### Author Topic: Formula derivation for finding arithmetic mean of an exploding die  (Read 2222 times)

#### Morgenstern

• Control
• Posts: 6108
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #15 on: November 12, 2012, 07:14:43 PM »
That's actually kind of what I mean. Most players don't even know what we're talking about here, let lone how LLS compares to ToR regarding deviation or thresholds. Just that both give more explosions. That's why people take both even though it doesn't make much sense to you.

*chuckle* Yup. My own petard. Grab a rope and pull, I'm going up.

It's probably as simple to fix as tack a special: entry on both of them, or somehow script the additional exploding face as a namd bonus (so named bonueses don't stack). Basically the Campaign quality would pretty much eliminate both of them as valid character options (no geat loss on LLS, small loss of semetry on TotR), but campaign qualities do that.
At your own pace: Do. It. Now.
How about some pie? - Heroes of the Expanse

#### Krensky

• Control
• Posts: 8142
• WWTWD?
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #16 on: November 12, 2012, 07:33:51 PM »
I'll have to figure out how to program the old High Roller core to see what it does to the numbers.

Looking forward to it .

Ok.... I THINK this is right.

For a d6, re-rolling the first 1 and exploding on 6s I get a mean of 4.84 and a deviation of 3.18. This is slightly tighter then a straight d6, but almost a whole pip higher on the mean.

http://anydice.com/program/198c
Code: [Select]
`function: highroller ROLLEDVALUE:n { if ROLLEDVALUE = {1} { result: d{} } else { if ROLLEDVALUE = {6} { result: ROLLEDVALUE  + [explode d6] }} result: ROLLEDVALUE }output [highroller d6] named "High Roller d6"output [explode d6] named "d6"`
For reference, I went back to the built in explode function since it's the same as default AD.
« Last Edit: November 12, 2012, 07:40:47 PM by Krensky »
We can lick gravity, but sometimes the paperwork is overwhelming. - Werner von Braun
Right now you have no idea how lucky you are that I am not a sociopath. - A sign seen above my desk.
There's no upside in screwing with things you can't explain. - Captain Roy Montgomery

#### Krensky

• Control
• Posts: 8142
• WWTWD?
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #17 on: November 12, 2012, 07:40:28 PM »
That's actually kind of what I mean. Most players don't even know what we're talking about here, let lone how LLS compares to ToR regarding deviation or thresholds. Just that both give more explosions. That's why people take both even though it doesn't make much sense to you.

*chuckle* Yup. My own petard. Grab a rope and pull, I'm going up.

It's probably as simple to fix as tack a special: entry on both of them, or somehow script the additional exploding face as a namd bonus (so named bonueses don't stack). Basically the Campaign quality would pretty much eliminate both of them as valid character options (no geat loss on LLS, small loss of semetry on TotR), but campaign qualities do that.

Well, since LLS is your whipping feat and thematically it adding to your swinginess doesn't make much sense (since the phrase isn't really used for when thing don't go your way) it could be switched to a pure bonus to the AD roll?
We can lick gravity, but sometimes the paperwork is overwhelming. - Werner von Braun
Right now you have no idea how lucky you are that I am not a sociopath. - A sign seen above my desk.
There's no upside in screwing with things you can't explain. - Captain Roy Montgomery

#### Morgenstern

• Control
• Posts: 6108
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #18 on: November 12, 2012, 08:31:27 PM »
Well, since LLS is your whipping feat and thematically it adding to your swinginess doesn't make much sense (since the phrase isn't really used for when thing don't go your way) it could be switched to a pure bonus to the AD roll?

*Points at Fortune Favors the Bold*

LLS is probably best off getting swapped to explode on 1 or removed entirely. TotR and the campaign quality implicity don't work together and may or may not need to explicity not work togeter. Whichever way you go with that carries over to the adjusted LLS.
At your own pace: Do. It. Now.
How about some pie? - Heroes of the Expanse

#### Krensky

• Control
• Posts: 8142
• WWTWD?
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #19 on: November 12, 2012, 08:38:39 PM »
I knew I was forgetting something.

TotR could be made a little more meta/narrative…

Maybe something like once a session/adventure spending x dice and giving a good story to make any failure really a success.
We can lick gravity, but sometimes the paperwork is overwhelming. - Werner von Braun
Right now you have no idea how lucky you are that I am not a sociopath. - A sign seen above my desk.
There's no upside in screwing with things you can't explain. - Captain Roy Montgomery

#### Morgenstern

• Control
• Posts: 6108
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #20 on: November 12, 2012, 08:42:35 PM »
I knew I was forgetting something.

TotR could be made a little more meta/narrative…

Maybe something like once a session/adventure spending x dice and giving a good story to make any failure really a success.

Sure. Something along those lines could be interesting, but ultimately it means you scrapped LLS (mechanically), and then picked the name off the wreckage to use for something else. It's a great name, so salvaging it is totally legit.
At your own pace: Do. It. Now.
How about some pie? - Heroes of the Expanse

#### Krensky

• Control
• Posts: 8142
• WWTWD?
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #21 on: November 12, 2012, 08:53:59 PM »
Actually, I meant Tales of the Rascal could be a little more Münchhausen.
We can lick gravity, but sometimes the paperwork is overwhelming. - Werner von Braun
Right now you have no idea how lucky you are that I am not a sociopath. - A sign seen above my desk.
There's no upside in screwing with things you can't explain. - Captain Roy Montgomery

#### Morgenstern

• Control
• Posts: 6108
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #22 on: November 13, 2012, 07:04:48 AM »
It took me a long, long time to appreciate that movie.

Hmm...
At your own pace: Do. It. Now.
How about some pie? - Heroes of the Expanse

#### Krensky

• Control
• Posts: 8142
• WWTWD?
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #23 on: November 13, 2012, 08:14:51 AM »
What I mean is it seems very flavorful for a feat called Tales of the Rascal would involve some storytelling to explain how what seems to be a failure is really a success. That the roll fails, but some narrative control currency and a tall tale makes it a success seems more Rascal than just mechanically twiddling the odds on your action dice, especially if LLS becomes explode on a 1.
We can lick gravity, but sometimes the paperwork is overwhelming. - Werner von Braun
Right now you have no idea how lucky you are that I am not a sociopath. - A sign seen above my desk.
There's no upside in screwing with things you can't explain. - Captain Roy Montgomery

#### Sletchman

• Control
• Posts: 4111
• Gentleman Scholar.
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #24 on: November 13, 2012, 09:35:39 AM »
I like that Krensky, but then I am very much in favour of action dice becoming an almost strictly narrative control mechanism.  So Chance feats would open up new / different avenues of narrative control or be strictly edge related.

#### Blankbeard

• Handler
• Posts: 781
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #25 on: November 13, 2012, 11:56:01 AM »
We gone around about this before.  I know it seems like what you're saying is true but the reality is that regardless of which number the die explodes on you're going reach the same result over the course of an evening or a career.  Exploding on a one doesn't get you to a total of five any more often than exploding on a six does.

Take a look at that statement again.

Target: 5 rolling a d6 without explosions. Chance of success 33% (2 outcomes in 6).
Target: 5 rolling a d6 with unlimited explosions on 6. Chance of success 33% (2 outcomes in 6).

Notice that while the average jumped considerably, the chance of success didn't move at all.

Target: 5 rolling a d6 with a single explsion on 1. Chance of success ~41.6% (15 outcomes in 36).
Target: 5 rolling a d6 with up to two explsions on 1s. Chance of success ~43.5% (94 outcomes in 216).
Target: 5 rolling a d6 with up to three explsions on 1s. Chance of success ~43.9% (569 outcomes in 1296).
Target: 5 rolling a d6 with 4 or more explsions on 1s. Chance of success ~43.98% (3420 outcomes in 7776).

I worked through the math on the sort of case you're looking at here.  Yes, in this case you get a higher chance of success if the target is smaller than the die size.  The trade off is that you lose any chance to succeed if you need a total of 10.  Is that better?  What if the target number is 8?  Exploding on a 1 gives you about a 1 in 1.67 million chance of success.  Is that better than exploding on a six?  So I guess the question becomes is it more important to be able to throw an action die when you need 5 more than when you need 8 more.

Since the choice on when to spend an action die rests with the player, they're generally going to spend them when they increase the chance of success.  I bet not too many of your players throw a d4 when they fail by 14.  But by 3?  Sure.  7 eh, not so much but maybe.  That's all the meatbag statistician at work.  Even if they never divide by 2 they're using knowledge of what that die is likely to throw.

The reality is averages don't reflect thresholds - because the results are not being summed. They are being converted in a binary operation to 0s (failures) and 1s (successes). In a more extreme case a series of ten results of 1, 1, 1, 1, 1, 1, 1, 1, 1, 11 yeilds an average of 2. Since the average is 2 and the threshold is 2, I succeeded half the time, right? Nope.

There's more than one type of average you know.  The median of your data set is 1 while the mode is just a bit higher, both characterizing the dataset as full of fail.  Traditional averages are best used with datasets that are normally distributed or flat distributions.  Die rolls fall into these categories.

Similar argument as you mentioned about crits, its not the average number of crits in a sample of 400 rolls that maters, its how many checks are required to give me a 50% chance that one has occured. When all results above X are equivalent to X, a result of 100 is no better than a result of X, and while a result of 100 pulls the average up, it does not intrinsically shift the likelyhood of success. Explosions are result replacements (instead of a six, I now had 1/6ths chances of a 7, 8, 9, 10, 11, or 12 replacing that result), and if they take place on results at or beyond the threshold they do not shift the likelyhood of achiving that threshold. If they takeplace on results below the threshold, they do.

What you're not taking into account is that there isn't just a maximum result, there's also a minimum result.  Reaching X+20 may not be any better than reaching X EXCEPT that my chance of success is now independent of my d20 roll!  Think about that.  If I have a +20 skill bonus, I can not fail on a DC20 task.  Even a 1 roll a 1 I still succeed.  If I get that certainty by spending an action die, I've effectively purchased a better than flawless ability for the low price of a single action die.  I really don't care if I end up beating the DC by 10 or 20 or 30 (except for bragging rights  ) but likewise I don't care if I fail by one or 10.  The tools I use to determine whether I want to spend that die are its range and its average.  Where the die explodes changes the range but it doesn't change the average boost that I'm going to get from spending dice over time.

Quote
I guess you're including rolls like autofire and the knife mastery trick in damage rolls.

Actually I mean straight linear conversions. Sword Supremacy may be the only one that survived into Fantasycraft.

I have a build that really wants to succeed by 9 or by 14 due to using Blade Fury and All-Out Attack.  With this sort of multiple threshold situation, I don't care about succeeding by 12, but I do care about 14.  Since I know what value to expect from an action die and I can decide if I want to use one after seeing the initial result, I know that if my action dice are d8 or larger I have a very good chance of bumping up the result to the next threshold.  Doing my knife damage again plus the eight from All-Out is a much sweeter deal than just rolling that die on damage.

Also damage and refresh rolls, opposed rolls, anything that counts effect.  In short, a lot of rolls depend on being able to get a large result.

Glass Half Full
Or empty if you prefer.  Players may select whether their action dice explode on the highest or lowest number the first time they roll one.  Once this choice is made, it may not be changed for a given player.

It's another thing to keep track of but it allows a choice between "anything goes" and "this won't suck."

#### Elvith Gent

• Recruit
• Posts: 30
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #26 on: November 13, 2012, 01:32:51 PM »
I have had a couple of requests to show how the mean value of an exploding die is calculated.

If you do not like math, stop reading now.

You have been warned.

(…)

Thus the mean value of a die with x sides and exploding on y faces can be found using the formula f(x, y) = (x + 1) / (2 * (1 - (y / x)))

Woah. That's simple. Congrats ! It's really, really clever while my calculations were kind of "brute-force". Now I don't think it will interest you that much, because it's essentially not succeeding into what you've succeeding

#### Blankbeard

• Handler
• Posts: 781
##### Re: Formula derivation for finding arithmetic mean of an exploding die
« Reply #27 on: November 13, 2012, 01:45:58 PM »
There's an even simpler way to find the same number.  Take the average of the die and multiply it by x/(x-n) where x is the number of sides and n is the number of faces the die explodes on.

So a six sided die would be 3.5 * (6/(6-1) = 3.5 * (6/5) = about 4.2