First to roll a six dice game (Probability)

Hi, Came across this Maths question:

In a game four players take it in turns to roll a normal unbiased six-sided die.
The winner of the game is the first person to roll a 6.
Find the probability that
a) The first player wins
b) The last player wins

I am wondering if my method/answers are right. I got 216/671 for the first answer and 125/671 for the last.
I did this by forming a geometric sum where a= 1/6 for the first player and r = 5/6 all to the power of 4 (4 players) is this thinking correct? I can explain my reasoning more if helpful! Many thanks!

the fact that you said "came across" suggests that this isn't formal so I am wondering if I am permitted to give my ideas for a conclusive answer. first I will give hints because they are allowed.
so the first one is simple. this is just the probability of an outcome (in this case the probability of rolling a 6) which is pretty simple so I won't touch that.
the last person winning is a little different. it is just asking what are the odds the 4th roll will be a 6. you can use a probability tree or use simple logic.

The first player wins if the first roll is a 6. The probability of this is just 1/6.

I think the question is asking more about the first player rather than the first roll!

Here is another question, bit more stuck on this one. I can do part a, but wondering if there is a way to simplify the rest of the question, rather than brute forcing!

Quite a lot on combinations/counting in problem solving, for instance
https://kheavan.files.wordpress.com/2010/06/paul-zeitz-author-the-art-and-craft-of-problem-solving-2edwiley20060471789011.pdf
Has a decent chapter and questions.

There are 1 line answers for each of (a), (b), (c), although for (c) at least brute force is going to be fairly painless.

Incidentally, this looks very much like an Oxford MAT question, in which case there are threads specifically dedicated to the Oxford MAT in the Maths Exams subforum.

so for b do you think systematic counting is a valid method

I am wondering if my method/answers are right. I got 216/671 for the first answer and 125/671 for the last.
I did this by forming a geometric sum where a= 1/6 for the first player and r = 5/6 all to the power of 4 (4 players) is this thinking correct? I can explain my reasoning more if helpful! Many thanks!

Agreed on the answers. There's actually a very short way of doing this. If you define A, B, C, D as the events that each player wins, it's easy to form expressions for p(B), p(C), p(D) in terms of p(A). Then use the fact that p(A)+p(B)+p(C)+p(D) = 1 to find p(A) and thence p(B), p(C), p(D).

Is it that each chance is 5/6 times smaller as you go from player 1 to 2 to 3 to 4?

Yes. There's a 5/6 chance that A doesn't win on the first roll, after which B is in exactly the same position that A was in terms of chance of winning. And so on down the line.

For the grid and counters, I get the following.