I'm given this sample question:
If a pair of fair dice is tossed six times, what is the probability that the sum on the sixth roll is not a repetition of the sum on any previous roll?
I figured I should know how to do it. Maybe simplify the question to 2 rolls or 3 rolls instead of 6. However I cannot solve it.
A very thorough explanation would be much appreciated. Thanks.
It's going to be the sum of probabilities for the $6$th roll being each one of $11$ possible outcomes times the probability of not getting that outcome on $5$ other rolls. The probability of getting a sum of $2$ or $12$ are the same so the calculation below pairs outcomes except for $7$.
$P = 2 * \frac{1}{36} * (\frac{35}{36})^5 + 2 * \frac{2}{36} * (\frac{34}{36})^5 + 2 * \frac{3}{36} * (\frac{33}{36})^5 + 2 * \frac{4}{36} * (\frac{32}{36})^5 + 2*\frac{5}{36} * (\frac{31}{36})^5 + \frac{6}{36} * (\frac{30}{36})^5$