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[Thanee]

I'm always wondering how this is supposed to work. (IMG:style_emoticons/default/smile.gif)

You can use Edge to negate a Glitch.

But can you also use Edge to gain extra hits, despite having rolled a Glitch (without negating it, of course)?

i.e. you roll 10 dice, 2 hits and a Glitch. Can you use Edge to re-roll 8 dice, in order to increase the number of hits you get (but still suffer the Glitch)?

Or does a Glitch automatically negate all other Edge-options (apart from negating the Glitch, that is)?

Bye
Thanee

[bannockburn]

Interesting. I always thought that you can negate a glitch by re-rolling dice without hits. Now I know better. I wonder if I'll keep that houserule. (IMG:style_emoticons/default/smile.gif)

[StealthSigma]

Interesting. I always thought that you can negate a glitch by re-rolling dice without hits. Now I know better. I wonder if I'll keep that houserule. (IMG:style_emoticons/default/smile.gif)

There's not enough evidence to prove one way or another.

[Draco18s]

There's not enough evidence to prove one way or another.

This. At my table, if you pick up failed dice and reroll them, they're no longer 1s. Glitches aren't determined until after the dice stop rolling.
So at our table you can "negate" a glitch by rerolling the dice.

Maybe it doesn't make sense "any other way" if the rule exists...unless it doesn't make any sense for the rule to exist.

It's like an if(false) { } statement in programming. It'll never execute, but it'll compile anyway!
(Most compilers will throw a warning that it won't ever execute, but there are valid reasons to do it).

(Similarly there's if(a != a) { }--read "if a value does not equal itself," which appears at first glance to always evaluate to false, but there is one case where it will evaluate to true: NaN.* All comparisons for equality to NaN return false. I.e. NaN == NaN will return false, so the inverse will return true).

*Special value, "Not A Number." Happens when you attempt to divide by 0 or similar.