I’ve done a little searching but can’t find a clear official answer for this. I understand that when you do Beginner’s Luck in a Versus roll that you have to double the other guy’s successes to beat him.
But how does that work when doing something like a Suppressive Fire vs. Direct Fire action in a combat where the number of excess successes of the winner matter. I see two ways of handling it. In both of these examples the Suppressive Fire action is the Beginner’s Luck roll.
Double the Direct Fire successes and subtract the Beginner’s Luck roll normally. So Direct Fire rolls 5 successes and the Suppressive Fire rolls 3 successes meaning the Direct Fire action has 7 successes in total.
Double the Direct Fire successes only for purposes of determining the Obstacle and only successes over the doubled value subtract from the original Direct Fire roll.
Example A: Direct Fire rolls 5 success and the Suppressive Fire rolls 3 successes. Since the Beginner’s Luck roll didn’t make it over 10 the Direct Fire scores all 5 successes (but not 7 as in the first case).
Example B: Direct Fire rolls 5 success and the Suppressive Fire rolls a whopping 12 successes! Meaning the Direct Fire action wins by 3. I rolled 2 over the doubled obstacle and thus reduces the original Direct Fire action by 2.
We half the unskilled successes and round down fractions. So in the first example, DF has MOS 4 over SF. Same as your calculation, actually.
Or you can double the skilled successes against the unskilled and count it like that – if I’m skilled and I roll 5 successes, you need to get 10 successes to tie me.
Remember that this doubling doesn’t generate extra successes for the side that is skilled. If I roll skilled Direct Fire with 6 successes and you roll unskilled Suppressive Fire with 2 successes, I come out of the exchange with a net of 5 successes, not 10 (which would be brutal and cruel). When using the “double skilled successes” method, make sure you halve the MoS (of either side) after the fact, otherwise you’ll end up with wonky test results. I think you’ll also need to round fractions up if the skilled person succeeds and round down if the unskilled person succeeds.
I like halving the unskilled successes before subtracting (Luke’s way really) for tests where MoS matters since it doesn’t have any weird math edge cases. When doing things where MoS doesn’t matter (a straight persuade-off for instance) I like to double the skilled successes before subtracting since it makes the math super extra easy and you never need to worry about rounding.
Yes. I see now.
Where Margin of Success matters it takes two unskilled successes to cancel one skilled success. That makes sense. Although I’m not exactly sure the book makes that clear.
Beginner’s Luck page 308 spells it out fairly clearly I think. The one thing I’m not sure of is if Beginner’s Luck MoS counts as full successes or half successes . For example, with the Direct Fire action, do you need an MoS of 1 or an MoS of 2 to buy a second shot opportunity for your squad?
That’s how we’ve been playing it (and the assumption I’d made when I answered above) but I realized afterwards that the language in the text could be interpreted either way, depending on if you consider it “using a success to do something” or if it it’s more of a post-roll modifier to your Ob. Thanks!