the majority guesses they match the minority. So for N prisoners, in some pre-arranged time period t, if you see t-1 people wearing different hats to the majority, make a guess that you match those people. We're only stuck in the vanishingly small (as N grows) 1/2^(N-1) chance that we all got the same colour hat, or our gaolers are serving different prisoners meals at differing intervals in our timeless dungeon cells. In which case the whole thing was just a cruel trick to give us hope.
i must have missed something in the question, because if the 3 prisoners can divise a strategy beforehand, they just pick one person to guess first, if they see 2 different coloured hats they choose blue, if they see two hats the same they choose red. based on that first prisoners choice, the remaining 2 prisoners know what hat they have.
I was thinking they could just pick 1 colour from the outset... Or pick "the colour that comes first in alphabetic order" With only 2 colours, if they all choose the same colour they should get similar results. (Presuming they were able to communicate a plan before starting...)
Before seeing the answer, with 2 colors and 3 heads, there's 8 combinations. 2 of them will be all blue or all red, leaving 6 more that will all be either 2 blue 1 red, or 2 red 1 blue. So a prisoner should only vote if they see 2 of the same color hats on the others heads and vote for the opposite color than what that they see. That would get 6 of the 8 correct. Guess I'll find out if there's a better solution.
It wasn't obvious to me when you first explained it that the prisoners didn't have to decide whether or not to guess until the hats were given out. I paused the video and spent 10 minutes thinking about it before playing the next 15 seconds and realizing i misunderstood the question. I would suggest stressing that bit more.
I miss understood the question and come up with a strategy where at least one of them will always get it right: Basicly prisoner A aleays guess the color of prisioner B hat, B guess C and C guess A. You can see it work because if it didnt you would get hat A color ≠ hat B color ≠ hat C color ≠ hat A color. Which means you have 3 different colors one for each hat, which is not true there are just 2
I have made an arithmetic operation that i want to share with you. It's called append operator, As name suggests it is used to append a number into the right end of another number. I have designed a symbol but it doesn't has its unicode so I will denote it as (+P) Eg: 17 (+P) 0 = 170 168 (+P) 78 = 16978 It has some properties, 1. (Add what number to get smallest next digit number) like 17(+P)0 = 170 so the addative digit is 170 - 17 = 153
Nice puzzle but IMO almost opposite of Monty Hall where one is revealed (here two). In any case, strategy has only two scenarios and it is, purely by probability, obvious in both scenarios to bet with any possible advantage, And, for doing better than that (choosing opposite in case of two same) is clearly no room.
My take for 5 prisoners. The strategy is as follows: Whenever you see at least 3 hats of the same color on the other prisoners, guess the opposite color. This works in 20 of the 32 possibilities and failes in 12. Specifically, if all hats have the same color (2 possibilities), everyone would guess wrong. If there are 4 hats of one color, and 1 of the other, four prisoners would guess wrong, and only the person with the odd color would guess right. If there are 2 hats of one color, and 3 of the other (20 possibilities), two prisoners would guess right, and the others would remain silent. So all in all its a 20/32 chance, which is better than 1/2. This strategy also guarantees, that each time, at least someone is guessing.
I don’t get the rules what do you mean ‘all players win if AT LEAST ONE guesses correctly’ but lose for ANY INCORRECT guess (emphasis added). Doesn’t that just mean that the FIRST and ONLY guess has to be correct, otherwise it’s lost? Or is the aim of the game really to get three correct guesses (despite already ‘winning’ on the first correct guess)… I’ve paused the video as want to know the rules before having a go, so hope you reply… Thanks
The real question here is what constitutes communication during the guessing period. Since they are allowed to strategize ahead of time, they can set up some protocols that do not include direct communication. For example. All three inmates put their duck bill to the side. The second guy enters the room and sees the first guy there. If the first guy is blue, the second guy puts turns his duck bill forward. If the first guy is red, the second guy turns his duck bill to the back. When the first guy sees the second guy he does the same for the second guy. The third guy comes in, the first and second guy switches their hat accordingly. The third guy does the same for the second guy, which confirms that the first guy isn't wrong/or lying about the second guy. Now all three should know exactly what color they are. The reason you want to start at the side, is because if someone forgets to do it once they are in the room, that would be a clue that you can't trust them. If they come in and its not to the side, you can't trust them. If they come and its to the side, but they never switch it, you can't trust them. Its a check to make sure that everyone is playing ball..
If we allow non-verbal communication, it makes the game trivial, there can be billions of ways to transmit this information. We have to assume that participants can see only the color of the others' hats and nothing else.
Considering this was a "math" competition, I'd suspect you wouldn't do too well by getting into a debate about what constitutes communication. Your proposals I think anyone would consider clear examples of non-verbal communication.
if there are 10 persons , guess immediately if you see all hats of the same color. If you see one of a different color, wait 10 secs, then guess. If you see 2 of the other color, wait 20 secs, then guess, and so on. if there are 5 hats of each color, everyone will see 5 hats of one color, and 4 of the other and all answer correctly after 40 seconds. They only lose if all hats are the same color. (Only works if the prisoners can choose when they guess, but the question is unclear about that.)
After contemplating about this video, I was considering showing the strategy you describe in my blog. But obviously, someone else had the same idea. Nice! There is another well known problem: Assume the rules are that each prisoner can go back to prison, or guess. If he guesses and is right, he goes free, else he dies. Can you make it that only one of the prisoners has to remain in prison?
@@ReneGrothmann you mean they have to guess one at a time and each one hear the answer of the previous ones ? or can they also delay their answer like in my solution for the first problem ?
@@ReneGrothmann If they can speak freely in any order, one solution could be to choose a prisoner as volunteer , if he sees an even number of blue hats, he goes back to prison, otherwise , he doesn't. From there, the other can look at his decision and the color of his hat to deduce the color of theirs, all the other will answer correctly. He still doesn't know the color of his own hat at the end though, so he'll still choose to go to prison after everyone else left. (wether this is allowed also depends on how exacly they can answer.) if the order of the answers is completely free, the other prisoners can even answer in an order that depends on the color of the hat of the first guy. so he also knows this information, and if he hasn't returned to prison yet (because he saw an odd number of blue hats) then he can still guess his own color and be free too. (in this case , 50% chance to have 1 guy in prison, and 50% chance that everyone is free)
@@isaz2425 The idea with even or odd works. As in the video, one of the prisoners who sees an even number of blue hats can go to prison. All others will know. Note: In case of an odd number of prisoners, there is always one that sees an even number of blue hats. For an even number of prisoners, you need to adjust the procedure to "odd". Now for hardcore mathematicians: Make it work for a countable number of prisoners, using the Axiom of Choice.
If the prisoners agree that if they see two hats the same colour they should stand to the left of the three prisoners starting position, then if all do this then they have the same colour hat and all can guess correctly. If one prisoner moves left, then the other two can guess correctly as they have the same colour hat.
It sounds like a paradox. Most will think of the probability of wearing a blue hat under the condition that the other two are red, which is indeed 1/2. But that does not apply here. Here, we have the probability that any one player is wearing a specific hat color under the condition that the other two are wearing a different one. I find your explanation much better, exploiting all equally likely outcomes of hats.
If there's some way for the prisoners to know who is going to volunteer to guess they can win every time. If they see two hats that are different colors they do not guess. If more than one volunteers to guess they guess their hats are all the same.
What about a rule where every prisoner picks the color their neighbor to the left has? Doesn’t that guarantee at least one correct answer in all cases?
The phrase, "no communication of any kind," can be circumvented by nodding. They have to be able to see one another to determine the color of the hat of their fellow prisoners so each prisoner's head movements could be easily seen by the others . There are only four possibilities. Tilt your head down if you see 2 blue hats. Tilt your head right if the prisoner on your right has a blue hat and the prisoner on your left has a red hat. Tilt your head left f the prisoner on your left has a blue hat and the prisoner on your right has a red. Don't move your head if you see two red hats. Each player can determine exactly what their own hat is by process of elimination. It bends the rules a little but these are convicted crooks so by nature they'll be looking to skirt the rules.
If the prisoners are allowed to coordinate then I feel the solution should be - 'If you see two hats of the same color, wear your hat backwards. If you see different colors wear your hat forward.' That would immediately tell an observer their own hat color.
Still kind of confuses me how the gambler's fallacy doesn't apply here. If I see two blue hats, it's still a 50/50 chance that my had is red, and it would be the gambler's fallacy to suggest that since I see two blues, it's more likely that mine is red than blue.
Because you're dealing with some selection here which is not in a vacuum. In advance, you decide the only person who will guess will be someone who sees two of the same color hat. By making this decision in advance, you now succeed in 6 of the 8 possible coin flip scenarios. It's really related to the Monty Hall problem in that sense.
So 1/4 of the time, we get three wrong guesses, all at once. The other 3/4 of the time, we get exactly one right guess. That's pretty cool. It means that on average there will be just as many wrong guesses as right guesses, but the wrong guesses are all concentrated together into the same scenario (EDIT: oops. I wrote this half way through the video and thought I was adding something clever. Then the video said basically the same thing)
My initial gut guess is only make a guess if you see two identical hats, and guess the opposite. Since there are less combinations of all same color than 2 and 1, you should escape 75% of the time. Unpausing. Edit: Yep, answer was as i thought, pretty straightforward. Fun thought experiment though.
Huh? I thought of the answer in a few seconds, then spent the next 10 minutes trying to come up with a better answer cause my initial one felt like the "obvious" one. Just the simple logic that the chances of all 3 hats being the same colour is lower than there being a mix of colours.
The combinatorics sound unfavorable with more than three. If you guess red only when you see at least X blue hats, then you lose whenever there are X+1 blue hats, and you fail to make a guess when there are X-1. I don't know if having a known prisoner ordering helps, but it's reasonable to assume they could know them by sight and choose an order. So if you are at the beginning of the longest color chain, then you could guess the opposite color, for example. Generally, a chain of about k will be the longest for a group of 2^k, with the inaccuracy I in the actual number (1+I)k shrinking with high probability for k large. I guess if there are two chains tied for the longest chain, you would win, with the prisoner priors prior to those chains guessing, and the prisoners at the start of the chains seeing any other ties for the longest as a reason not to guess. But if there is just one longest chain, then the prisoner at the start and the prisoner before that both guess, and the prisoner at the start is wrong. (The chance of one big chain is 2 out of 2^n and negligible, unlike the cass with three prisoners)
I think the chance that the longest happens twice is probably reasonably large. It's not quite rigorous to treat different chains as independent, but it should be close enough. Suppose you have two or more chains that are currently L long. You win if they all die on the next flip, and you keep playing if at least two survive.
The question is a little bit unclear on whether you're forced to all guess simultaneously - if you have to guess simultaneously, then you could potentially still get a 3/4 chance just by ignoring one of the players (just designate one player to never guess at all and have everyone ignore the 4th player's hat colour). It's probably still possible to do a bit better than this with a very convoluted strategy (because you can strategize beforehand, you can have each person using a different set of instructions which I think with the right setup can probably marginally improve your odds, but I'd need to spend a lot more time thinking about it). If guesses don't have to be simultaneous, then you can do a lot to improve it I think. If you imagined a case with for instance 7 players, you could say something like this: "If anyone sees 6x of the same colour, guess the opposite in the first 10 seconds, then if nobody has guessed in the first 10 seconds, if anyone sees 5x of the same colour guess the opposite in the 11th-20th seconds, then if nobody has guessed after 20 seconds, whoever sees 4x of the same colour guesses the opposite". I believe that kind of strategy only ever loses if all players have the same colour, so it should be 1/(2^n-1) chance of losing (if n>1 anyway) EDIT: Actually, in retrospect, you can do even better than that if you don't have to guess simultaneously (including the case with only 3 players!). Instead of choosing when you see "6 of the same colour" you instead change that to "if you see 6x red, guess blue, wait 10 seconds, then if you see 5x red, guess blue, wait 10 seconds, then if you see 4x red, guess blue...", and obviously not guessing if someone has already guessed. That way you only lose if all players are red instead of losing when they're either all red or all blue. That way you have a 1/2^n chance of losing.
@@asdfqwerty14587 Probably counts as communication. But if not, counting n seconds and then guessing red if you see exactly n red hats wins, unless there are all blues hats.
I do feel like the wording "can see the other two" mildly suggests that they can see them and the person attached to them as a clearly identified, ordered entity.
I don't fully understand the rule at 0:56. If 1 prisoner guesses correctly, then a few minutes or hours later another prisoner guesses wrong, do they lose? If not, i have a strategy: Prisoner A, B, C. - Imagine all 3 are red: if B and C sees A=red, they should wait 2 min. After 2 minutes of waiting, A guess red. - Otherwise (if A=blue), B should guess before 2min. If C sees B=red & A=blue, C should wait 1 min. After 1 min, B should guess red. - Otherwise (if B=blue), C should guess before 1min. C will just guess red (50% chance). So they all lose if all of them are blue, 1/8 chance.
Maybe it would be better stated like this: all 3 prisoners see others hats and can't communicate at all. Then they go their separate way and make a guess or don't while others can't see
@@swenji9113 A couple of us seem to have come up with equivalent 7/8 solutions. It is not necessary that the prisoners communicate after they see the hats, as long as the game ends as soon as any prisoner makes a guess.
@@VidkunQL I agree but if the game ends when a prisoner makes a good guess, then i'd argue that this is a way to communicate between prisoners. Anyways those are 2 different problems, you can talk about one or the other, but I believe the one that ends early is not very interesting intellectually (only my opinion) because the "any incorrect guess makes all prisoners lose the game" part is made instantly useless by the "prisoners give answers in a given order" trick that does not revolve around logic
@@swenji9113 I beg your pardon, but it most certainly _does_ revolve around logic. And the prisoners are bound to find out whether they win or lose, so if that constitutes communication, then the game can't be played. And anyway, if the game _doesn't_ end after the first guess, then when does it end? Must the game continue until either someone guesses wrong or all three prisoners have guessed? If so, then the solution(s) given in the video won't work.
@@VidkunQL There are elements of logic in this solution, but my point is that it uses other physical elements : time and the capacity of humans to learn to count it. When i see a riddle like this one, here is what I think: the point is not to free real prisoners knowing they are real humans. This is just a setting to turn a maths problem into a more accessible question. Here my interpretation of the maths problem is the following: Find three maps f_1, f_2, f_3 from {1,2}² to {0,1,2} such that the probability that "f_1(b,c) € {0,a} , f_2(a,c) € {0,b} , f_3(a,b) € {0,c} AND at least one of the previous 3 images is not 0" is maximal. I'm sure everyone interested in maths can see that this problem is interesting in itself and how using information such as "when are f_1, f_2, f_3 computed?" changes its whole nature. To answer the question about "when does the game stop?" well there are many ways you could explain it. For example say that everyone guesses at the same time except that their guess can be "I don't know", but I believe that finding the exact way to phrase it is not the point here. The point is that the answers of the 3 prisoners have to be automatic and use the colors of the 2 other hats they see as their only parameter, you can't use time nor can you use the answers of other prisoners nor can you use the facial expressions of the other prisoners nor can you use the smell of the hat to determine its chemical composition and therefore its color. Of course I'm joking there but hopefully you can see what I mean. This is only 1 interpretation of the riddle. You can always use your own and go with it, there's nothing wrong there. I simply believe that with experience one can see when a problem can be interpreted as an abstract maths problem (that's what I meant by "logic" in my last message, my bad for not being precise enough) and when it is what's expected, but I didn't want to make it sound like this interpretation is better than another, this is just the one I and the video are most interested in, and it's purely subjective!
I have a strategy that can win 7/8 of the possible outcomes, EVERY PERSON CHOOSES BLUE ALWAYS, the only way that they will all get wrong is that if their hats are all red
I thought of this strategy, but it occured to me that the description of the rules is rather vague. What counts as communication? In particular, how do the players actually make guesses? Can I see when another player guesses? In this case, we apply the same strategy but agree to guess exactly 5 seconds (or any fixed time period) after the game starts. If everyone starts guessing simultaneously we choose the color of the hats we see rather than the reverse color. Thus we win 100% of the time.
0:52: "once the hats are placed, there is no communication of any kind [between prisoners]. The prisoners won't even know if anyone else has guessed" I took that to mean that you only have three options, guess blue, guess red, or don't guess. And I took it to mean that you have to make that choice purely on the basis of the two hat colours you can see, no other information
@@douglaswolfen7820 How will you know when the game is over? Are they in the same room? How are they seeing the hats? What is the actual procedure for making a guess. If I move my head (clearly visible since my hat must be visible) does that count as a signal? If I can't move or speak, how do I guess? I am just hesitant to accept vague rules like "no communication" and prefer to have the actual procedure explained so I can verify for myself what kind of strategies are actually possible. After all "no communication" really means "don't get caught communicating" since prisoners playing for their freedom are likely willing to bend the rules. It is likely possible to construct a procedure that closes all the loopholes, but I need to see the procedure to believe that the game maker succeeded at that goal.
@@harmonicarchipelgo9351 that makes sense if you're trying to imagine real people playing this out as a real scenario. But it's not intended to be taken that way, it's a logic puzzle, not a lateral thinking puzzle, or a trick question
@@harmonicarchipelgo9351 working back from the intended solution, I'll propose a simple modified version of the game that might be easier to work with. In this version, you cannot see your fellow players at all, not even their hats. You are all in separate rooms. Instead of seeing their hats, you are simply _informed_ which colour hat each of them is wearing. You are then required to select exactly one of three options: guess that your hat is blue, guess that your hat is red, or make no guess at all. Until you select one of these options, absolutely no new information will enter the room. After you select one of the three options your selection will be reported to the game organiser, and you will not be allowed to withdraw it. The other two players will go through the same procedure. You will be informed of the results of the game after the organizer has reviewed an answer from all three players
@@harmonicarchipelgo9351 to make it even more rigorous: you're actually supposed to come up with a rule for each of the three players, determining what they'll do based on the colours of their teammates hats. Their choice has to be exactly one of "guess red, guess blue, don't guess", and it has to be entirely determined by the colors of their teammates' hats
I paused at 0:43 for a while to see if I can remember how I solved the problem more than twenty years ago. I can't! lol But I do have a hunch how it might be solved mathematically... let's see Edit after watching: I was thinking in terms of Karnaugh maps. ...
Nobody has to guess anything, if they use this 100% succes strategy: Prisoner A stands next to B. They know the other’s colour, but not their own. Prisoner C looks at their colours. Option 1: C sees two different colours. In that case he goes in the middle. Then A knows he has a different colour than B and vice versa so two prisoners know their own colour. Option 2: C sees the same colour. In that case he stands next to them. In that case both A and B know they have the same colour. In the end, only C doesn’t know his own colour.
To make it 100% correct. Additional to your deduction can be just raise your hand (hint that you are guessing) if you are seeing two same color given you are the only one who is raising hand then you guess opposite color and if all of you are raising hand then you all are the same color.
Good thinking, but you're exploring a different problem than the one in the video. He made it pretty clear that there's no communication between the players _at all_.
If you can use tactics like "raising hand"/"not raising hand" to communicate binary information like "I'm about to guess"/"I'm not about to guess" then there's an easier solution: Designate Alice as the person who's always going to answer, and then use the hand signal to communicate "Alice, your hat is blue"/"Alice your hat isn't blue"
Couldn't the prisoners just come up with a subtle signal before the game starts to reveal what hat each has? Like standing with hands on hips means I'm telling you you're wearing red, or folded arms means blue?
There are actually only 4 possible combinations. 3 blue hats and 0 red hats, 3 red hats and 0 blue hats, 2 blue hats and 1 red hat, and 2 red hats and 1 blue hat. If I see 2 red hats I know that the answer cannot be 3 blue hats and 0 red hats or 2 blue hats and 1 red hat which means that the remaining possible answers are 3 red hats and 0 blue hats or 2 red hats and 1 blue hat. This means that there is only a 50% chance of guessing correctly with the strategy you suggested which is also the best chance and no different than one person just blindly guessing.
almost... there are 3 times when there are 2 blue hats and 1 red hat.... 1) Person A=Blue, B=Blue, C=Red; 2) A=Blue, B=Red, C=Blue; 3) A=Red, B=Blue, C=Blue and there are 3 times when there are 2 red and 1 blue so there are a total of 8 combinations. If you see two of the same colour and you guess that colour, you will be right 1/4 of the time, if you guess the other colour, you'll be right 3/4 of the time
Wait i didnt understand the goal. How can it be that they win if "at least one guesses correctly" and also "no incorrect guess can be made?" The second condition nullifies the first condition. Doesnt that mean they must all guess their hat correctly?
The win conditions are: a) one guesses correctly and the other two abstain from guessing, or b) two of them guess correctly and the third abstains from guessing, or c) all three of them guess correctly The lose conditions are: a) all three guess incorrectly b) two guess incorrectly and the third guesses correctly c) one guesses incorrectly and the other two guess correctly d) two guess incorrectly and the third abstains from guessing e) one guesses incorrectly and the other two abstain from guessing f) one guess incorrectly, one guesses correctly and the third abstains from guessing The game has not yet ended condition is: a) all three have thus far abstained from guessing
@MrDannyDetail Ah okay. That cleared it up. The rules in the video (at least before he says to pause it) didnt make it clear that the prisoners could choose to abstain from guessing.
I don't buy it. It's actually a paradox. Once you've observed the other 2 hats, you've narrowed the set of possible outcomes to 2 in which case you have a 50/50 chance of either being the case and you have no better odds at guessing correctly
almost... there are 3 times when there are 2 blue hats and 1 red hat.... 1) Person A=Blue, B=Blue, C=Red; 2) A=Blue, B=Red, C=Blue; 3) A=Red, B=Blue, C=Blue and there are 3 times when there are 2 red and 1 blue so there are a total of 8 combinations.
@@andrewhughes8687 you are correct, but once you get into the situation where you know 2 of the other hats are blue, you still have a 50/50 chance that yours is red because you've eliminated 6 of the 8 permutations and are now dealing with only a single binary between two of them.
@@mikelewis495well yes in that case its a 50/50, but you don't have to be the one that chooses. In 4/8 cases the other prisoners choose and you still live so its 6/8
@@mikelewis495 Art Bill Chad 1 Blue Blue Blue 2 Blue Blue Red 3 Blue Red Blue 4 Blue Red Red 5 Red Blue Blue 6 Red Blue Red 7 Red Red Blue 8 Red Red Red 1, you all say red, you're all wrong, you all lose => lose 2, Chad says Red, Art and Bill stay quiet - no wrong guesses, one correct guess => win 3, Bill says Red, Art and Chad stay quiet - no wrong guesses, one correct guess => win 4, Art says Blue, Bill and Chad quiet, no wrong, one correct => win 5, Art says Red, B&C quiet, 0 wrong, 1 correct => win 6, Bill says Blue, A&C quiet, 0 wrong, 1 correct => win 7, Chad says Blue, A&B quiet, 0 wrong, 1 correct => win 8, you all say blue, you're all wrong, 3 wrong, 0 correct => lose 1 and 8 lose = 2 2,3,4,5,6 and 7 win = 6
@@boiii2148 Thing is, as long as you make sure only one person guesses, it doesn't matter who guesses. Your chances are 50/50. You can use the "only guess if you see 2 matching hats" strategy or you can pick someone at random. It doesn't matter. At the moment that you guess, the chances of guessing correctly are still 50/50 and they never get any better than that.
If participants stand in a circle, there's a strategy which always guarantees a win: Guess the same color as the guy standing on your right. Think about it...
Wait. The rule doesn't restrict a prisoner to announce the hat's color of other prisoner, so why the prisoners don't exploit this glitch to make sure they win?
The premise of this puzzle has a contradiction! If a fair coin is flipped, the outcomes are not equally likely. A fair coin will land with the same side up 50.8 percent of the time.
@@the_peacemaker002 The way to ensure a fair coin flip would be to flip it four times - two times with heads and two times with tails being up at the time it is flipped - and only use the result then. Or let a computer randomize it with non-deterministic input.
Interesting video and very well explained. But to me you sound a little robotic and monotonous at times. Which makes it difficult to follow the entire video sometimes. I'm trying to be constructive and not offend, I hope your channel has the growth it deserves!
