How can you solve this probability problem using combinatorics instead of diagrams? A game involves rolling $2$ $6$-sided dice, Another identical die is then rolled, and the game is won if the number rolled on the third die is in between the previous two rolls. What is the probability that a player has no chance of winning before the third die roll? And what is the probability that the player wins the game
I tried a method of eliminating the sample space ie identical numbers are banned, consecutive numbers are banned. I'm still having trouble keeping track of all the cases though - is there a general method for these types of problems? I also want to know of a way that uses combinatorics $6C1$ but I'm having trouble with thinking of one up myself. I also know I could draw a sample space diagram but I even have trouble with that! I drew a $6$ by $6$ grid but I don't know how to consider the fact that it has to be "between" - it seems I need a $3$D table!
Edit: I have solved part one of the problem using a table. $P = \frac{4}{9}$. I still welcome other solutions using other methods
 A: There are six outcomes in which the first two rolls are identical.  There are five pairs of consecutive numbers.  Each of these five pairs can occur in two ways, depending on which of the dice has the higher outcome.  Hence, there are
$$6 + 2 \cdot 5$$
outcomes which make it impossible to win on the third roll.  Thus, the probability that it is impossible for the player to win on the third roll is indeed
$$\frac{6 + 2 \cdot 5}{36} = \frac{16}{36} = \frac{4}{9}$$
as you found.
For the player to win, there must be three distinct outcomes on the three dice, which can occur in $\binom{6}{3}$ ways, and the third roll must be the middle value, which can occur in two ways, depending on whether the first or second die shows the highest value.  Hence, there are
$$2\binom{6}{3}$$
favorable outcomes among the $6^3$ outcomes of the three rolls.  Hence, the probability that the player wins the game is
$$\frac{2\dbinom{6}{3}}{6^3} = \frac{40}{216} = \frac{5}{27}$$
Addendum:  An alternate way of phrasing the above argument is that there are $6 \cdot 5 \cdot 4$ ways of obtaining three distinct outcomes in a sequence of three rolls, one third of which have the third roll as the middle value, giving
$$\frac{6 \cdot 5 \cdot 4}{3} = 40$$
favorable outcomes.
