Only clear thing I could find about Plain Chocolate: chocolate with a slightly bitter flavour and dark colour compared to milk chocolate, but I think that is not the question here (although it might have been a harder question)

Yeah, meant dark (I doubt chocolate which doesn't taste of anything wouldn't sell that much).

Moros, I saw that question somewhere (might have been a friend who gave it to me) and I don't have the answer. My and my mate both got the same as you, so I assumed you are right.

Okay, here's a new problem.
How many armies do you need on a territory in Warlight to be 51% sure it survives three attacks of four armies each? And how many would you need to be 100% sure it survives? 75% luck, weighted round, standard offence and defence settings.

Moros, my friend and I made a mistake with that problem which we think you did too.

Look at the second thing you put. It should be a third not a half.

A third? In your post you clearly say a half.

Moros, do you have any kind of solution to the problem you posted? I know how to solve it but I won't even begin to try and calculate it, it is more than extremely annoying when it comes to calculation (the 51% part that is).

Okay, change 51% to any percentage over 50%, but as close to 50% as possible if you understand. Like 50,34%, or 52%, but not 64%.

Question:

You see an open game for 6 players, 3v3 random teams.
There is 1 open seat.
There is 1 existing player that has already joined that is on your blacklist.

Part 1: If you join, what are the chances the one person on your blacklist will be on your team?
Part 2: What are the chances the person on your blacklist will ruin the game for everyone whoever's team he is on.
Part 3: Do you even bother joining?

Simple one:

Two players (A and B) pick digits (from 0 to 9) by turns (A as the first one). Each player have to pick 3 digits, the digits can't repeat.
Selected digits written in the order they were selected create certain number: 1A*100000 +2B*10000 +3A*1000 +4B*100 +5A*10 +6B*1
Player B wants the obtained number to be a prime number, while player A want to prevent it.
Is there a winning strategy for any player? Prove it.

after the junk of a number created by the first five numbers person b just adds a number that makes it prime because there are infinite prime numbers so there will always be one greater than it

Seahawk, that is not always true. Here's another way to put it:
The number is AXBYCZ. Player 1 picks A, B and C, and player 2 picks X, Y and Z. Together they form a single 6-digit number.
The only winning strategy I could think of for player 1 is to memorize all 6-digit primes that start with one digit (a 7 for example), and pick digit C so there's no Z that can make the whole a prime number. As prime numbers are farther away from each other as numbers get larger, this won't be hard.

http://pdfcast.org/pdf/solution-to-widzisz-s-problem-1

Aww, the "proofs" above are so cute :)

didnt realize it was 0-9

Yea, Piotrek got it right. Glad you liked it :)