Simple Math Problems with Hard Solutions

[@ThoughtThrill365]

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[@blakejhonshen2710]

Why on Earth would pirate E get thrown overboard? According to the rules, it's the most senior pirate that would die if their plan is discarded. E is never in any danger, and would never accept a plan from pirate D that involved pirate D getting all the coins because E could simply vote against D, D would no longer have the majority of the vote, and D would die. Unless you mean to say that a 50/50 goes in the favor of the person who proposed the plan, but even in that scenario E would still not be in danger of dying. Did you mean to say that ALL pirates who vote against the plan are thrown overboard? Or did you mean to say that E would NOT accept the plan from pirate D in the 2-pirate scenario? Edit: Here are the relevant rules as stated: 1) The most senior pirate proposes a distribution 2A) If the majority (including the proposer) agree, the coins are distributed accordingly. 2B) If not, the *proposer* is thrown overboard and the next, most senior pirate makes a new proposal. These imply that you have to have MORE than 50% of the vote in order for your plan to be enacted, and that the only person being thrown overboard is the one making the proposal. I think you misspoke a couple times in the pirate riddle.

[@longlivethe9989]

Pirate E won't be thrown overboard but is guaranteed to get zero nonetheless, because in a D, E scenario, D proposes 100 for themselves and E is powerless

[@megaclpb2623]

He didn't just mispoke he also forgot to explain how it works in case of a tie, but fortunately there's a wikipedia article on the problem en.wikipedia.org/wiki/Pirate_game

[@DEMEMZEA]

​@@longlivethe9989no, bc E would deny it and kill D

[@Og.Just_aGamer1593]

You guys understood that that ? 😑

[@cubing7276]

If ties means the proposer would get thrown overboard then there would be another answer which is interesting to me

[@redgamer821]

You didn't follow the rules of the pirate problem, why would E get thrown overboard??

[@WaffleAbuser]

He did state it incorrectly. Pirate E would not get thrown overboard. He also didn’t explain how it works in a tie: in the case of a tie, the proposer gets the tie-breaking vote. Therefore, the real reason pirate D would keep everything for himself is that he only needs his own vote. Wikipedia page: en.m.wikipedia.org/wiki/Pirate_game

[@DEMEMZEA]

​@@WaffleAbuserexaclty

[@willy8285]

@@redgamer821 hes ai

[@Ramp4ge28]

But, if the poison takes 24h to do effect, you can give all the bottles to one prisioner, and count for each bottle, the time in wich the prisioner died gives you the bottle that has poison

[@minirop]

if you mean spacing the drinking of each bottle by 1 minutes and then see when the prisonner dies it doesn't work since you have to give your answer after 24 hours (at the same time the poison kills in a way) otherwise, yeah, you only need 1 person.

[@Ramp4ge28]

@@minirop yes, thank you

[@ieo8446]

"You have 10 prisoners *and 24 hours* to identify the poisoned bottle."

[@Ramp4ge28]

@@ieo8446 that does not have any sense: The solution in the video still takes time, and my solution can be done in an arbitrary time (because you only need the time to check when the prisioner has died)

[@ieo8446]

@@Ramp4ge28 No, your solution takes longer than 24 hours by design, whereas the binary solution takes 24 hours exactly, with strategy and logistics obviously not being considered.

[@qc1okay]

The solution process to the pirate gold puzzle is wrong. The puzzle isn't stated correctly. The key error is that when the vote is tied, the proposer wins (he has highest seniority and casts the deciding vote). Thus, the description at 2:40 in the video is completely wrong and is the crucial part of the reasoning. There is never any scenario of there being only one pirate left, since if the process made it all the way to just two pirates, then of course the senior one would distribute 100-0, and nothing would allow him to throw the least senior one overboard. (The concept of "throw overboard" is based on there being more pirates doing the throwing than pirates being thrown!) So the reasoning is: 2 pirates left means 100-0; 3 pirates left means 99-0-1 (since proposer just needs one vote from the other two, so simply offer 1 coin to the pirate who would otherwise get 0); 4 pirates left means 99-0-1-0 (since proposer just needs one vote from the other three, so simply offer 1 coin to the pirate who would otherwise get 0); so 5 pirates means 98-0-1-0-1 (proposer offers 1 coin to each pirate who would otherwise get 0).

[@HeinrichDixon]

A comprehensive solution, well explained! 99-0-0-1 also works for four pirates. 🍌🙂

[@jossdeiboss]

@@HeinrichDixon Actually not. This is because another rule that was forgotten is that pirates prefer to overthrow as many pirates as they can. So, if B offers 99-0-0-1, E would vote against and overthrow B. This is because he will get "1" from C as well (99-0-1), so he prefers to get rid of B.

[@HeinrichDixon]

@@jossdeiboss The 'rules' of this problem as given in the video are VERY poorly defined and incomplete! With your additional 'overthrowing' rule, only 99-0-1-0 is correct, as you said. 🍌🙄

[@MrDawn]

Thank you! I was so confused, like why would E ever get thrown overboard by the rules stated? The video contradicts itself a lot

[@willy8285]

@@qc1okay bro is ai that's wht

[@axiezimmah]

They're more logic puzzles than math puzzles. Also, if Pirate E hates Pirate A he'd be able to deny him the vote and still get his coin from B. In fact I am pretty sure you got the details wrong on the puzzle

[@konuralpyldzkan1495]

1 coins are alternating actually. each pirate give nothing to the pirate that is one below him and it alternates. so e has to vote for a, otherwise b won't give anything to him

[@Trombinha_jooj]

was stated that the pirates act logically, not emotionally

[@AngelShrimpChips]

Bro might be the worst explainer or maybe I’m just sleep deprived

[@Γιώργος-ε6τ]

You are sleep deprived

[@drakegrimm1287]

@@Γιώργος-ε6τ no, this guy totally blundered pirate game explanation, the way people are performing kills in Josephus problem should be a part of problem statement, and couple of problems were described very rapidly

[@willy8285]

@@Γιώργος-ε6τ worst explainer

[@bladdnun3016]

Options not mutually exclusive.

[@Γιώργος-ε6τ]

@@bladdnun3016 exactly it could be both

[@PaulOSullivan-ng6tk]

I’ve already seen these answers on ted-Ed and I still don’t understand after this man tells me

[@scores.22]

The solution to the wine problem is pretty funny if you think about it in reality. Each prisoner would have to drink from about 500 different bottles of wine, which isn't completely unrealistic if you give them about a few droplets per each bottle. 1mL per bottle would add up to about half a liter so they could probably drink it before alcohol gets to them. The bigger problem would be actually distributing the liquors. The most effective way would be giving each prisoner a syringe or a pipet of some sort. Then each prisoner would take their empty bottles which they would drink from and go around the thousand bottles to take a few droplets from each of their assigned bottles. If taking some drops of wine from one bottle and moving to the next one takes about 3 seconds, the whole process would take 25 minutes for each prisoner. You could make each prisoner start on a different bottle to ensure (assuming each prisoner moves at the same rate) they could all simultaneously do this gathering without any two prisoners having to meet up and wait for the other to finish. Explaining the method, arranging the bottles so that they could execute the method efficiently, taking the droplets, and drinking what would probably be at least 10 shots of wine, would probably take under 40 minutes if done without any error, but thinking about the workhouse-like scene is kind of funny in itself.

[@abderrahim552]

5:30 I'm so confused it's just better to just eat some charcoal as it's an effective counter poison

[@durarda5711]

my answer for 1000 bottles (i will look at the answer after) divide the bottles into 2 groups of 500 give the first 500 bottles to the first guy so we know if the poison is in the first half or not divide the 500 groups to 250 groups and give the second prisoner 500 250 from first groups 1 half and 250 from second groups first half so now we know tich 250 group the poisoned bottle is continue with splitting the groups in half and giving the next person 500 bottles to test from ensuring you get 1 part of every group only prisoner 1-500 possible bottles prisoner 2-250 possible bottles prisoner 3-125 possible bottles prisoner 4-63/62 possible bottles prisoner 5-32/31 possible bottles prisoner 6-16/15 possible bottles prisoner 7-8/7 possible bottles prisoner 8-4/3 possible bottles prisoner 9-2/1 possible bottles prisoner 10- only 1 possible bottle and best part is you can make them dring at the same time without loosing any information because every person drank from diffrent grups only requiring 1 day

[@JimmyVermeer]

Now if one of those prisoners drinking the poison just happened to die of some other cause, you'll make an incorrect conclusion about which bottle is poisoned.

[@shaunnotsean4308]

Half of these are from ted ed and one of them isnt even explained correctly

[@axiezimmah]

I think the poster doesn't even understand them himself, he's telling them very poorly and sometimes outright wrong.

[@michaelmiller2210]

TedEd does not invent these riddles lol, they just showcase them.

[@redgamer821]

​@@michaelmiller2210You could say popularized or introduced to a larger audience

[@kaspervanoverbeek4064]

For the handshake problem, Can't you just use the formula: 1 + 2 + 3 . . . + (n-1) + n? As the first person has n amount of unique handshakes the second person has n-1 and so on and on?

[@whatitmeans]

In the 2 envelopes problem: when taking the expected value, Should't one have to substract your current amount of money since you will lose it if you change? this will leads just to x/4 so you just stay with the envelope you have right now

[@antondobrovolskiy]

Video's approach is wrong, but if we believe it, if you want to know a profit, calculate relative profit, 50%*(+x)+50%*(-x/2)=x/4 (your average profit, so you "should" change) But this approach just wrong, because you have 50% chance pick 2$ envelop at the start and 50% chance pick a 1$ envelop. So if you switch, you have 50% chance to gain $ and 50% chance to lose a dollar, depends of the starting envelop, so it is actually doesn't matter, you are switching or not. Maybe if you hold some amount of money like 0.77$, you can be sure it is not a double envelop and should always switch, but I assume this is not the case

[@Rokubeee]

Actually, for the 1000 Wine Bottles of King, you can test 2^n+1 bottles instead of 2^n as mentioned, for n prisoners. If no prisoner dies, the "+1" bottle is poisoned.

[@ayaankhan6320]

Most of these are ted ed riddles 😂😅

[@ThoughtThrill365]

ted ed is not the originator of any of these riddles.

[@themadvirus613]

@@ThoughtThrill365 you're right, ted still did him first

[@LorenzoBorromeo-bb8ve]

@@themadvirus613Who’s Ted and why is he the first at it?

[@themadvirus613]

@@LorenzoBorromeo-bb8ve www.youtube.com/@TEDEd

[@luketaylor1257]

Is 50 percent considered a majority in the pirate problem? I would think it wouldn't be so if Pirate D only got 50 percent of the vote. And why would pirate E get thrown over board if they reject D's offer? I feel like it would be the other way, D gets thrown overboard because E rejects his offer and therefore D has no majority? That's the rules? I'm so lost.

[@flying_cookie7200]

The traditional wording is that any tie is decided by the person with the highest seniority. Thus pirate E never has any choice if it makes it to pirate D

[@Kaznodzieja1]

@@flying_cookie7200 so why does B need 3/4 votes?

[@HeinrichDixon]

@luketaylor1257 Just what I thought! 🍌🤔

[@huntercovington913]

The real solution is to be pirate E and keep convincing the others to mutiny until it's just you and D. After that, you simply have to hope you can win against pirate D

[@PMA65537]

4:26 Mentioned bottle 0 when they have been numbered 1-1000.

[@themadvirus613]

I think your prisoners are going to suffer from alcohol poisoning if you actually tried this. wine has a deceptively high alcohol content.

[@professorhaystacks6606]

Important point on the Josephus problem: The deal was the last 2 people would attempt to kill each other simultaneously. Josephus convinced the other guy that they were the last two standing for a reason and so the Lord must want them to live. Mind you, our only source for this story is Josephus himself, so take with a grain of salt.

[@Vengemann]

The sponsor problem was the most critical to be honest

[@JaroslavStastny-ob1yj]

As for the poisoned wine bottles problem, I’m afraid the prisoners would die from alcohol poisoning even sooner than from the poison itself if you gave them that much wine to drink.

[@gavingoebel1423]

Besides the pirate problem, it was a very interesting and enjoyable video 👍

[@evrimaltaylar3465]

Solved all of them except the last one, how am i supposed to understand that they get in a circle when you say they kill each other in a fixed order and the pirate question had some explanation issues but it didn't bother me because i already knew that one. Other than that great vid!

[@fatih3806]

Harbi birbirlerini nasıl öldürdüklerini açıklayıp “oturulacak en mantıklı yer neresi?” diye sorma olayını Problem dedikten sonra açıklamak yerine Solution dedikten sonra açıklamış

[@fatih3806]

İngilizcem var bu arada üşendim sadece

[@jakeaustria5445]

Thank you

[@gabrote42]

2:47 I hadto replay this thrice because the videomaker forgot to say the senior pirate has the tiebreaker vote. Bad video

[@raphdm3776]

You may not be very smart then...

[@gabrote42]

@@raphdm3776 Why would I guess that there was a tiebreaker vote when logic problems are supoosed to be taken at face value?

[@blank4502]

1:53 - 1:56 "If the majority, including the proposer, agree, ...

[@Deadbushfan1618]

If you are trying to find the poisoned wine and are at the disposal of prisoners just treat them better and they won't poison your wine

[@ThoughtThrill365]

😂😂😂

[@foxymetroid]

Who said they poisoned it?

[@Deadbushfan1618]

@@foxymetroid 'Extremely strictly speaking, who said that 'they' referred to the prisoners, it could have been a completely irrelevant 'them'' Bro, it is no simple math problem with hard solutions that it's a joke so honestly, congrats on ruining the joke for yourself :)

[@--fishiiki-]

I still don't quite understand the first one even after watching it multiple times

[@Vaaaaadim]

We can pretend the first person doesn't exist, and instead you have a line of people with black and white hats, and you're told whether there is an even or odd number of black hats (that is the information the first person gives). In this alternative version, the 1st person in line sees how many hats in front of them are black, and if it doesn't match the parity of what there is supposed to be (even or odd), that must mean their hat is black, so they announce that. The 2nd person knows how many black hats are behind them (based on whether person 1 said black or white). They can add up how many black hats are behind them, and how many they can see in front of them. If that sum doesn't match the parity of what it is supposed to be, that must mean their hat is black, so they announce that. And so on. The kth person knows how many black hats are behind them, they can see how many black hats are in front of them, if the sum of those two things don't add up to the correct parity, then their hat must be black, but if the parity matches then their hat must be white. Let's consider this example of the original problem. Where initially it isn't know if there are an even or odd number of black hats. ?BBWWBWWB 1 9 Person 1 (the ?) sees an even number of B. So they announce W. Person 2 knows that Person 2 to Person 9 have an even number of B. But they see 3 B's. So Person 2 knows they're B, and announces B. Person 3 knows that Person 2 is B, they see 2 B's in front of them. Adding up to 3 B's, but there is supposed to be an even amount. So they're B, and they announce it. Person 4 knows that Person 2,3 are B, they see 2 B's in front of them. Adding up to 4 B's, so Person 3 can't be B otherwise there would be an odd amount. So they're W, and they announce it. Person 5 knows that Person 2,3 are B, they see 2 B's in front of them. Adding up to 4 B's, so Person 5 can't be B for the same reason, so they announce W. Person 6 knows that Person 2,3 are B, they see 1 B in front of them. Adding up to 3 B's, so they must be B to make an even amount of Bs, so they announce B. and so on...

[@iamadooddood4331]

2. Pirate B only needs one more vote other than his own. He just needs to give pirate D one coin. Since pirate D will get nothing under pirate C if he rejects pirate B's proposal, he will have to accept. So pirate B can get away with 99/0/1/0. 4. Whether having the fastest guy escort all the rest, or having the slowest two guys go together, is faster, depends on whether the fastest guy and third fastest guy each going once is faster (the former), or whether the second fastest guy going twice is faster (the latter). If person B took 5 minutes instead of 2 minutes, the former would actually be faster. 5. If the amount the envelops contain are X and 2X, the expected amount is 1.5X regardless of whether you switch. There shouldn't have been a paradox here.

[@metalim]

3:06 that's not correct. Pirate B needs ONE more vote, not 2. Half is enough, as he gets the casting vote. Otherwise solution for 2 pirates (D and E) with D taking al the money would not work. All following explanation is complete BS. WTF?

[@Trombinha_jooj]

2 more besides him

[@wiener_process]

The envelope problem is one of my all time favorites.

[@surtech5]

First puzzle falls apart if the only black (or white) hat is on the first guy and everyone else ended up with the same color.

[@michaelmiller2210]

I didn't watch the solution because it seemed like a pretty easy problem, but here's what I came up with: If the first person to guess their color sees an odd number of black hats, they say black. If they see an odd number of white hats they say white. This works in your proposed situation where the last nine people all have the same colored hats. Say for instance they were white. Person 1 says white since he sees an odd number of whites (9). Person 2 says white because he sees an even number of whites (8) and since person 1 saw an odd number of whites, they know they must have a white hat to make the total odd. Person 3 says white because they know that person 2 saw an even number of whites, but person 3 sees an odd number of whites. This repeats for the remaining prisoners, and 9/10 of them will get it right. Which is all they need.

[@ciCCapROSTi]

Okay, how is the handshake problem hard?

[@pislerpaul8984]

Sounds like Dwight from the Office

[@mohamedgalal8873]

just filter out odurless bottles then filter out tasteless bottles then let the prisoners test them

[@niom-nx7kb]

The handshake problem isn’t a hard solution

[@FulltimeSlacker]

For the first one, couldn't you just say the color of the person in front of you so that everyone knows what the color of their hat is except the first guy ? Makes much more sense to me and just seems overall simpler. Edit: They never said you couldn't say black or white without guessing meaning you can first tell the guy in front of you and then guess

[@White-ul3dl]

You need to guess the colour of YOUR hat

[@FulltimeSlacker]

@@White-ul3dl yeah, just tell the guy in front of you the color of his hat so that he knows it, that way everyone except for the first guy knows the color the of their hat and they are assured to win. My logic stands

[@brandonm8901]

Because the person in front can't do anything with that information if they also need to say the colour of the hat in front of them. Everyone will KNOW the colour of their hat, but they will SAY the colour of the hat in front of them. This will still have a 50% chance of each prisoner being correct

[@FulltimeSlacker]

@@brandonm8901 you should read the edit on my comment

[@brandonm8901]

@@FulltimeSlacker I think it's clear it means you can only say "black" or "white" once. Of course you can't tell the person in front of you and then make a guess for yourself

[@jimmea6317]

what is it with math riddles all being formulated by complete sadistic psychos

[@ommapodi]

Pirate gold problem is stupid

[@FaisalAbdullahi-c3y]

Simple math problems*

[@meedrowh1809]

I mean, I don't get the solution for the first problem. Just... high-pitch, low-pitch for black and white on the following prisoner. No need for 'odd' and 'even', that's just overcomplication.

[@kargoncoppercoin2093]

You don't need a formula for the handshake one. Its just n-1 factorial

[@michaonski8617]

This is probably the worst riddle explaining video ive ever seen... my god

[@WonderOfU24]

Thankgod i left maths

[@AlexTrusk91]

9:30 you didn't explain it mathematically. here's at least a try: [long post warning] assign a real value with one envelope, like 1$, and a real value with the other, 2$, then calculate the whole probability tree. And sínstead of an initial envelope, name a Person 1 and a Person 2 (that gets the 2nd envelope you didnbt't choose. Ask: what's the probability that Person 2 is better off? With absolute values it's jsut fine. Expected value is 1.50$ (or 150% of the lower value) Working with relative values seems to be the flaw. you can't stand at two branches of the probability tree at once. what's really the brainfuck here is that you get to know the content of one envelope. if you see 100$ yyou think "either I loose 50 or gain 100$ more, the odds are 50/50. But the odds can't be 50/50. you need to add different branch levels together, giving a different result, when you really should add up all the possibilities of the same branch level. then the expected value stays fixed at 150%. of the lower value, or 75% of the upper value in the example in the video, X stands for two different values at the same time, namely "1$ and 2$" simultanously. the expected value is the average, 1.50$, making a switch to the 2nd envelope useless, because the expected value there is also 1.50$. Only when you assume that X is 1$ and X is 2$ at the same time you get weird results. It doesn't even make sense. if the envelopes just contain cards saying "2x" or "x/2", what does it mean? what x exactly? can't be one or two, beacuse then it breaks down. and theres no 3rd value in the middle. 1.50$ being the expected value for the 1st envelope, the second can have either +1/3 (more) the that (2$) or -1/3 (less) (1$), you jsut don't know if your are currently above or under the expedted value, you can't be on it, bc you only got two options. the scenarious are: 2/3*0.5 * 4/3*0.5 (you get the lower value first, than switch) or 4/3 * 50% * 2/3 * 50% 3/3 is the expected value and you are either 1/3 over or below it. X can't be what"s in the 1st envelope, it's either less or more, but not the first value. and you don't know with it is, so both scenrious are equally likely- cool video tho, loved it

[@willy8285]

@@AlexTrusk91 bro is ai that's wht

[@willy8285]

I hate AI generated content

[@ThoughtThrill365]

me too.

[@harleyspeedthrust4013]

all of these videos are voiced by AI. they're not even correct in many cases; they make serious errors of omission. you would be better off learning from a proper textbook or an online class.

[@willy8285]

@@harleyspeedthrust4013 haha ik. it's so obvious and the creator is a fucking fake. i wanted to see if they were conscious of somebody telling them about somebody conscious, but i don't blame him for thinking that i'm just as stupid as everybody else. ironically though, he probably does genuinely hate ai content. scum of the earth, but i mean hey, anything for that dollar. I would get it if u broke and need money, but the creator is prob a bum

[@ThoughtThrill365]

this video wasn't voiced by AI. c'mon guys, i know ai is getting popular but it doesn't mean everyone uses it.

[@willy8285]

@@ThoughtThrill365 it literally is, wdyn

[@chezb1367]

you didnt even describe the final problem, that they stand in a circle and the order in how they kill and also for literally every "problem" your explanation is just: "this is the solution because if n=(a^2) then number 1 always wins, if its n+(2x+1) then the last person always wins like STOP SPAMMING MATHS FORMULAS OUT OF UR 🍑 AND EXPLAIN WHAT YOURE TALKING ABOUT PLS IMPROVE FOR NEXZ TIME

[@Nightlife_Offical]

Why doesn’t the prisoner just say the colour of the hat in front of them? 😂

[@CieraHeadrick]

Cuz it wouldn't work

[@CieraHeadrick]

You have to say your own hat color, you can't say your own and the one in front.. what if one is BLK, and one is white..

[@axiezimmah]

Because then everyone would die

[@bird_obsession]

Doing this will lead to exactly one prisoner knowing their hat color. The goal is for 9/10 to guess correctly, not 1/10.

[@HeinrichDixon]

That was my first - stupid - thought, Nightlife. 🍌🙄