average result from an open ended dice roll

[Nifft]

Using Nifft's open-ended version, does this bring the average result of 2d6 up to 8? 3d6 to 12?

Or is the result more complicated?

Nope, that's exactly right.

The expected value of each d6 rolled is exactly 4. Makes it really easy to replace static bonuses with dice -- every +4, cut off the bonus and add +1d6.

Cheers, -- N

 

[Nifft]

Here's another problem for the maths-heads.

Tunnels & Trolls used an open-ended rolling system for saving throws. Roll 2d6, if you got a double (any double, even snake-eyes), you got to roll again and add the two rolls together.

And as long as you rolled doubles, you could carry on rolling and adding.

What would be the average roll for that situation?

You get a reroll 6/36 times, so it'd be 7 + 7/6 + 7/(6^2) + 7/(6^3) ...

Works out to about 8.4

Cheers, -- N

 

[silentounce]

You get a reroll 6/36 times, so it'd be 7 + 7/6 + 7/(6^2) + 7/(6^3) ...

Works out to about 8.4

Cheers, -- N

Heh, this is all much easier than you guys are making it out to be.

http://www.rpg.net/columns/rollthebones/rollthebones2.phtml
Scroll down to open ended rolls.

M = M0/(1-p))

Where M is the average you're looking for. M0 is the average without rerolls and P is the probability of rerolling.

So, the 2d6 rerolling all doubles looks like this M = 7/(1-1/6)

You are correct it is 8.4, but not "about", it's exactly 8.4.

We've done all the averages in depth in the Vorpal Weapons thread.

d4 3.33
d6 4.2
d8 5.14
d10 6.11
d12 7.09

As you can see, the larger the die size, the less chance of explosion and the lesser the increase in average.

Finally, about the save. You'd do it like this: twenty possibilities, 9 of them result in one check. So, the average amount of checks after one roll is .45. The reroll probability is also .45 becaues you'll only reroll on a failure.

M = .45/.55 = ~82% or in this case that means .82 checks or .82 rounds of effect.

 

[Wulf Ratbane]

Nope, that's exactly right.

The expected value of each d6 rolled is exactly 4. Makes it really easy to replace static bonuses with dice -- every +4, cut off the bonus and add +1d6.

Cheers, -- N

Thanks.

Yeah, that's really sweet.

As long as I have your eyeballs here, can I hit you for a couple more?

How does "drop the lowest" figure into the calculations?

For example, 4d6 open ended (assume when I say open ended, I mean your version), drop the lowest, and add the remaining 3d6.

Not an example chosen at random, obviously-- but I would be interested in seeing the calculations.

 

[silentounce]

Thanks.

Yeah, that's really sweet.

As long as I have your eyeballs here, can I hit you for a couple more?

How does "drop the lowest" figure into the calculations?

For example, 4d6 open ended (assume when I say open ended, I mean your version), drop the lowest, and add the remaining 3d6.

Not an example chosen at random, obviously-- but I would be interested in seeing the calculations.

http://www.rpg.net/columns/rollthebones/rollthebones5.phtml

Scroll down a bit to Adding a subset of the rolled dice.

 

[Wulf Ratbane]

http://www.rpg.net/columns/rollthebones/rollthebones5.phtml

Scroll down a bit to Adding a subset of the rolled dice.

And then what? That section doesn't answer my question.

Seriously, thanks for pointing out that thread, it's an interesting read, but please do me the favor of actually reading this thread and the issues raised here before immediately pointing folks over there.

 

[Nifft]

http://www.rpg.net/columns/rollthebones/rollthebones5.phtml

Scroll down a bit to Adding a subset of the rolled dice.

Did this answer your question? It seems light on details.

Basically, when you roll two dice and take the maximum value, you're allowing large values to cover small values, which slants the distribution towards the high end.

Consider the case of max(2d20). You have 400 possible results, but if either die is a 20, the other die won't matter -- a roll of 20 "covers" all nineteen lesser values. This means there are 39/400 cases where the end result is 20, as opposed to 1/20 cases normally -- in percentage terms, you can expect max(2d20) to yield a 20 result 9.75% of the time, as opposed to 5% of the time.

Now, 4d6 drop lowest is more complicated to visualize because there are more thingies to track, but the slanting works similarly.

Cheers, -- N

 

[Wulf Ratbane]

Did this answer your question? It seems light on details.

Not at all. It told me this:

Selecting the lowest dice will slant the distribution towards low values and selecting the highest dice will slant it towards the high end of the range.

Believe it or not, I figured that out myself without doing any math.

Basically, when you roll two dice and take the maximum value, you're allowing large values to cover small values, which slants the distribution towards the high end.

Yes... Without having a strong foundation in probability, my solutions all tend toward brute force calculations. (Seriously-- some really ugly Excel sheets full of matrices of results.)

I was hoping for a formula approaching the simplicity of (N+1)/2 (though, clearly, not so simple).

 

[silentounce]

And then what? That section doesn't answer my question.

Seriously, thanks for pointing out that thread, it's an interesting read, but please do me the favor of actually reading this thread and the issues raised here before immediately pointing folks over there.

I DID read this thread, all of it, several times. I even commented on several of the earlier posts, did YOU not read that? So, PLEASE, do ME the favor of actually reading the thread and then realize that I already read it. I really feel like flaming here, I'm in a bad mood I guess, but I'll resist. I just misunderstood your question. And that's not the same article I referenced earlier if you even saw my earlier post.

 

[Wulf Ratbane]

I DID read this thread, all of it, several times. I even commented on several of the earlier posts, did YOU not read that?

Yeah, I read where you said, "The guy who said it adds .5 is wrong."

But I notice you edited that out.

Maybe after you actually read the thread.