Kogorous Riddle Thread and eworms Deduction Thread

[Edogawa4869]

My guess:

Spoiler: Probably Wrong

99 of them are guaranteed freedom for a weekend.

Each time it is someone's turn, he takes note of the color of the hat in front of him.  If it were one color, white, for example, the prisoner would immediately answer when it was his turn.  If it were the other color, black, he would pause before answering.  I feel bad for the person in last though.

[Kogorou]

HA
*lights cigarette*
Prepare to be astonished.

Spoiler:

Every Prisoner counts the number of white hats he sees.
Let me explain with 10 Prisoners.
w-s-w-s-s-w-w-s-w-w
1:0
2:1
3:1
4:2
5:2
6:2
7:3
8:4
9:4
10:5

Well the last one is really in a bad situation because he has to guess because he doesn't know how many there are.
With this strategy you can save up to {amount of prisoners}-1.

[eworm]

Well, I came up with something a bit similar to Edogawa4869's deduction.

Spoiler:

The prisoners agree upon a signal meaning "your hat is the same color". It may be a cough before answering, a change in intonation or something like this. Then it's only the first answering who has to guess - with 50% success rate anyway. After that every prisoner knows their hat's color with the signal or lack of signal in the previous prisoner's answer. They say that color and if the color in the front of them is the same they give the signal too. That way 99 prisoners are saved - or even all of them if the first manages to guess right.

[Ankor]

Edogawa (the same for eworm) - the number is correct, but there is a way to save 99 without any type of signal, just by saying "black" or "white" 
Kogorou - so everybody counted white hats and what's next? what do they answer?

P.S.
I am at fault here  I forgot to mention "No signals allowed. The only thing prisoner can do is to say black or white".
Without this rule specified Edogawa's answer is considered correct. And changing rules during the run is quite bad IMHO.
So please decide whether we count it as a correct answer or we continue guessing the answer that doesn't use any kind of signals.

[Kogorou]

Well Edogawa was the first one and when you say it's the right answer he has to continue

But mine was way cooler because it has math in it

[Ankor]

Kogorou: Your answer is incomplete, as You just said about some math so what are prisoners going to say after calculations?

[Kogorou]

Everybody but the last person can go home in this case it would be 99 in total.
If they can't say numbers they should say the color of the hat in front of them.

Look at my small explanation for 10 Prisoners it works out because my way too but they have to say numbers

[Ankor]

There is a way using simple math to guaranty that 99 people go home using only only words "black" and "white"and not using any signals 

So everyone as I previously said: do we change the rules and continue till riddle is solved (or eworm's riddle is complete), or should we name Edogawa's answer correct and wait for a riddle from him (as it didn't brake any original rules and saves maximum number of prisoners that is possible), I'll also name the expected answer

[eworm]

I'm almost done with my newest riddle that is going to be so awesome because of 

But you can still continue - however you like. I'm kind of curious what the method to free the prisoners without any signal is...

[c-square]

There is a way using simple math to guaranty that 99 people go home using only only words "black" and "white"and not using any signals 

I'm really having a hard time figuring this out...

Spoiler: my thoughts

So, to guarantee 99 people go home, that means 99 people must each say the color of his or her hat.  Since the first person can receive no clues, there is absolutely no way the first person can be sure to say his or her own color.  Therefore, it must be the remaining 99 people who answer correctly. 

Now, the hats can be distributed in any order and in any amount.  They could all have black hats, they could all have white hats.  They could have some black and white hats either mixed up, or all black first followed by all white.  Since everyone else must say the color of his or her hat, the only added information can come from the first prisoner.  The first prisoner can only say 'black' or 'white', which amounts to 'true' or 'false'...  Hold on a second, I think I just realized something...

Okay, so the first prisoner counts the number of white hats he sees.  If it is an odd number, he answers 'black'.  If it's an even number, he answers 'white'.  The second prisoner then counts the number of white hats she sees.  If she counts an odd number and the first person said 'black', then her hat must be black because the first person also saw an odd number of white hats.  If she counts an odd number of white hats and the first person said 'white', then her hat must be white so that the first person would see an even number of white hats.  Similarly, if she counts and even number of white hats and the first person answered 'black', then her hat must be white, and if she counts and even number of white hats and the first person answered 'white', then her hat must be black.

Each prisoner in line keeps count of how many 'white' responses (s)he hears after the first and compares that to the number of white hats he or she sees.  Then each prisoner uses that information, with the information of whether or not the first prisoner saw an odd or even number of white hats, to determine the correct color of his/her hat.

Nice riddle! 

[eworm]

Hey, c-square, long time no see!
I think I'll post the third riddle today... For me it means at most five hours with minutes.

[Ankor]

c-square
that was the original answer 

[KainTheVampire]

No guts, no glory.

Plus I might be a tad insane.  It's not been proven yet, though.

*proves*

[eworm]

My riddle is done... I may post it but it's c-square's turn now... What to do?

[Edogawa4869]

No guts, no glory.

Plus I might be a tad insane.  It's not been proven yet, though.

*proves*