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So... do I have this down correctly? Regardless of where a re-roll a 1 occurs (hit vs. wound), it's going to increase your average success rate to 7/6 what it would otherwise be?

Hitting on 6s: 1/6 -> 7/36 increase of 7/6
Hitting on 5s: 2/6 -> 7/18 increase of 7/6
Hitting on 4s: 3/6 -> 7/12 increase of 7/6
Hitting on 3s: 4/6 -> 7/9 increase of 7/6
Hitting on 2s: 5/6 -> 35/36 increase of 7/6

Yes, the maths checks out.

That depends on what you mean by "what it would otherwise be".

But in general, yeah it amounts to +1/6 the number of expected successes for that roll (so if you were expecting 6 hits, now it's 7). So in general you want to use it either on lots of dice, or on stuff that already has a high success rate.

Note that all other things equal though, re-rolling hits is better than re-rolling wounds. Because hits have a larger dice pool, which gives more opportunities to re-roll, which pushes more total dice down the line. Re-rolling wounds is for things that auto-hit, or that are already re-rolling hits (because you can't re-roll the same roll twice).

But if your wound-roll has a significantly higher success rate than your hit-roll (like 4+ to hit and 2+ to wound), you'll probably want to re-roll 1s on the 2+ because it'll multiply the already-high success rate. If you have a full re-roll available you might want to put it on the 4+, because it'll benefit more from being able to re-roll 2s and 3s, and push more dice to the wound stage.

 ross-128 wrote:
That depends on what you mean by "what it would otherwise be".

I meant that, say there's a 1/4 average chance that a given attack forces a save. A re-roll of 1 to hit or 1 to wound will in both cases move that average to 7/24?

Yeah, multiplication is commutative. So unless there's some sort of extra effects like cool things happening on a wound roll of a 6 or bad things happening on a 1, it doesn't matter which you re-roll.

Cool, just wanted to be sure I wasn't missing anything obvious. Thanks!

Normally your statistic of getting a roll is X/6 where X is the number of successful results. for example in a 3+ you have 4 numbers that are a success 3, 4, 5, and 6. So the statistic is 4/6 or about .667.

The chance of rolling a 1 is always 1/6. If you make that 1/6 you then have a X/6 chance of getting a success. So your statistic is 1/6(x/6). For example for a 3+ you have a 1/6(4/6) chance of success after rolling a 1. Or .111.

You can add these two together giving the equation (X/6) + (1/6)(X/6). For our example of 3+ you have (4/6) + 1/6(4/6) which is about .778.

You can simplify the equation. (X/6) = (6/6)(X/6) when you put that in the equation you get (6/6)(X/6)+(1/6)(X/6) = (7/6)(X/6)

So we see that the chance of a success when rerolling 1's is equal to (7/6) times the normal success rate.

I suppose I was overthinking it a bit, because I was thinking "you can't roll a fraction of a die, so you'll probably want to front-load your bonuses to turn those fractions into whole dice".

But in general, yeah, it's close enough.