Jostero + Cruise Missile

By MrSunol, in X-Wing Rules Questions

Capitain Jostero
"Once per round, after an enemy ship that is not defending against an attack suffers damage or critical damage, you may perform an attack against that ship."

Cruise Missile:
"You may roll additional attack dice equal to the speed of the manuever you performed this round, to a maximum of 4 additional dice."

If another ship receives a damage before Jostero reveals his dial and moves, and Jostero chooses to attack that ship with Cruise missile, does he:
- A: Not roll any additional dice because he has not performed a maneuver yet?
- B: Roll dice equal to the speed of the last maneuver he performed (being the previous round)?
- C: Reveal his dial and add dice equal to the speed of the maneuver he is going to perform this round?
- D: Other?

My assumption is on A, but just wanted to post the question.

I would say A also. Cruise Missiles specifically says "this round".

Edited by Dr Zoidberg

Immediately after an enemy ship suffers damage (not from an attack) is the timing window. The timing window is the important part here.

A: If you haven't executed a manoeuvre yet, then there are no dice to add.

B: Cruise Missiles specifies "this round" (as per Dr Zoidberg's post), so you can't use the previous round's manoeuvre.

C: If you decide to wait until you reveal Jostero's dial, you've now passed the timing window for his ability, and can't use it.

So the answer is most definitely A.

A.

Arguably if you've not performed a move yet the speed of the move you performed is not a number so the calculation fails and there's no value for the number of dice to roll but i don't think that's how it would be ruled.

7 minutes ago, thespaceinvader said:

A.

Arguably if you've not performed a move yet the speed of the move you performed is not a number so the calculation fails and there's no value for the number of dice to roll but i don't think that's how it would be ruled.

But, there is a default number of dice to roll with Cruise Missiles, and that's a single attack die. If you haven't executed a manoeuvre yet, then it's 1+0=1 attack die only.

It's not 1+0 though. That would be firing them after a stop. E.g. with Inertial Dampers.

It's 1+null.

1 minute ago, thespaceinvader said:

It's not 1+0 though. That would be firing them after a stop. E.g. with Inertial Dampers.

It's 1+null.

Either way, it still leaves you with just one die.

7 hours ago, thespaceinvader said:

It's not 1+0 though. That would be firing them after a stop. E.g. with Inertial Dampers.

It's 1+null.

A distinction without a difference.

12 hours ago, sharrrp said:

A distinction without a difference.

Depends on the mathematical system or programming language you're using. In some systems, 1 + null = 1. In other systems, 1 + null = null. In still others, 1 + null is undefined and non-deterministic in its results. And in yet more others, 1 + null causes an immediate error and will crash everything if you don't handle the exception.

All that said, I'm on the side of A. Given that rolling additional dice is a may, not a must, I'd say you attempt to execute the may clause, discover that you can't because it results in a null value, and roll back to the decision point just like you do when discovering that the target you wanted to target lock is not in range. The end result is not even 1 + 0 or 1 + null; it's just 1 because you can't execute the may clause.

I'm on the side of A as well. Just so I'm clear.

D. is that Jostero decide to skip that opportunity, then activates, performs his maneuver and gets another shot later, now with bonus... Don't forget that the ability is optional and you do not have to resolve it immediately, important i.e. if your opponent is rolling for bombs damage and there is Jostero waiting with his shot - you do not have to use your first opportunity.

@EdgeOfDreams It is not 1+null, because you may add additional dice, while in that case you may not add dice because you did not perform any maneuver (condition for adding dice is not met), hence it is just 1 = 1.

On 16.08.2017 at 5:05 PM, sharrrp said:

A distinction without a difference.

Without difference in the end, though still fun to analyze in theory :) at least I find it fun...