Donjon: Clarification on determining successes.

[The Goat]

Ran my first session last night, it did however prompt this clarification.

If a player rolls 9 dice, and I am rolling 3, we compare and they beat all of my rolls does that mean they get 9 successes?

ex.

player rolls

20, 19, 18, 15, 15, 10, 3, 2, 1

I roll

10, 5, 2.

20 beats 10 one success, 19 beats 5 second success, 18 beats 2 third success, and since I don't have 6 more dice do they get all nine as successes?

Am I doing this right?

[Luke Sineath]

No.  I'm pretty sure they only have 5 successes.  They win because their highest is greater than your highest.  They get 5 successes because 5 of their dice come up greater than your 10.  The 6th number, 10, ties your highest, but tie-breaking rules only come into play for the highest (first) numbers.

[Christopher Weeks]

I agree with Luke.  And to highlight what he wrote there at the end, if you added a 20 to each player's roll you'd still have the five successes because one 20 from each player would cancel and you'd figure it as if they weren't there.  If, instead, you added a 19 to each player's roll, there would only be one success because the one 20 would be greater than the opposing 19.

Chris

[Trevis Martin]

Luke and Chris are correct.  In practice, my group usually rolls then compare highest die.  Whoever has the highest die wins, and the number of dice the winner has that show a value higher than the losers highest reflect the magnitude of success.  If for some reason both parties have the same number on their highest die, then you look to the next highest and so on until one player is the winner.

There is a little tweak to the donjon dice that a lot of people miss.  Once you have figured out the winner, every die that is higher than the die used to determine the win or loss counts as a success.

Example:

20,18,14,11,10,10,9,5 rolled by Bob

19,17,11,10,10,5,4 rolled by Sam

Bob wins (look at the highest die, Bob beats Sam.)  Bob only has one success (he only has one die higher than Sam's highest die.

Example 2:

19,18,15,14,11,11,6,5,1 rolled by Bob

19,18,12,10,10,9,8,6 rolled by Sam

Bob wins again. Highest die is tied at 19, so we look at the next pair.  It is also tied at 18, so we go to the next pair where Bob's 15 beats Sams 12.

Now for the tricky part. Every die of the winner that is higher than the die used to determine the victory counts as a success.  In other words Bob takes both his normal successes that are higher than Sam's 12 (14 and 15) and also his tied dice count as successes.  

So in this case Bob wins with 4 successes.

Hope that helps.

Trevis

[Christopher Weeks]

Seriously?  I thought it was just like Sorcerer.  I wonder if my copy is packed...

Chris

[Mike Holmes]

Nope slightly different than Sorcerer as Trevis correctly points out. I'm certain of it because it was the result of some of the playtesting that resulted in the rule working that way. In fact, it's very important, otherwise you wouldn't get the number of successes that you really need to drive the system.

Mike

[The Goat]

Thanks guys, after posting this I re-read the book and it's pretty clear that this is how it works.  I guess I was running it like sorcerer.  I thought something was hinky when the spell caster was routinely getting 8+ successes on his gather power rolls and I smashed a PC twice, didn't seem right.

Thanks for the clarification.

Note to self, improve reading comprehension :)

[Christopher Weeks]

I guess I was running it like sorcerer.

Um, no.  Scanning back up to your example:

player rolls

20, 19, 18, 15, 15, 10, 3, 2, 1

I roll

10, 5, 2.

20 beats 10 one success, 19 beats 5 second success, 18 beats 2 third success, and since I don't have 6 more dice do they get all nine as successes?

That's not how Sorcerer resolves the dice, either.  In the case of your example, the two systems are the same.  If there had been a tie at the top, then the Donjon rule would provide _more_ successes than the Sorcerer rule.

Chris