Dice

[clmckay]

The new distribution is:

1 Skull
1 double Skull
1 White Shield
1 Double White Shield
1 Black Shield
1 Double Black Shield

[QorDaq]

Thanks Chris, I appreciate all your efforts today.

The new yellow die looks to offer a nicely balanced distribution. Very cool.

Makes me wonder if there are other future dice in the works. Or, at the very least if the new site is any indication that the potential is there for future additions.

Thanks again Chris.

maj!

[Count Mohawk]

They changed the orange die too, though. The new distribution is just two doubles for each type of face, which is literally just a doubled Blue die. They should have left it alone as [2Skull, 2Skull, Skull, 2Shield, 2Walrus, Walrus] for a more unique distribution.

[clmckay]

I think he just has the dice oriented wrong. If you look at the pic for the orange dice you can see the BS.

[QorDaq]

Finally got to take a look at the new site...

Looks like I could get a couple of the assortment sets (one of each color w/yellow) and a set of white dice for around 35 USD (Based on the website's shipping and the included vat). That would be two of each of the non-standard colors and 6 white dice. Roughly 1.95 total per die, which is WAY cheaper than the cost of trying to have dice made.

So, that said, anyone care to chime in on whether or not two of each die is sufficient? I cannot imagine needing more than that, but I am certainly curious to hear from those who have used these dice.

I am also curious if anyone has ever come up with a standard use for them or if everyone just kind of does their own thing with the non-standard dice?

Thoughts and opinions welcome...

maj!

[clmckay]

I think there's several different ways to use them. I've seen people here mention using them as a bonus die or two in addition to a normal roll of the white dice. I've made a small handful of custom artifacts that use them instead of the regular. For example: An enchanted broadsword that rolls 3 purple (or a super-duper one that rolls 3 black/orange). Same can be done with armor for a defense roll. Some big bad monsters or boss could roll Blue Dice. I have a couple cards for (but the heroes have never purchased them) potions/spell scrolls that grant a roll of a certain color. Lots of possibilities.

I've been careful not to get too many different colored die in play at the same time. It can get confusing. So if I've got a race of monsters in play that roll purple, I won't have a boss or other monster group that rolls blue (or anything other than blue and regular).

Thankfully it's not been an issue with the heroes. It's rare when they use anything but a single color (typically white and one other, depending on weapon/armor) and each player only controlling 1 hero has limited the chance for confusion there.

I got them in sets of 6, that way there are enough no matter how they get used.

[clmckay]

Apparently, I've forgotten how to use Excel for basic probability.

I've seen some true spreadsheet wizards around here, maybe you cna give me a quick pointer to the right function.

I want to calculate the odds of rolls on the dice referenced in this thread. Soooo.... Odds of at Least 1- 6 Skulls, White and Black Shields with 1-6 dice being rolled.

I thought I'd use the binom.dist function to figure that, for example:

=1-BINOM.DIST(2,6,.5,TRUE)
In this case, the odds of at least 2 Skulls on 6 dice being rolled. This one is givng me odds of a tad over 65%. Now......if I manually calculate it I come with something closer to 87.5%

So either it or I am wrong....Am I using the wrong formula? Perhaps my Stats have left me completely over the years!

[Count Mohawk]

Apparently, I've forgotten how to use Excel for basic probability.

I've seen some true spreadsheet wizards around here, maybe you cna give me a quick pointer to the right function.

I want to calculate the odds of rolls on the dice referenced in this thread. Soooo.... Odds of at Least 1- 6 Skulls, White and Black Shields with 1-6 dice being rolled.

I thought I'd use the binom.dist function to figure that, for example:

=1-BINOM.DIST(2,6,.5,TRUE)
In this case, the odds of at least 2 Skulls on 6 dice being rolled. This one is givng me odds of a tad over 65%. Now......if I manually calculate it I come with something closer to 87.5%

So either it or I am wrong....Am I using the wrong formula? Perhaps my Stats have left me completely over the years!

You're almost correct. If you use BINOM.DIST with Cumulative set to "TRUE", you get the odds of rolling less than or equal to that many skulls. So the formula "=BINOM.DIST(2,6,.5,TRUE)" indicates the probability of 2 or fewer successes out of 6 trials, each with a 50% chance of success - or in layman's terms, rolling 2 or fewer Skulls with 6 white combat dice. To get 2 or more skulls out of 6, you need the first term to be 1 less than the minimum number of skulls you want to roll; i.e. "=1-BINOM.DIST(1,6,.5,TRUE). This formula works out to 89.06%.

Now, if you wanted to calculate the number of undefended skulls, that would be an entirely different kettle of fish, and I would need to direct you to another thread in this forum. Be warned before clicking - some math nerds went crazy in there!

[clmckay]

DOH!

Thanks!

Any thoughts on how to calculate this for the dice sets with 2 Skulls, Shields?

I'm thinking this:

=1-BINOM.DIST(0,6,(2/6),TRUE)*(1-BINOM.DIST(0,6,(1/6),TRUE))

That would be the odds of getting at least 1 Skull with six dice (purple, in this case). It's distribution is: 2Skulls, 1D-Skull, 3-White/Black Shields

[Count Mohawk]

DOH!

Thanks!

Any thoughts on how to calculate this for the dice sets with 2 Skulls, Shields?

I'm thinking this:

=1-BINOM.DIST(0,6,(2/6),TRUE)*(1-BINOM.DIST(0,6,(1/6),TRUE))

That would be the odds of getting at least 1 Skull with six dice (purple, in this case). It's distribution is: 2Skulls, 1D-Skull, 3-White/Black Shields

Oh, you want the odds with multicolored dice now? Well, trying to find the odds of N skulls using purple dice is no longer a binomial function, so you can't use BINOM.DIST at all for this one.
If there's an easy, simple statistics function that can emulate what is effectively a 'trinomial' function, I don't know it, so we're stuck doing this the long way. Fortunately, the long way isn't actually all that long, as I will describe below.

The proper formula for calculating this is:
(Odds of rolling a Double Skull)^(Number of Double Skulls rolled) * (Odds of rolling a Single Skull)^(Number of Single Skulls rolled) * (Odds of rolling a non-Skull)^(Number of non-Skulls rolled) * (number of unique combinations for that set of die faces)
_ Also, the way to calculate the number of unique combinations for a set of 6 faces (i.e. 4 Skulls, 1 Double, 1 non-Skull) would be:
_ (nCr, where N is the number of dice rolled and R is the number of Single Skull faces) * (mCs, where M is (N minus R) from the previous step and S is the number of Double Skull faces)
...where nCr = N! / ((R!) * (N-R)!).

To the Microsoft Excel!