Somnova Studios

HEY HEY HEY

It's puzzle time!

43.

Ravenous wrote:43.

Wow, that was fast! Nice work.

Next one coming up!

Next puzzle is up!

You will boil me in oil.

brainy_kevin wrote:You will boil me in oil.

CAN NOTHING SATE YOU?!?

Nice, nice...

Okay, new one is up. Have fun~

He shouldn't shoot anyone.

sky wrote:I am now officially smarter than ILY.

:3c

#1 Offensive post of the day

If he doesn't shoot anyone, he doesn't have any chance of surviving.

Now, who are Mr. Gray and Mr. White going to shoot when it's their turn? If they always try to shoot Mr. Black, then Mr. Black has to shoot Mr. White first to have any chance of surviving.

But if they shoot randomly, then Mr. Black has a higher chance of surviving if he shoots at Mr. Gray first.

(Or, I might have completely misunderstood the question...)

Yev wrote:If he doesn't shoot anyone, he doesn't have any chance of surviving.

I was only referring to his first turn.

Ah, you're right. Dang it.

ILY wrote:He shouldn't shoot anyone.

sky wrote:I am now officially smarter than ILY.

:3c

#1 Offensive post of the day

Correct! Mr. Gray has a good chance (2/3) he'll hit his target (Presumably Mr. White, who always hits his target). Mr. Black's odds increase to 1/2, giving him a much better chance at offing Mr. Gray on his next turn.

I'll be posting later - busy end of the week!

WOAH THAT WAS TOO LONG.

BUT HEY, GUESS WHO'S GOT A NEW BATCH OF PUZZLES~!

He has a red hat as he can see the other two are wearing black hats.

I think.

I assume that A is in the back and C at the front: If B and C both wore black, A would know he has red, so B and C must have at least one red hat between them. If C had a black hat B would see this and know that B has the red hat, so C must have a red hat. And no, I don't think we do know, it seems that A and B could have any combination of red and black.

I think.

It should be mentioned that my answer takes advantage of information that is missing that usually accompanies this logic puzzle.
I just occam's razor'd it instead.

Since Kevin answered the main question, I thought I'd try to tackle the bonus one.

This is all just probability
If B also has a red hat, the chance of A having a red hat is low. That leaves me with the option that B likely has a black hat, and A has a red hat

A doesn't know whether he has a black hat or a red hat, and he can't see those of the other two.
B can see A's hat, but he doesn't know which hat he wears. As knowing the color of one hat is insufficient to conclude a hat color, he can't say anything.
C can see both of their hats, and knows the distribution of the colors. Since he knows what hat he's wearing, the only conclusion to draw is that A and B are wearing hats that led him to figure out the color of his own.
If A and B were wearing red hats, C would not know whether he was wearing a red hat (from 1 left over) or a black hat (from 2 left over).
If A and B were wearing 1 red and 1 black hat, C would not know whether he was wearing a red hat or a black hat, using similarly logic.
If A and B were wearing black hats, C's only option is that he is wearing one of the 3 red hats instead of any of the black ones.
So, A is wearing a black hat, B is wearing a black hat, C is wearing a red hat.