This could very well be clear to everyone else, but I thought it worth posing to the DH crowd.

How does one calculate Degrees of Success for the winner of an Opposed Test (p24)? I'm wanting to double-check that the DoS has nothing to do with a comparison between the results of the winner and the results of the loser of the Opposed Test.

Do you simply calculate the DoS per the winner's result compared to what his Target Number was?

EDIT: Just comparing successes between participants is enough for most situations. Only when the number of successes matter do you subtract the victor's [Degrees of Success] from the loser's DoS, such as seeing how effective a [Disarm] attempt is. The result is the number to use for the winner's DoS.

If you need degrees of success on an opposed test you take the degree of success of the winner minus the degree of success from the looser, always having at least one degree of success.

So if you have 3 degrees of success on your Focus power action for your force sword and your opponent have 1 degree of success you score 2 times 1d10 damage. Note degrees of failures does not count.

At least that is how we are used to play the opposed roll. However, the rules does not really state anywhere that the degree of success from the looser should be subtracted, so I guess we could use a clarification.

Appreciate the ideas - I've been using the difference as well, but then decided to double-check whether that's actually how it's done and couldn't find anything on it in the RAW.

I don't think it's in there, but if anyone has any RAW citations on Opposed Test DoS, please post them....

Do you have specific effects that you want to resolve? Even [Disarm], which I thought implied subtraction, simply states that the bonus condition requires three or more DoS. Similar, but not quite the same as scoring three DoS over the opponent. Take an Acolyte with 70 WS that rolls a 49 on his test. He would be able to steal the opponent's weapon even if he won by one or two DoS since he reached the three DoS to satisfy [Disarm]. Maybe subtraction isn't a thing after all.

P. 24.

Whoever succeeds wins the test.

If both succeed at the test, the one with the highest DoS wins .

If the number of DoS is equal, the one with the highest characteristic wins.

If still tied, the one who rolled the lowest wins.

Should both fail, it ends in a stalemate, or nothing happens, or both parties reroll until there is a winner. GM's discretion.

Question, why would you need to calculate DoS by comparing?

Edit: that is - comparing against the other party in the opposed test.

From my understanding, the original post's question is referring to how to calculate DoS on Opposed Tests. As an example, an Acolyte scores 3 DoS while his opponent scores 2 DoS. Then, for the purpose of determining DoS, is it more right to accept the Acolyte's full 3 DoS or should it only be 1 DoS to account for the opponent's tests? I believe this is the issue, but I am not sure when this difference would become relevant off the top of my head.

An example would be using the Interrogation Skill, page 105.

The Interrogator makes an Interrogation Skill test, based on Willpower, against the detainee's Willpower in an Opposed Roll. If the Interrogator succeeds, they get the answer to a number of questions equal to the DoS.

It's a good example, because for Interrogation the number of DoF of an Opposed Test come into play, as well.

If you went straight from RAW, the only DoS/DoF calculation method is to simply use the DoS of the winner period - or the DoF of the loser period (which would come into play with Interrogation Opposed Tests). I cannot find anywhere in the rules that the DoS is calculated from the difference of two opposing rolls - if anyone can find such a section, please post here.

In general, here are some possible ways of calculating the DoS/DoF from an Opposed Roll:

1. Calculate it per RAW, using the DoS of the winner only.
2. Use the DoS of the winner minus the DoS of the loser. Disregard loser's result if they had no DoS.
3. Add the DoS of the winner to the DoF of the loser.

My point exactly, I've been over the book several times and by RAW it is "option" 1.

As far as RAW goes, and as far as I know from the developers, if both are succesful, the lower DoS are NOT subtracted from the higher DoS, but the winner takes it all with his full DoS.

I think it can rather be seen as a kind of saving throw.

Question, why would you need to calculate DoS by comparing?

Edit: that is - comparing against the other party in the opposed test.

It is clear who wins, but once you won do you have the original DoS as your success or do you have the difference as your success. This matters for instance when using a force sword with opposed willpower check, where each success equates a d10 damage.

See above.

While i figure i might be a tad late for the party i just wanted to chime in since i haven't seen anyone post the errata statement from OW which also deals with DoS on opposed tests:

"Question: How does a character calculate his final Degrees