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46709 Posts in 5588 Topics by 13297 Members Latest Member: - Shane786 Most online today: 109 - most online ever: 429 (November 03, 2007, 04:35:43 AM)
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Author Topic: [Sorcerer] A weaker opponent gives me fewer rollover dice?  (Read 3562 times)
macmoyer
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Posts: 6


« 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!
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Christopher Kubasik
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Posts: 1159


« Reply #1 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
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"Can't we for once just do what we're supposed to do -- and then stop?
Lemonhead, The Shield
macmoyer
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Posts: 6


« Reply #2 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."
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greyorm
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My name is Raven.


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« Reply #3 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.
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Rev. Ravenscrye Grey Daegmorgan
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Ron Edwards
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« Reply #4 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 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
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macmoyer
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Posts: 6


« Reply #5 on: September 02, 2008, 02:34:57 PM »

Thanks, Christopher and Ron! And Raven, too, for taking a shot.
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macmoyer
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Posts: 6


« Reply #6 on: September 02, 2008, 02:40:43 PM »

Oh, and the exact, rabbit-influenced words were, "highly reminiscent of Risk."
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Ron Edwards
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« Reply #7 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
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greyorm
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My name is Raven.


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« Reply #8 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.
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Rev. Ravenscrye Grey Daegmorgan
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Ron Edwards
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« Reply #9 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


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rabindranath72
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Posts: 26


« Reply #10 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. ?-
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rabindranath72
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Posts: 26


« Reply #11 on: September 03, 2008, 03:58:45 AM »

No ok, my bad. I misread the sequence.
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Ron Edwards
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« Reply #12 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
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rabindranath72
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Posts: 26


« Reply #13 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
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rabindranath72
Member

Posts: 26


« Reply #14 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.
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