Welcome to the forums!

There's actually a Mathematics 2U Board you might want to check out a bit later (to post further questions, and to view other questions that you may have that have already been asked! otherwise there are more resources there

)

For part i), consider this: in what cases do you have no chance of winning before the game starts? Essentially, this maps out a bit like a game of bullshit; if you roll a certain number on the first die, and you roll one up, one down or the same on the second, you instantly have no chance of winning. Drawing a 6x6 table with each outcome helps visualise this best, but you can also list them out case by case

Try doing this yourself, and then check the spoiler for the solution

Spoiler

(1, 1), (1, 2), (2, 1), (2, 2), (2, 3) ... (6, 6) = 16 cases out of a possible 36, ie 4/9

ii) This is a little trickier. There are a few different ways we can consider this;

a) By the number of numbers in between each pair

b) case-by-case (which takes a lot longer)

a) is the easiest option, but you can have a look at b) if you have a lot of time.

You can have anywhere from 1-4 numbers in between each pair ie. for 1, an example (2, 4) and for 4, an example would be (1, 6).

There are eight cases with 1 number in between: hence, we multiply 8/36 (the chance of getting this case) by the chance of hitting a number in between on the third die, which is 1/6.

There are six cases with 2 numbers in between: repeating, we have 6/36 x 2/6

There are four cases with 3 numbers in between: repeating, we have 4/36 x 3/6

There is one case with 4 numbers in between: and lastly we have 2/36 x 4/6

Adding all these probabilities will give you the answer!

Also, if I recall correctly, there was a similar question in the 2018 HSC, you might want to check that out! (unless this is from the 2018 HSC?). If you have any further questions, please enquire further!

Hope this helps!!