Dice math.

[ichabod]

Here's a puzzle that I labored over quite a while ago and gave up on, but maybe someone here can solve?

Using only multiples of a single type of die per element (xd1, xd2, xd3, xd6, xd8, xd10, xd20, (optional xd30), xd100) and +/- modifiers, construct a set where the average roll begins at n<-[0,6] and increases linearlly to infinity, while the maximum possible roll increases at a greater rate.

That means increasing the number or value of dice over time and not just picking 1d4+1, 1d4+2, 1d4+3, ..., 1d4+n

I'm sure the value of such a set would be obvious to roleplayers (and rollplayers alike)

Well, you could do this easily with fudge dice: df, 2df+1, 3df+2, 4df+3, and so on. The average increases by one each time, and the maximum increases by two. I'm not sure if that quite fits your restrictions, but it works.

Let's see if I can fit your restrictions better, using commercially available dice: 0=1df, 1=2df+1, 2=1d3, 3=1d5, 4=1d7, 5=2d4, 6=2d5, 7=2d6, 8=2d7, 9=2d8, 10=4d4, 11=2d10, 12=3d7, 13=3d7+1, 14=4d6, 15=6d4,... Of course, this has a non-decreasing maximum, did you mean a strictly increasing maximum?

It works much better if you allow combinations of different dice. Otherwise you have to us +1's alot with the higher prime numbers.

 

[Altamont Ravenard]

If your funds are unlimited, why are you bothering gambling?

lol

The "winning" system is called a Martingale, IIRC. It "works", but casino's usually impose a maximum bet and will probably kick you out if you try this.

AR

 

[Cheiromancer]

Well, you could do this easily with fudge dice: df, 2df+1, 3df+2, 4df+3, and so on. The average increases by one each time, and the maximum increases by two.

How are the sides of a fudge dice labelled? From your description I am guessing [-2,-1, 0, 0, 1, 2].

 

[ichabod]

lol

The "winning" system is called a Martingale, IIRC.
AR

No, martingale is the mathematical term for that sort of game: a random process whose value is expected to decrease. A losing game, in other words. A submartingale is a random process whose value is expected to increase. There is a mathematical proof that if you have a finite stopping time, it is impossible to create a betting strategy that will turn a martingale into a submartingale, or something close to that effect. I'd have to check my notes to make sure I have the qualifiers right.

 

[ichabod]

How are the sides of a fudge dice labelled? From your description I am guessing [-2,-1, 0, 0, 1, 2].

Sorry, thought that had been covered. Fudge dice are labelled [-1,-1,0,0,+1,+1]. Although really they're labelled with two "+" sides, two "-" sides, and two blank sides, but that's how you read them.

 

[Ferret]

Why not highest+lowest/2. On an even die the other opposite numbers equal the highest+lowest, so it make no odds. This works with multiple ds. Watch:

10d10, 10+100/2=55(20+90/2=55, 30+80/2=55 etc.)

 

[Wolfspirit]

It's threads like these that allways make me sad when I look at my sheet showing how it took me 179 (? Might have to recount) rolls on a d20 to roll a 20.

I'd been recording my dice rolls in game for about a semester, with about 600 rolls recorded (changing out the die after 100), when the rest of my group convinced me to stop, since they allready acknowledged that I was unlucky. Sigh, I guess I should just bite the bullet and excel them out sometime so that I can do some statistics on them.

 

[Chaldfont]

Here's a question for you guys:

Instead of keeping track of wand charges, I would rather roll dice every time the wand is used. If some number is rolled, that was the last charge on the wand.

I have been making my players roll d100 and saying that a result of 01-02 means that the charges have been depleted--but this mechanic is based on just gut feeling. What is the most statistically accurate way to do this?

 

[MerakSpielman]

I don't like it. A crafted wand has a 2% chance of having only 1 charge before it's used up. Sure, if the party uses tons of wands, on average they'll last for 50 charges each, but if there aren't that many wands in the campaign, this could be funky.

 

[wocky]

Roll when they find the wand and keep track of the number

Your method is not good, since the wand has 1:50 chance of having 1 charge, but succesively smaller chances of having more. Anyway, without an in-depth analysis I'm drawn to say that there's no simple method that would give you the uniform distribution you're probably after.