First, let's compute a couple of easy probabilities - we'll use them to check our work in a moment. If I was one space from the target, I only hit the target on a roll of 1. Any other roll takes me past it and I miss it.
So the probability I hit the target from one space away is 1/6.
If I was two spaces away, I hit the target with a roll of 2 (probability 1/6); but I can also hit the target by moving one space (1/6) and then hitting the target from there (times the 1/6 we already worked out a second ago). Total probability = 7/36.
Let's say P(i) is the probability of hitting the target space, T, from space i.
From the current space, i, the probability you hit the target is the probability you roll 1 times the probability you hit the target from one space further ahead, plus the probability you roll 2 and hit the target from 2 spaces further ahead, and so on up to 6 spaces.
That is, we get the recursive formula
P(i) = 1/6 [P(i+1)+P(i+2)+...+P(i+6)].
Let's look at that for a second. If I know the six probabilities immediately ahead of the current space, I can work out the probability from this space.
Now, let's look at the target space, T. Obviously, we're there already, so P(T)=1.
How would I work out the probability for the previous space?
P(T-1) = 1/6 [P(T) + P(T+1) + ... + P(T+5)].
Wait. What's P(T+1)? Well, from past the space I have no chance. It's 0. Same with P(T+2), etc.
P(T-1) = 1/6 [ 1 + 0 + 0 + 0 + 0 + 0 ] = 1/6 (checks out)
P(T-2) = 1/6 [1/6+ 1 + 0 + 0 + 0 + 0 ] = 7/36 (checks out)
[It's not hard from this to show that, for the 5 spaces immediately before the target, the hit-the-target probability for the NEXT space further away is 7/6 of the probability for the current space. Consequently, for those spaces, P(T-i) = 7i-1/6i. So from 6 spaces away, the probability is 75/66 = 16,807/46,656. However, I'm going to take a lazier-but-more general approach for computing probabilities.]
We can compute all the probabilities numerically, working one by one back from the target. Indeed, with a neat recursive formula like that one above, a spreadhseet is a handy tool (I happened to use Excel). Write the probabilities for P(T) to P(T+5) in a column (1 and 5 0's), and write the formula for P(T-1) in terms of those values:
Okay, instead of 1/6*sum(B5:B10) I lazily used the average function, but you see the idea. Copy that formula and paste it down as many rows as you like.
At large distances, the number converges to 2/7 ~= 28.57% (if the average roll is 3.5 spaces, it's perhaps not so surprising that the average probability you hit a space is 1/3.5).
In Excel, you can do neat plots to see what's going on:
In short, if you have some influence on where you try to land on the target from (such as the ability to change your roll by one, for example), the best spots (in order) are 6 spaces away, 5 spaces away and 11 spaces away, and of those, 6 is by far the best. Try to avoid being fewer than 4 spaces away, and 7 isn't much good either.
I used the approach outlined here to look at rolling two and three dice (2d6 and 3d6) as well; the direct calculation approach here adapts quite easily to these situations, yielding probabilities with a minute or so of effort.