It can be simplified greatly by considering what events must occur in each of the 4 cases.
For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.
Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)
In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.