Probability of 6 die rolls

Question

If two fair $6$-sided dice are tossed six times, find the probability that the sixth sum obtained is not a repetition.

The solution given to me is not very helpful in explaining the steps, and my attempt at it is far away from the final answer. This question appears to be different from the standard die questions I am used to. Intuition and explanation would be appreciated.

The Solution: $2*\frac{1}{36}(\frac{35}{36})^5+2*\frac{1}{18}(\frac{7}{18})^5+2*\frac{1}{12}(\frac{11}{12})^5+2*\frac{1}{9}(\frac{8}{9})^5+2*\frac{5}{36}(\frac{31}{36})^5+\frac{1}{6}(\frac{5}{6})^5\approx0.5614$

Answer 1

So this is the possible sums we can have in each trial.

\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&2&3&4&5&6&7\\ 2&3&4&5&6&7&8\\ 3&4&5&6&7&8&9\\ 4&5&6&7&8&9&10\\ 5&6&7&8&9&10&11\\ 6&7&8&9&10&11&12\\ \end{array}

What the solution appears to be doing is to

1. Choose which sum to be the sixth sum.
2. Have the other sums appear $5$ times.

Which means we have required probability $p$ such that \begin{align} p&=\sum_{i=2}^{12}{P(X=i)P(X\ne i)^5}\\ &=\frac{1}{36}\left(1-\frac{1}{36}\right)^5 +\frac{2}{36}\left(1-\frac{2}{36}\right)^5 +\frac{3}{36}\left(1-\frac{3}{36}\right)^5 +\frac{4}{36}\left(1-\frac{4}{36}\right)^5 +\frac{5}{36}\left(1-\frac{5}{36}\right)^5\\ &+\frac{6}{36}\left(1-\frac{6}{36}\right)^5 +\frac{5}{36}\left(1-\frac{5}{36}\right)^5 +\frac{4}{36}\left(1-\frac{4}{36}\right)^5 +\frac{3}{36}\left(1-\frac{3}{36}\right)^5\\ &+\frac{2}{36}\left(1-\frac{2}{36}\right)^5 +\frac{1}{36}\left(1-\frac{1}{36}\right)^5\\ &=2\times\frac{1}{36}\left(1-\frac{1}{36}\right)^5+2\times\frac{2}{36}\left(1-\frac{2}{36}\right)^5 +2\times\frac{3}{36}\left(1-\frac{3}{36}\right)^5\\&+2\times\frac{4}{36}\left(1-\frac{4}{36}\right)^5 +2\times\frac{5}{36}\left(1-\frac{5}{36}\right)^5+\frac{6}{36}\left(1-\frac{6}{36}\right)^5\\ &=2\times\frac{1}{36}\left(\frac{35}{36}\right)^5+2\times\frac{1}{18}\left(\frac{7}{18}\right)^5+2\times\frac{1}{12}\left(\frac{11}{12}\right)^5\\&+2\times\frac{1}{9}\left(\frac{8}{9}\right)^5+2\times\frac{5}{36}\left(\frac{31}{36}\right)^5+\frac{1}{6}\left(\frac{5}{6}\right)^5\\ &\approx0.5614 \end{align} which is exactly your solution.

Answer 2

If you toss two dice, the resulting sum can be anywhere between 2 and 12.

$$\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&2&3&4&5&6&7\\ 2&3&4&5&6&7&8\\ 3&4&5&6&7&8&9\\ 4&5&6&7&8&9&10\\ 5&6&7&8&9&10&11\\ 6&7&8&9&10&11&12\\ \end{array}$$

For me, one helpful mental trick here is considering the probability that the first roll of the dice is not a repetition. This must be the same as the probability that the last roll of the dice is not a repetition, because order does not matter.

So what we do is we consider each possible sum the first dice might have. For example, the sum might be 6. Out of the 36 outcomes in this table, five outcomes result in 6, and thirty-one outcomes result in something other than 6. Hence the probability of getting a 6 on the first roll of dice, then never getting a 6 again is:

$$\frac{5}{36} \cdot (\frac{31}{36})^5$$

We can compute the probability that the first roll of the dice results in $k$, then the next five rolls all result in something other than $k$. If we do this for all sums in the range $k=2\ldots 12$ and add up the probabilities, we will have our result.

The answer is:

$$P = \underbrace{\frac{1}{36} \left(\frac{36-1}{36}\right)^5}_{k=2} + \underbrace{\frac{2}{36} \left(\frac{36-2}{36}\right)^5}_{k=3} + \ldots + \underbrace{\frac{5}{36} \left(\frac{36-5}{36}\right)^5}_{k=6} +\underbrace{\frac{6}{36} \left(\frac{36-6}{36}\right)^5}_{k=7} + \underbrace{\frac{5}{36} \left(\frac{36-5}{36}\right)^5}_{k=8} + \ldots +\underbrace{\frac{2}{36} \left(\frac{36-2}{36}\right)^5}_{k=11}+ \underbrace{\frac{1}{36} \left(\frac{36-1}{36}\right)^5}_{k=12}$$

In each case, we have a term that looks like $\frac{r}{36}\left(\frac{36-r}{36}\right)^5$, where $r$ is the number of times the given sum appears in the table. It's the probability of getting that sum on the first roll, then getting some other sum in each of the five subsequent rolls.