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Average of exploding dice?

InVinoVeritas

Villager
Agamon said:
Okay, smarty pantses, what are the odds your die will explode when you roll it?

For those that don't know, I mean blow up good.
My dice explode every time I roll them. I don't know what's wrong with yours. :cool:
 

Nifft

Penguin Herder
My pet peeve with exploding dice is the holes in the distribution. For example, you can't roll exactly 6 on an exploding d6 -- or any multiple of 6.

I came up with something which I've been calling "d5+1" -- on a natural 6, you get 5 + a re-roll. The expected value of any roll is exactly 4, and all results are possible.

Cheers, -- N
 

Christian

Explorer
Nifft said:
My pet peeve with exploding dice is the holes in the distribution. For example, you can't roll exactly 6 on an exploding d6 -- or any multiple of 6.

I came up with something which I've been calling "d5+1" -- on a natural 6, you get 5 + a re-roll. The expected value of any roll is exactly 4, and all results are possible.
Although it's less intuitive to think about, the math is actually easier, cause you're always adding a number from one through five to a multiple of five, which is simple in our base ten counting system ...
 

brehobit

Villager
freyar said:
Huh, how does that second term work? I must think in too straightfoward a manner. ;)
Think of it this way:
The expected value is equal to the expected value of a single (non-exploding) roll plus 1/6th of the expected value (you're just doing it again one sixth of the time).


Mark
 

brehobit

Villager
InVinoVeritas said:
It's freyar's result, without the geometric series.
Yep, and the same as mine other than you separated the non-exploding and exploding "first" roll and I put them together. (Plus you _solved_ the equation unlike lazy me :)

It's always fun to see how people think about things like this...

Mark
 

Lanefan

Victoria Rules
Nifft said:
My pet peeve with exploding dice is the holes in the distribution. For example, you can't roll exactly 6 on an exploding d6 -- or any multiple of 6.
Agreed.
I came up with something which I've been calling "d5+1" -- on a natural 6, you get 5 + a re-roll. The expected value of any roll is exactly 4, and all results are possible.
That's exactly how we do it; our term for it is "open-ended roll". We worked out that on any size die the average goes up by 0.5. We used this a lot in one game where critical hits were scaled back to near nothing and instead all damage rolls of any kind were open-ended.

Lanefan
 

brehobit

Villager
Nifft said:
My pet peeve with exploding dice is the holes in the distribution. For example, you can't roll exactly 6 on an exploding d6 -- or any multiple of 6.

I came up with something which I've been calling "d5+1" -- on a natural 6, you get 5 + a re-roll. The expected value of any roll is exactly 4, and all results are possible.

Cheers, -- N
Another trick is to have the reroll number be one less than the max.

The average stays the same, but the distribution tightens a little and the holes go away.

Not sure if it's easier, but it works.

Mark
 

Nifft

Penguin Herder
brehobit said:
Another trick is to have the reroll number be one less than the max.

The average stays the same, but the distribution tightens a little and the holes go away.
Holes go away? How do you end up with a value of 5?

All you're doing is pushing the holes around. There are still dips in the PDF, they're just not all zero any more.

Cheers, -- N
 

Blackrat

Villager
Agamon said:
Okay, smarty pantses, what are the odds your die will explode when you roll it?

For those that don't know, I mean blow up good.
Well this page answers that quite good in peasant terms http://plato.stanford.edu/entries/qt-quantlog/ . Ofcourse I'm so incredibly smart that I couldn't explain it in terms anyone here would understand so I had to post a link to this page :lol: :lol: :lol: :lol: :cool: .

To give a numerical answer: The probability that your dice blows up is 50%. It either does or does not. :D
 
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interwyrm

Villager
Nifft said:
Holes go away? How do you end up with a value of 5?

All you're doing is pushing the holes around. There are still dips in the PDF, they're just not all zero any more.

Cheers, -- N
I think the suggestion was...

1
2
3
4
5
6+0
6+1
6+2
6+3
6+4
6+5+0
etc.
 

Gez

Villager
interwyrm said:
I think the suggestion was...

1
2
3
4
5
6+0
6+1
6+2
6+3
6+4
6+5+0
etc.
Which accounts to about exactly the same thing.

If they roll 6 then 1,
Nifft has (6-1) + (1)
Brehobit has (6) + (1-1)

Whether you substract one from the exploding die or the reroll die doesn't change anything to the total sum...

(Now, it would be different for a system like Ars Magica's where it's a multiplication rather than an addition.)
 

brehobit

Villager
Nifft said:
Holes go away? How do you end up with a value of 5?

All you're doing is pushing the holes around. There are still dips in the PDF, they're just not all zero any more.

Cheers, -- N
You win. 5 is missing, so is 10, 15, etc.

Ah well, thinking isn't my strong suit.

Mark
 

Nifft

Penguin Herder
brehobit said:
You win. 5 is missing, so is 10, 15, etc.
Don't feel bad. I didn't notice about 6 (and 12, 18, etc.) being missing until someone else pointed it out. :)

Seeing the bumps & valleys in the distribution is really helpful.

Cheers, -- N
 

Gez

Villager
First I inferred the same thing as interwyrm.

But what brehobit meant was that if you roll 1, 2, 3, 4 or 6 it's what you got; while if you roll 5 you reroll a die and add it to 5. So you can get:
1
2
3
4
6
5+1, 5+2, 5+3, 5+4, 5+6
5+5+1, 5+5+2, 5+5+3, 5+5+4, 5+5+6
5+5+5+1, etc.
 

Grakarg

Explorer
Holy Thread Resurrection Batman!

Ok fellow math-masters, I need your help on this, and I can't seem to find a clear concise answer anywhere...

What is the average result for exploding dice when you're rolling multiple dice?
Example, whats the average result when rolling 2d6 vs 4d6 when open ended dice are involved?
What about when different die types are included, like 2d6+1d8+1d12?

Thanks for your input!

If you care, here is why I want to know:
I'm running a Super Hero game using Savage Worlds. I'm having trouble gauging the hero's power level vs opponent's toughness and damage dealing potential. Mainly because you start to stack multiple dice and dice types that can all explode. If I can ballpark a character's average, I can set targets that will provide a more interesting encounter, instead of the opponents getting blasted apart like wet tissue paper by the heroes; or destroying the heroes with a sideways glance by my villain.
 

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