Dice...

Ok, I'm just bored, so I'm just gonna talk some crap here.

Here's a question. You have 2 dices and you're asked to roll them. Out of the 11 possible numbers that can appear, which number has the highest chance of appearing? (11, not 12, because it's impossible to get 1 from 2 dices. The least you can get is 2.)

Right, there are 2 dices, right? And each dice has numbers ranging from 1 to 6. So, there will be 6x6=36 possible combinations.

Ok, let's list out the possible combinations for all 11 numbers.

1 -> None (unless 1 of the dices have a face with no spots!)
2 -> 1,1;
3 -> 1,2; 2,1;
4 -> 1,3; 2,2; 3,1;
5 -> 1,4; 2,3; 3,2; 4,1;
6 -> 1,5; 2,4; 3,3; 4,2; 5,1;
7 -> 1,6; 2,5; 3,4; 4,3; 5,2; 6,1;
8 -> 2,6; 3,5; 4,4; 5,3; 6,2;
9 -> 3,6; 4,5; 5,4; 6,3;
10 -> 4,6; 5,5; 6,4;
11 -> 5,6; 6,5;
12 -> 6,6;

As you can see, there are 6 ways you can form a 7 from 2 dices. So, there's a very high chance (16.67%) for 7 to appear whenever you roll 2 dices.

OK, we have finally answered that question above. Now, what about this: suppose that you are forced to get a double, like 1 and 1 or 3 and 3 and so on. What is probability of getting a double?

There are 6 possible ways to get a double, 1,1; 2,2; 3,3; 4,4; 5,5; and 6,6. All together, there are 36 combinations. So, the chances of getting a double is 6/36*100%=16.67%

From there, we can conclude that the probability of getting a double is the same as the probability of getting a 7.

Well, I shall end here. I wanna go eat some chocolates!!