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From what I understand from the rules, each participant in the contest would cancel each other's Mastery, leaving each with a skill of 1. Therefore, each side would be attempting to roll 1 on d20, any success effectively being a critical. This also seems like a colossal waste of time, trying to roll 1s on a d20... Is this true, or am I being dense?

I've read through many examples on many sites and rules, and all of them always have an example such as 10W vs 15W dropping down to 10 vs 15...

Thanks for the insight...

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2 Answers 2

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There's an opposed resolution table on page 28. For a simple contest, if both participants fail, the lower roll has a marginal victory; otherwise it's a tie.

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  • \$\begingroup\$ Ah, yes, forgot about that. Okay, if I am understanding correctly, even in an Extended contest, each failure vs failure, the lower roll wins with one marginal success or one RP. I can live with that. Thanks. \$\endgroup\$
    – Hans My Hedgehog
    Commented Sep 12, 2010 at 15:30
  • \$\begingroup\$ I still find it a bit odd, however, to have the only way of a success also be a critical - a role of 1 on d20. \$\endgroup\$
    – Hans My Hedgehog
    Commented Sep 12, 2010 at 15:36
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    \$\begingroup\$ Even skill is the least useful comparison in HQ. It helps to just remember that each mastery is 20 levels of skill; 1W1 vs 1W1 is not 1 vs 1, but 21 vs 21, and each is, in fact getting a success or a critical, but their skill cancels out. \$\endgroup\$
    – aramis
    Commented Sep 12, 2010 at 16:47
  • \$\begingroup\$ Good point, aramis. Thinking of it like that, maybe in terms of Pendragon with two skills of 21 vs 21, makes much more sense. I guess my brain was on holiday. \$\endgroup\$
    – Hans My Hedgehog
    Commented Sep 12, 2010 at 17:33
  • \$\begingroup\$ I believe, also, that functionally it doesn't make a difference if you subtract less than 20 from both participants, as long as (a) you remove an equivalent number of masteries for both sides, and (b) you subtract the same total value from both. The contest, mechanically, should function similarly if you give both sides a value of 11 or 16. All this will do is turn the widest range of interpretation from simple failures (2-19) to simple successes (2-16, in the case where you give both sides 16). \$\endgroup\$
    – Viktor Haag
    Commented Sep 13, 2010 at 12:52
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Working from memory... If it still works like the older edition, look at the opposed resolution table; if both fail, the lower roller still gets a partial success. So the roll still matters, it's just not as obvious.

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