What are the Odds of rolling doubles?

Answer 1

Answer 1:
The odds are very easy to calculate. Simply divide the number of "valid" rolls against all possible rolls. For ease, you can write down all possible combination for the 2 dice.

1-1; 1-2; 1-3; 1-4....and so on, remember 1-4 and 4-1 are different rolls

There are 36 unique possible combination, and 6 of them are doubles, so that's 6/36 chances (and since 6 goes into 36, 6 times, this reduces to 1/6) or about 17%

Answer 2:

Another way to look at this problem, generically, is to assume we have an 'n' face dice. In most cases, dice have 6 faces (1, 2, 3, 4, 5, 6). But why not create a solution that works for any number of sides? Well, if we are trying to calculate the probability of rolling two dice (dice-1 and dice-2) of 'n' sides at the same time and having them turn up as doubles, only one of the dice really matters. Here's why. Dice-1 is guaranteed to land on a number 1-n. This will happen every time (on a fair dice, disregarding freak incidents). What we are trying to calculate is the probability that dice-2 will land on the SAME number as dice-1. Dice-2 can only land on one of 'n' values: 1, 2, 3, 4, 5, 6, ... , 'n'. For you non math folks, this just means it must land on a number from 1 to 'n' where 'n' is the number of sides on your dice. Out of all of the sides that dice-2 can PHYSICALLY land on, one of the sides MUST necessarily have the same as the value that dice-1 landed on. That is to say, if dice-1 landed on the value 3, there must be some chance that dice-2 will also land on the value 3. The probability of this occurring on a fair die is 1 divided by the total number of possible outcomes, which would be 'n'. So, really, there is a 1/n chance that dice-2 will land on the same number as dice-1. Thus, our probability for rolling doubles is simple 1/n. For our 6 sided dice example, our dice-1 lands on some value between 1 and 6 and there is a 1/6 chance dice-2 will match it.