Riddle me this,

[Rendril]

@Eldrad's 100 Pirates Riddle

Let P = Total plunder (100 coins)

Let C = Captain's share of the plunder, where 0 < C <= 51.

Let D = P - C

Now, the captain chooses D pirates and gives them each 1 coin.

It all depends on how much plunder the captain chooses for himself, so at most, he will take 51 coins for himself and give 49 other pirates 1 coin. That way the attitude of the pirates would be 50:50 = 1:1 (including captain's vote)

and a mutiny would be avoided.

As for the analysis of the riddle, I assume that by "if it means they get the same amount of money they'd rather kill the captain just to see some blood" you do not mean that no two pirates may have the same share, otherwise you would need at minimum 4950 coins in order to give each one a different share.

[Metal Bunny]

I shall ignore the blatant insult to my persona, via the ignoring of my 2nd quickie and post a bigger, more literate version in which riddles may appear, you math freaks . (my math exam is over, so my mind is a dulled blade again )

How quickly can you find out what is so unusual about this paragraph? It looks so ordinary that you would think that nothing is wrong with it at all, and, in fact, nothing is. But it is unusual. Why? If you look at it, study it and think about it, you may find out, but I am not going to assist you in any way. You must do it without coaching. No doubt, if you work at it for long, it will dawn on you. Who knows? Go to work and try your skill. Par is about half an hour. So jump to it and try your skill at figuring it out. Good luck --don't blow your cool.

***edit***

Sorry, didn't see that Rendril already gave an answer to the previous riddle, and yes:

towel is the correct answer

So just concentrate on that paragraph

[Rendril]

@Metal Bunny's paragraph riddle

I'm not sure if this is the answer but there are no e's, the most common letter in the english alphabet and it is extremely verbose.

Riddle me this:

What has a voice, goes on four legs in the morning, on two legs in the afternoon, and on three legs in the evening?

[Metal Bunny]

Lol, rendril, that has to be one of the oldest riddles ever, seriously. It was asked by the sphinx on the road to every passerby, and she would eat every person who got it wrong. Oedipus (Poor poor guy) got it right and went on to fulfill one of the most famous Greek tragedies.... The answer is:

a human

Got another riddle for you guys, a more practical one.

Assuming you are awesomness like me, although I sadly have not experienced it, and you are about to have the intercourse with 3 beautifull women, or men, but then you wouldn't be awesomeness like me, because I like women more, you'd be more of a female version of awesomeness... or ... ugh whatever, depends on your taste . You are not doing them at the same time, they have some sort of issue with that, like totally lame.

Anyway, the problem is that an unknown person has a STD (remember, can't spell stud without std . It could be you, could be first woman, or the last one. Luckily you have condoms, sadly enough, you only have 2.

What do you have to do, so that you will have the all holy fornication with all the nice ladies, without risking an infection, or transferring it? Remember, weirdly enough, getting pregnant in this riddle is not a big issue.

But of course, in real life, you'd be long gone, I mean... one of you has a std, get out, get out, get out, go awaaaaaaay!

Anyway, what do you have to do? ( And no cheating, like burning a tire into a mold of your phallic glory.)

[Eldrad]

Rendril, you're on the right track but your answer is not right.

What I mean by "if it means they get the same amount of money they'd rather kill the captain just to see some blood" is that if after a mutiny they expect to get the same amount of money as they are currently getting they will still vote for a mutiny. Remember that a mutiny does not mean that the money is lost simply that the next pirate down the line will distribute it.

[Rendril]

Continuation of Eldrad's Pirate Riddle:

Then the captain has to convince the crew they can't get any more money than what he gives them. All that needs to be understood is that to members who get a coin, will not get any more coins if they were to mutiny but WILL lose that coin.

I forgot that the captain himself is also a pirate and therefore functions under the same pirate priorities. That means he can only choose to take 51 coins because that is the most he can take while still staying alive.

If the crew should have a mutiny the replacement to the captain will do the same but take a coin more due to there being less pirates to divide between.

The crew would have a mutiny once again, yeilding the same result.

This would carry on until there are 2 pirates left and in that case the replacement captain takes all coins for himself and it becomes 1:1 again but the other pirate gets nothing.

Basically, all the pirates who ARE getting a coin must realise they will get nothing if they mutiny, to make sure this comes to pass, the captain must deal out coins in an alternating pattern starting by giving a coin to the lowest ranked pirate, skipping the pirate above him in rank, giving a coin to the next one, skipping the next one , etc.

This way he assures himself 49 other pirates' votes and gets the maximum plunder he can while staying alive.

Whew, I hope that make sense

PS My first riddle remains unsolved.

I know that Sacosphilz got it but he didn't actually post it.

[Eldrad]

Closer.

Remember these are perfect rational actors so there's no need to "convince" you just need to prove that your statements are correct (which at the moment they don't happen to be, but even if they were you would need to prove them).

[Rendril]

Hmmm

By "convincing them" I meant to prove his arguement.

By explaining that the pirates currently getting a coin will lose that coin if they mutiny (since it ultimately boils down to a single pirate getting the coin), they will by rational deduction see their only chance of both living and keeping their coin is to vote for the captain.

The reason he must deal the coins in an alternating manner is to ensure those pirates lose the coin by mutiny but keep it by supporting.

That's 2 pirate priorities fullfilled, overiding the third (bloodlust) and being rational, 2 out of 3 would be more convincing than 1 out of 3.

[Eldrad]

You're close, if you attempt to come up with a proof you'll probably get the answer fairly quickly

[Rendril]

Bah, I'm successfully out of time right now. I'll have to ponder it another time.

I'm not sure what you mean by

attempting to prove it, I would have thought the proof is that the pirates would not mutiny out of bloodlust because their 2 other proirities are then no longer fulfilled.

[Eldrad]

You're correct that if their priority of money is filled (not tied) they vote not to mutiny. However you haven't proven that they won't get more (or the same amount of money) with the method you suggested, in fact with the method you suggested the first Captain would be killed.

[Sacosphilz]

@ 100 pirates riddle:

I think I'm making it more difficult than it really is but.... heck, I'd post something anyway.

The strange conclusion I got is: The first captain will always get killed no matter how he chooses to divide the coins, so will the 9 next captains. The 11th captain will be able to divide the coins in the manner that can satisfy enough pirates to prevent mutiny, and he will get 4 coins for himself. (I'm leaving out the other details for now since the division choice doesn't explain itself. Please remind me to post it if this happens to be the right method.)

I can't post all the steps needed to get the answer here because it's too long and possibly unreadable, but here are the first steps to give some general idea:

Let's name the pirates #1 through #100, where #1 being the lowest-ranked pirate, and #100 the highest (the first captain).

If there are 2 pirates left, #2 will take all 100 coins for himself and #1 can never mutiny since his vote won't matter. Considering this situation alone, their expectations would be:

#1 expects 0 coins.

#2 expects 100 coins.

From here onward, I kept analyzing the situations with 1 more pirate each time and updated the pirates' expectations for each situation. The assumptions are:

1. If by killing the current captain, a pirate will face certain death afterwards, he will not vote for mutiny no matter how much coins he gets, even zero.

2. If a pirate gets more coin than he can expect from the situations where the current captain gets killed, he will not vote for mutiny.

3. If 1. and 2. aren't satisfied, the pirate will vote for mutiny.

Let me post the few next steps as examples:

If there are 3 pirates left, #3 will only have to satisfy either #2 or #1. Since he needs 101 coins to satisfy #2, which is impossible, he only needs to go beyond #1's expectation and offer 1 coin. #3 takes 99 coins.

There expectations would be updated to:

#1 = 1 coin

#2 = 100 coins

#3 = 99 coins

If there are 4 pirates left, #4 will have to satisfy only one of #1, #2 or #3, and the best choice is clearly #1, by offering 2 coins. #4 takes 98 coins.

#1 = 2 coins

#2 = 100 coins

#3 = 99 coins

#4 = 98 coins

If there are 5 pirates left, #5 will have to satisfy two of the other 4 pirates, which is simply impossible. (he needs at least 102 coins for #1 and #4 together)

#1 = 2 coins

#2 = 100 coins

#3 = 99 coins

#4 = 98 coins

#5 = certain death (or "-1 coins" if you will)

If there are 6 pirates left, #6 will have to satisfy two of the other 5 pirates, and the best option would be to offer #1 3 coins and keep 97 coins for himself. #5 will side with #6 even if he gets no coins, because if #6 gets killed, so will #5.

#1 = 3 coins

#2 = 100 coins

#3 = 99 coins

#4 = 98 coins

#5 = 0 coins

#6 = 97 coins

With the same method, the next steps would be:

7 pirates:

#1 = 3 coins

#2 = 100 coins

#3 = 99 coins

#4 = 98 coins

#5 = 0 coins

#6 = 97 coins

#7 = certain death

8 pirates

#1 = 4 coins

#2 = 100 coins

#3 = 99 coins

#4 = 98 coins

#5 = 1 coin

#6 = 97 coins

#7 = 0 coins

#8 = 95 coins

And so on, and so forth

I kept going at this on paper until 30 pirates step, where things start getting out of pattern and very messy, and had to start writing a program to simulate it.

The logic for each iteration is:

- Order the pirates (except the current captain) by their expectation from low to high.

- Choose a number of pirates, from the one with lowest expectation and up, which is sufficient to prevent mutiny, then try to satisfy them with 100 coins.

- If they can be satisfied with 100 or less coins, the current captain takes the rest, and their expectations are updated accordingly.

- If they cannot be satisfied with 100 coins, the current captain's expectation is set to "-1 coins".

If this is confirmed as the right method (which I still doubt), I might attach an Excel file later if it would be useful. The source code (with an embarrassing makeshift Bubble Sort function) probably won't be useful whether you can read it or not.

[Trahern]

It could be you, could be first woman, or the last one. Luckily you have condoms, sadly enough, you only have 2.

Anyway, what do you have to do? ( And no cheating, like burning a tire into a mold of your phallic glory.)

They make tires that big? :lol:

If from the quoted spoilered sentence we're to assume this means that the second woman is safe, then you only need to use one condom for the first woman, and the second for the last one. That seems too simple though unless, A) It's a "read carefully" riddle, B ) It's an unsafe assumption, or C) I'm missing something.

If it's A, what color was George Washington's white horse?

And for the paragraph puzzle:

I actually answered it back here before you posted it.

[Eldrad]

Pi, you're not making it too hard

proof by induction is exactly the way to go

... but your answer happens to be incorrect. If you'd like I can point out where you went wrong or let you figure it out yourself.

[Grido]

@Trahern about their ans to 'Phallic' riddle

that shouldnt work, because you might have the STD which would mean you'd then give it to the second girl

[Yoric]

@Metal Bunny's paragraph riddle--

I guessed the answer about halfway through and skimmed the rest to make sure ;-) I'm pretty sure I've seen something like that before, though.

Also @MB-- I think the word you were looking for in your

shadow

riddle was "baleful".

@Rendril re: marble riddle--

The main point here is that *all* of the labels are now wrong, so you know that you're going to have to switch *all* of them. You choose the box that is now labelled "Black & White" and pull a marble out of that. If the marble is white, you know that that box has two white marbles in it because the label must be wrong. Therefore, the box labelled "White & White" contains two black marbles (because if it contained a black and a white marble, you'd only be switching two labels), and the box labelled "Black & Black" has a black and a white marble. I can give a more detailed answer if necessary ;-)

[Yoric]

@MB's sexytime riddle--

This one isn't possible. The common formulation of the riddle is if the man is healthy (and knows he is), what can he do with three women and two condoms. The answer is to wear one condom for the first woman, then the second condom for the second woman, and then put the first condom inside-out over the second condom for the third woman. The reason this doesn't work if the man may have an STD (an unknown STD) is that some STDs remain alive in seminal fluid for quite some time, and the third woman will be exposed to seminal fluid.

Another formulation would be to take the STDs out of the picture and have them avoid an unwanted pregnancy, three women, two condoms, vaginal intercourse only. The answer is similar.

[Sacosphilz]

Pi, you're not making it too hard

proof by induction is exactly the way to go

... but your answer happens to be incorrect. If you'd like I can point out where you went wrong or let you figure it out yourself.

Yes, please.

[Metal Bunny]

Dangit, was about to answer the marbles in the box with the labels riddle...

Anyway, so far, the answers to the condom riddle are all wrong, and yes it is solveable...

Oh and Yoric, thanks for the spelling, but try not to double post

[Yoric]

@MB-- Oh, sorry for the double posting. And you're right about the sexytime:

Both condoms at once for the first woman. Take off the outer condom for the second woman. Put the outer condom back on, inside out, for the third woman. Obviously IRL what you'd end up with is a lot of broken condoms, but hey! It's sex!

I'm sorry that I doubted you; I was so unfair.

[Eldrad]

Pi,

Your 2 pirate case is perfect. The results you get from your 3 pirate case are correct, the captain gives 1 gold to the lowest pirate and keeps the rest himself... this will always happen any time there are 3 pirates, so there is no way that the middle pirate will ever get 100 gold meaning his expectation of what he'll get if it drops down to 3 pirates is 0. So when there are 4 pirates he knows if the captain is killed he will get 0 gold. Correcting that you should have the answer and proof in a couple seconds.

[Morgana Le Fey]

i have a riddle for you guys it is easy if you think about it

it is a what am i riddle

i am around more in the day and less at night

in some places i am found all the time

i am always in more places than one

everywhere you go i am felt differently

where there is light, where there is clothes

i am there helping you

WHAT AM I

[Sacosphilz]

@Eldrad: Ah, so that's where I went wrong.

The answer would then become much simpler:

3 pirates:

#1: 1 coin

#2: 0 coins

#3: 99 coins

4 pirates:

#1: 0 coins

#2: 1 coin

#3: 0 coins

#4: 99 coins

5 pirates:

#1: 1 coin

#2: 0 coins

#3: 1 coin

#4: 0 coins

#5: 98 coins

...and so on...

So the best option for whoever is the captain is to give 1 coin to every second next pirate (every other even-numbered pirates if the captain is also even-numbered; or every odd-numbered pirates if the captain is also odd-numbered)

For 100 total pirates, that means the captain should give 1 coin each to #2, #4, #6, ... #98, a total of 49 coins, while keeping 51 coins for himself.

[Grido]

@morgana, i was going to say light till the line with it in came up, hmm what about

Life

?

i've got one, and because i'm not usually too inventive like this i've taken it from a famous person, and if you can't guess at least who the famous person is from the number *shakes head*

Look at this picture

The answer is simple, 42. What is the question?

note: there are 6 possible answers to find, one is really easy. With one of the answers it helps to have read his book.

note 2: don't google it, takes the fun away, if you do google dont post the answer.

*EDIT*The answers all have something to do with the picture

[Morgana Le Fey]

nope sorry but you were closer on your first guess though it was in the riddle

by the way i made this up in about 30 mins so it shouldnt be to hard

and if you need a clue i will give you all one tomorrow aswell though icant think ofhow to figure out your color ball puzzle lol