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Title: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 01, 2008, 10:09:43 PM
I just picked up Sorcerer, and I'm reading it through for the first time. I'm really pleased with it, and excited about starting a campaign, but there's a little effect of the basic dice method, as I'm interpreting it, that has me scratching my head.

If I have more dice than my opponent, it seems that the number of rollover dice I can win (assuming I get a really dominating roll) is limited to the number of dice my opponent has. Right? Consequently, if my opponent has fewer dice (is worse at the action), I'm able to win fewer rollover dice for the followup task.

So if I have five dice, and my opponent has four, and I beat the heck out of them on the roll, I can get as many as four rollover dice. But if the opponent has two dice, I can only get two rollover dice. The advantage I can get over a weaker opponent is smaller. This seems counterintuitive.

Am I missing something important? Am I misinterpreting something? Is this somehow intended in a really clever way I'm not getting yet? Is there a story-oriented way of ironing out this little wrinkle?

It seems like this is something that would come up a lot. Players (and the GM!) will want to get an edge on an important task by starting off with a task they can really dominate, to load up on rollover dice. It seems like the system is intended to support this (which I like a lot), but this kind of undercuts it.

I can't be the first person to notice this, so I must be missing something. Please tell me what!


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Christopher Kubasik on September 01, 2008, 10:32:20 PM
Hi Macmoyer,

The number of Victory Dice is equal to the number of dice the winner rolls over the highest, untied value of the loser.  The number of dice the loser rolled has nothing to do with it.

So:

If I roll 8, 7, 7, 4

and my opponent rolls a 4 and a 2 on two dice...

Then I have three Victory Dice.... and all of them can be rolled into the next, connected roll.

Does that help?

CK


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 02, 2008, 08:43:53 AM
Ah! That does help! That's exactly what I was missing. I was thinking you continued to pair-and-compare the dice, like in Risk. I think I read too many reviews that said, "The dice mechanic is just like Risk."

However, you've added the word "untied" to your interpretation that I don't see impied in the rulebook. The rulebook says, "The winner's degre of success is determined by the Victories: the number of dice that show higher values than the highest of the loser's dice."

If I roll 10 - 8 - 4 - 1, and you roll 10 - 7 - 4 - 2, I would win the roll, with zero Victories, according to the rulebook. I have zero dice that show higher values than your highest die, a 10.

But you say, "The number of Victory Dice is equal to the number of dice the winner rolls over the highest, untied value of the loser." In the example above, I would win with two Victories. I have two dice, my 10 and 8, over your highest, untied die, the 7.

So, are dice from high ties discarded before counting Victories? This seems intuitive, and also seems to be supported by the second example on p18: 8 - 7 - 6 - 1 gives two victories (not three) over 8 - 5 - 5 - 2 because "the first two 8's cancel out."


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: greyorm on September 02, 2008, 09:05:35 AM
I wish my copy of Sorcerer weren't packed away as I could point you to the pages with the rule spelled out, but ties are discussed in the rulebook: all ties are ignored; a die that ties does not count for counting victories. So your first example would result in one victory rather than two, and your sense about them being "discarded" is accurate. However, don't just remove ties from the table, as someone who cancels their action may roll for defense instead and the previous tied roll would be moot.

Here's quick, loose example for you:

Fred and Barney are fighting, taking swings at each other.
Fred has a Stamina of 5, Barney has a Stamina of 3.

Fred rolls: 8, 7, 5, 3, 3
Barney rolls: 8, 2, 2

The 8s tie, so they are ignored. Looking at the highest numbers (7 and 2), Barney would lose the roll and Fred would have four victories against him (7, 5, 3, 3).
Ouch. Barney decides to cancel his action and roll defense instead.
Barney rolls: 6, 5, 3

Barney still loses the roll (8 versus 6), but Fred only gains two victories in this case (8, 7).

Note how the 8 was ignored as a tie originally, but isn't now because Barney scooped the action that roll was for. No more tie, so it counts again against Barney's defense roll.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 02, 2008, 09:22:13 AM
Owww.

Raven, I can't say it any more nicely. What you wrote is not correct and will cause huge confusion.

If Fred and Barney are taking swings at one another, then their two rolls are not compared. At all. The two 8's do not tie, because they are not compared. Macmoyer, please ignore Raven's post in full. It is not correct.

Let's stick with the basic issue. First, yes - this dice system is not like Risk. If some reviewer wrote that it's "just like Risk," then that reviewer clearly had just stuck a lubricated Rabbit (http://en.wikipedia.org/wiki/Rabbit_vibrator) into his ear and switched it on, thereby distracting him from actually reading the rules.

Second, you asked,

Quote
are dice from high ties discarded before counting Victories?


The answer is yes. In your example, though, neither of your readings is quite right. The way to do it is to discount the tied 10's entirely, leaving behind the high 8 and the high 7. So the 8 beats the 7 with one victory.

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 02, 2008, 02:34:57 PM
Thanks, Christopher and Ron! And Raven, too, for taking a shot.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 02, 2008, 02:40:43 PM
Oh, and the exact, rabbit-influenced words (http://www.rpg.net/news+reviews/reviews/rev_1726.html) were, "highly reminiscent of Risk."


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 02, 2008, 03:33:38 PM
As Ralph just cuttingly reminded me by private message, there is one important thing that Raven was referring to, and I'll get to that when we discuss more complex resolution. Later.

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: greyorm on September 02, 2008, 08:28:32 PM
If Fred and Barney are taking swings at one another, then their two rolls are not compared. At all. The two 8's do not tie, because they are not compared. Macmoyer, please ignore Raven's post in full. It is not correct.

Hrm. Ok, please PM me or give me a call tomorrow sometime, Ron, because I am sitting here scratching my head and completely not grasping what/how I've screwed up the above example. And apologies, Mac, if I've caused any confusion. I'll stay out until Ron and I have had a chance to talk.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 03, 2008, 01:25:41 AM
Actually, I'd like to talk with Macmoyer for a while. I think the result will shake out into what you're looking for.

Macmoyer (I need a real first name, actually), we're good with the basic dice comparison process, right? Double-check (first is A, second is B):

9, 9, 8, 8, 7, 4, 3, 2, 1, 1 vs. 10, 8, 9, 4, 4
B wins with one victory

9, 9, 9, 7, 7, 6, 5, 5, 4, 2, 2 vs. 9, 8, 7, 6, 3
A wins with two victories

10, 9, 9, 9, 8, 7, 5, 2, 1, 1 vs. 10, 9, 9, 9, 7, 5, 4, 3, 3
A wins with one victory

And just to nail down the original question for good:
6, 6, 5, 4, 3, 3, 3, 1 vs. 2, 1
A wins with seven victories

How's this look? Making sense? Questions, comments?

Best, Ron




Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 03, 2008, 03:54:26 AM
10, 9, 9, 9, 8, 7, 5, 2, 1, 1 vs. 10, 9, 9, 9, 7, 5, 4, 3, 3
A wins with one victory



Hi Ron,
shouldn't the above example be 3 victories for A? The 10, 9, 9, 9 cancel.
Then we match 8 vs. 7, which is one victory for A. But then, we need to look next to see if B "stops the fall" of victories, and we find 7 vs. 5, for another victory for A. Then, 5 vs. 4, another victory for A. Then 2 vs. 3, so the process stops. ?-


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 03, 2008, 03:58:45 AM
No ok, my bad. I misread the sequence.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 03, 2008, 04:15:28 AM
It seems like you're still doing Risk, or reminiscent thereof. I should state it right out:

Once you've hit the relevant high value for the losing side, stop.

Does that help?

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 03, 2008, 04:27:25 AM
I guess I have it right.
The algorithm would be:
1) Sort the values for both sides from highest to lowest
2) Discard ties
3) One of the two sides has an highest die; that side is the winner.
4) Count the victories: count the number of dice that the winning side has which are larger than the highest die of the losing side.

So:
A = 10, 9, 9, 7, 5, 1
B= 10, 9, 8, 1

Remove ties:
A = 9, 7, 5, 1
B= 8, 1

A wins, with one victory (only 9 is larger than all the dice remaining to B)

Is this correct?

Cheers,
Antonio


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 03, 2008, 04:33:02 AM
A wins, with one victory (only 9 is larger than all the dice remaining to B)

Well, actually the sentence above is redundant, we only need check the highest die rolled by B.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 03, 2008, 04:57:36 AM
You got it, Antonio.

In practice, leave the high tied dice on the table. So where you say "discard," I say "ignore."

This is because sometimes a roll is being compared against more than one defensive roll. There are some other reasons when other rules or situations are operative, but that's not important here.

Macmoyer, how's it looking for you?

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 03, 2008, 05:11:48 AM
Ok, thanks! I have not had the opportunity to play, and I am glad the issue has been discussed.

I guess it may be interesting to show a case of Total Victory. By slightly modifying one of the above examples:

6, 6, 5, 4, 3, 3, 3, 3 vs. 2, 1

A wins with 8 victory, i.e. a Total Victory.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 03, 2008, 05:23:59 AM
Right. And so is:

8, 7 vs. 5, 5, 4, 4, 4, 3, 3, 2, 1, 1

In this case, A wins with 2 victories, also Total Victory.

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 03, 2008, 09:15:20 AM
Macmoyer (I need a real first name, actually), we're good with the basic dice comparison process, right?

Yup, all of those examples make perfect sense to me now.

My first name is Mac. Pleased to meetcha.

The algorithm would be:
1) Sort the values for both sides from highest to lowest
2) Discard ties
3) One of the two sides has an highest die; that side is the winner.
4) Count the victories: count the number of dice that the winning side has which are larger than the highest die of the losing side.

I think the sorting step is unnecessary. Just find the highest die, and go from there:
1) Grab your highest die, and compare it to your opponent's highest die. If one of the dice is better than the other, that player wins! Note the value on the loser's die, and skip to step 3.
2) If the highest pair of dice is a tie, set aside both dice. They don't count any more. At all. Go back to step 1.
3) Count how many of the winner's dice are better than the loser's best die. Don't count any that were set aside in step 2. That's how many victories the winner has.

Essentially, steps 1 and 2 decide who wins, and set a difficulty score for determining the degree of success in step 3. I can see this working pretty fast once my group gets it down (I'm planning to use six-siders):
"Roll 'em."
"I got a six."
"I got a six, too."
"I got another six."
"Dangit, I got a four."
"I've got three dice that beat that, so I've got three victories."

And I like that. It quickly gives you a degree of success, which guides narration, without somebody having to figure out a difficulty level before every roll. And the degree of success can mechanically influence followup actions, which gives you a real, mechanical motive to play up your strengths to get an edge before you try to do something you're not as good at... and that makes good story sense. That's how people really operate, and it's how fictional characters ought to operate. Especially demons and diabolists, who I see as defaulting to the dishonest approach to any situation. Sorcerer rewards coming at your opponent at an oblique angle, and that feels right.

I like to get some nuance like this out of a dice mechanic, but I get pretty impatient with dice mechanics that take a long time to resolve. It's rare to get both, and I hate it when a lot of dice-fidgeting interrupts a good, dramatic scene. I think the Sorcerer dice mechanic is going to give us what we want from it, and then get out of the way, and that's exactly what I'm after.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: macmoyer on September 03, 2008, 09:17:48 AM
I like to get some nuance like this out of a dice mechanic, but I get pretty impatient with dice mechanics that take a long time to resolve. It's rare to get both....

Ahem. I mean, it's rare to get one without the other.


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 03, 2008, 10:39:22 AM
Hi Mac and Antonio,

It looks like we've arrived. Mac, that phrasing and those comments about the system exactly match my priorities as a designer back in the early Sorcerer days, and I'm confident that you're really going to like it in play.

Anyway, we aren't quite done yet. There's a new step to take.

So far, we've only been talking about oppositional rolling. One guy is trying to do something, and the other guy may be (a) trying to stop him, (b) trying to avoid the effects, or (c) trying to do it first. In all cases, if the second guy succeeds, the first guy can't do it, or can't do what he wants with it. It's a zero-sum situation; success on either's part cancels the possibility of success for the other.

You know pretty much all there is to know about this kind of conflict, using Sorcerer mechanics, at this point. However, there is another kind of conflict, which I naively called "combat" back when writing the game, which is better termed orthogonal. If you have two or more characters acting at cross-purposes, and both, one, or neither might conceivably succeed, that's orthogonal. Timing is often important because the effects of success might knock a character out of the running before his or her action happens to.

Sorcerer mechanics are the fucking bomb when it comes to orthogonal conflict, and the fun part is that it's just as easy as oppositional; you just have to know how.

OK, A is now Abigail, B is now Bartholomew, and there's a new player, Hortense, as well. Abigail is punching Bartholomew, Bartholomew is punching Hortense, and Hortense is punching Abigail. Let's say they're all kung fu experts whch means it's OK to hit girls 'cause they're bad-asses too.

In orthogonal conflicts, defense doesn't have to be announced; it's implicit. It's only important to mention it if the only thing someone is doing is defense, in which case their presence in the conflict has shifted to oppositional. So in this case, no one is doing that, and defense is considered to be present without announcement.

Everyone rolls simultaneously. Abigail gets 10, 8, 8, 6, 6; Bartholomew gets 10, 10, 9, 2; and Hortense gets 9, 8, 6, 5, 3, 2 (I've set the pools at slightly different sizes). At this point, these are the rolls for the punches. They are not compared in the same way as all those rolls we've done so far.

However, their relative values do play a basic roll: order of actions. This is considered without reference to degrees of success. Bartholomew's punch is the fastest, Abigail's second fastest, and Hortense's last. First key point: do not touch the dice. Leave them right there where they lie. Everything we do now will have to use more dice.

OK, since Bartholomew's punch is connecting first, we turn to Hortense. She has a choice. Either she uses one die to defend, in which case she can leave those attack dice on the table where they are, or she aborts her standing attack, removing the dice on the table, to defend with full dice. The risk in the first place is that she might fail and her attack will either be canceled or weakened anyway. Of course, to avoid that risk and have a better chance at defending, she'll have to give up her attack. That's the choice.

Hortense's player can see Bartholomew's dice right where they are, and that's two 10's and a 9. Nasty. One dice is a risky choice in this case.

Anyway, whatever it is that Hortense chooses, she rolls the relevant amount of dice (one or six), and we see what happens to her. Now we move to Abigail's punch, which happens to land, oh, about right now. What happens to Bartholomew? He is in a better situation than Abigail was because his punch is already resolved. He simply defends with full dice. (In Sorcerer, once you've gone, you can use full dice to defend each time something comes at you, even if it's ten attacks or however many.)

After we see what happens to Bartholomew, finally, Hortense's punch is considered. That is, if she decided to keep it. If not, then the round is over. But let's say the following happened: she chose to defend with a single die and miraculoulsy pulled off a 10. So that means she lost by one victory (see the rules for one-die ties in the book, which give the low-dice person a bit of leeway). OK, what does that mean for her punch at Abigail?

First, let's look at the damage. Bartholomew was using a Fists attack, so Hortense took one temporary penalty and one lasting penalty from his punch, for a current penalty of two dice. Her attack dice are still sitting on the table, now technically penalized by two dice. Here's how to do it. Abigail rolls her full five dice to defend, plus two dice to represent the penalty, for a total of seven. Now we can see how well the punch works out.

Going into the next round, Hortense will have a single die penalty because the temporary one has been applied. Let's say her punch did hit Abigail for two victories (Abigail must have rolled really shitty on her seven dice). That means Abigail would be going into the next round with three penalties, one lasting and two temporary, based on the table for Fists attack.

So - how's that look, guys?

Best, Ron


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: rabindranath72 on September 04, 2008, 01:02:34 AM
Still trying to wrap my head around the whole process (I am still reading the books!), but I like that strategy has a definite weight in the final outcome, not only chance.
For essentially abstract games like RPGs, strategy should definitely have a place at the table, instead of tactics and miniatures.
Once I have a clearer picture of the rules in my head, I will come back.


Thanks and cheers,
Antonio


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Joel P. Shempert on September 04, 2008, 04:50:26 PM
I've got a couple of questions from the sidelines--how does "(c) trying to do it first" in Oppositional Conflict differ from an Orthagonal Conflict of, say, people punching each other? Is it only Orthagonal (and thus using the "abort to defend" rule) if the targets of actions are assymetrical?

And also, how do you know when to apply the Damage rules (that is, Lasting Penalties as opposed to just Rollover Bonuses)? To all Orthagonal conflict? To any conflict of either sort where physical "combat" is present? Or what?

Peace,
-Joel


Title: Re: [Sorcerer] A weaker opponent gives me fewer rollover dice?
Post by: Ron Edwards on September 04, 2008, 08:38:37 PM
Hi Joel,

"Doing it first" was stated in that case to mean that once it's done, it's done, in a zero-sum fashion, and not only that, but it will be done. If A does it, B can't, and vice versa; furthermore, there's no chance that neither will do it, and there's no reason to imagine that either would give up in the middle of trying.

A more nuanced or more complicated situation, in which both succeeding could conceivably happen (i.e .grabbing something before it can be set off), either might, or neither might, would be orthogonal. In some cases the zero-sum issue can seem valid, but timing is still important, especially when more people are involved with all of their actions too, for timing purposes at the least, and in case aborting seems involved (and when more than two people are involved, that can be more likely than one might think). It's a safe bet to say that whenever more than two agents are doing things at the same time, use the orthogonal approach.

Best, Ron