posted 01-27-2005 02:20 PM            

Started running a new pen and paper role playing game recently. It's got an interesting dicing mechanic that has kind of baffled me.

Not that I don't understand how to use the mechanic. The problem is that I can't 'feel' the odds very well in my head. In many game systems I could easily gauge how difficult I want to make a task to achieve.

e. g. I want a skilled thief to be able to open this lock 1 attempt out of 4, say, so I assign a difficulty to the dicing mechanic to set the math right. Knowing that a skilled thief-at a given level of play-will have +10 added to his skill roll, I just say the difficulty of the lock is say 25. Player must exceed 25 on 1d20+10 to open the lock. This will leave 15 possible rolls that won't open the lock(1-14), and 5 that will (15-20).

Bleh, enough background. Suffice to say, it would be nice to be able to guage the basic probabilities for this mechanic.... but got no clue. Perhaps one of the far more statistically/probability minded folk can help.

Ok, here's the system in a nutshell.

There is only one die type. The d20 (20 sided die, numbered 1 - 20).

There are two variables involved in a roll. All rolls are contests between a player and the GM.

Pd - the number of dice the Player is rolling.
Gd - the number of dice the GM is rolling.

Once the dice are rolled, player and GM compare highest die. The person who has the greatest value as their 'highest' die has succeeded, and generates a number of 'success points' = to the number of dice they have showing values greater than the loser's greatest valued die.

example:
Player rolling 5 dice gets: 19, 16, 12, 10, 5
GM rolling 4 dice gets 14, 12, 10, 10

In this case Player succeeds, and has two dice that count as successes. (The 19 and 16, as they both exceed the GM's 14).

In the case of a tie, you compare each persons next greatest die counting down until you find a winner. All tied dice are awarded to this winner as 'successes'. If a person does not have a die remaining, they are considered to have rolled a 0/zero for that comparison.

So... for given values of Pd and Gd... how can I get a rough feeling for how often a player should succeed, and also important, how many success points should be expected for given values of Pd and Gd?

I'm utterly lost on how to even start on that.

(understand completely if you delete this topic as it's hardly junk to debunk... but I thought it was an interesting problem to place before the panel.)

Dale
(edit, added bottom disclaimer about deletion)

posted 01-27-2005 03:02 PM            

quote:

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This will leave 15 possible rolls that won't open the lock(1-14), and 5 that will (15-20).

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posted 01-27-2005 03:43 PM            

quote:

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The problem is that I can't 'feel' the odds very well in my head. In many game systems I could easily gauge how difficult I want to make a task to achieve.

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Ok, you are playing the d20 system.

The Feel for how successful you will be is simple. Subtract all your bonuses for a particular task from the DC... that number determines the number on the d20 you must roll (or better) to succeed.

Remember that on a percentile, 1 pip of the d20 is 5%. So, if you have a modified DC of 15, that is a 70% fail rate (1-14), and a 30% success rate, since ties succeed. This is for each individual roll of the d20.

If you are rolling a highest die against highest die mechanic, the task is the same, except the DC in question is simply the highest die roll of the GM, and not some arbitrary number. All dice are calculated independantly.

posted 01-27-2005 04:21 PM            

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Originally posted by Quintain:
Ok, you are playing the d20 system.

The Feel for how successful you will be is simple. Subtract all your bonuses for a particular task from the DC... that number determines the number on the d20 you must roll (or better) to succeed.

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Nope, shouldn't have included the d20 example at all. Was just to show the sort of thing I'd like to be able to stick in my head with the Donjon mechanic (the business with the greatest die comparisons, etc)

quote:

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If you are rolling a highest die against highest die mechanic, the task is the same, except the DC in question is simply the highest die roll of the GM, and not some arbitrary number. All dice are calculated independantly.

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And therein lies the rub. I'm not at all sure how to set a difficulty.

Say, I'm generating an encounter where the party must open a locked door. I'd like the thief to fail an attempt or two on it to make it a dramatic scene. Hordes of critters coming down the hall, or water filling the passage, or some other mcguffin making opening the door a 'tense' moment.

So I want to be able to set the difficulty high enough that the players will see opening the lock as an important 'yay!' type event.

Now, I have access to the character sheets and can say... hmm.. the thief has a total # of 8 dice to roll on a 'pick lock' roll.

Now how many dice of difficulty do I assign to the lock to create a 'tense' but not 'impossible' event?

posted 01-27-2005 04:23 PM            

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Originally posted by Casper:
Think about that for a moment

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Yeah, brain damage... should be 15 rolls that won't open the lock (1-15, cause a tie doesn't succeed) and 5 rolls that succeed (16-20).

posted 01-27-2005 04:25 PM            

Each die is equally likely to contain the highest number. That means that the probability of winning is the relative number of dice being thrown by that player (5/9 in your example).

posted 01-27-2005 04:27 PM            

quote:

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Originally posted by dwargon:

Now how many dice of difficulty do I assign to the lock to create a 'tense' but not 'impossible' event?

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Sheesh, a GM with a concience. Pah.

posted 01-27-2005 04:42 PM            

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Originally posted by Victor:

Each die is equally likely to contain the highest number. That means that the probability of winning is the relative number of dice being thrown by that player (5/9 in your example).

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Ok, that makes a nice bit of intuitive sense that I'd not seen.

So for just determining a victor of the dicing contest, a simple ratio of 'dice you are rolling' to 'total dice being rolled' should be a rough estimate. Bah, thats a bit easier than I was thinking. Sometimes the simple stuff is what breaks your head. :-P

So with a player rolling 8 dice of a lock picking skill:

I can:
**Assign 4 dice of difficulty to make the task something he should expect to be able to do 2/3rds of the time.

**give it an 8 dice rating for a 50/50 chance each round, or

**assign 16 dice to the difficulty, and make him sweat but still have an ~1/3 chance of succeeding?

However, I'm wondering if ties won't be significant enough to throw this estimate off. And, what - if anything - can I expect to see in the way of success generation?

The successes generated by a dicing test are important, so that would be a useful thing to have a feel for as well.

Dale

posted 01-27-2005 04:49 PM            

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Originally posted by Victor:
[BEach die is equally likely to contain the highest number. That means that the probability of winning is the relative number of dice being thrown by that player (5/9 in your example).

Each of the remaining dice is equally likely to have the second highest number. The probability of scoring a second point is therefore 4/8 or 0.5. Calculations for the remaining points are similar.[/B]

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posted 01-27-2005 04:58 PM            

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Originally posted by tralfaz:
One more thought: With large #'s of dice or more successes needed, the ties become more important & math gets quite ugly. Write a program & run some experiments with a random number generator. Screw the math. Take the easy way out.

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Good part is that the majority of the time, you only really *need* 1 success.

More successes give you bonuses on future actions and an ability to manipulate the game environment somewhat. Thus, while I'd really like to have a feel for # of successes, the simple ratio (doh, how'd I miss that) is fairly good for most actions.

I do plan on simulating out rolls of up to 20 dice per side though, and seeing how significant ties become on the higher end, though.

Dale

(edit - someday, I'll put on the screen whats in my head without multiple revisions.... nah)

posted 01-28-2005 01:03 AM            

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Originally posted by Sam Mc Kee:
NERDS!

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posted 01-28-2005 08:00 AM            

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Originally posted by KGB:
Jealous?

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Nah. He's just trying to assuage his own dorkiness.

Clicky

No, you can't hide, too many people with good trigger memory. And, as I pointed out in another thread, no-one here is not a nerd/dork/geek, etc.

posted 01-28-2005 12:01 PM            

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Originally posted by LaneH:
Nah. He's just trying to assuage his own dorkiness.

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I vote we start calling him Rose.

posted 01-28-2005 03:07 PM            

If you really want to assign difficulties to various RPG tasks, and maintain a "fun gaming" atmosphere, then you should throw out the dice and use beer bongs and a stopwatch.

The really difficult stuff can be decided with tequila shots and a dart board.