Stat-rolling math problem

[UndeadArcanist]

I was wondering what the chances are of rolling an 18 (3 sixes) in 4d6 best 3 stat rolling. I initially thought it would be 1/6 X 1/6 X 1/6, but I realized the possibility of the fourth die being a six complicates things. Could you please explain how to calculate this?

[JBPuffin]

I looked it up...apparently it involves statistics formulas, as shown here (http://stats.stackexchange.com/questions/130025/formula-for-dropping-dice-non-brute-force), to do it algebraically. No thanks.

Preston Poulter (http://prestonpoulter.com/2010/10/19/the-mathematics-behind-4d6-drop-the-lowest/), however, notes that there are 21 dice combinations that result in three sixes - 6666 and (666N)x4 b/c there are 4 dice the not-6 could land on, x5 b/c there are 5 non-6 numbers for it to be. 21/(1296=6^4) is the probability.

So really, it's more logic than algebra (still math, tho! :smallbiggrin:).

[Stan]

AnyDice will give you the full distribution for dice combos.
http://anydice.com/

Here's best 3 of 4d6 graphed out:
http://anydice.com/articles/4d6-drop-lowest/

[UndeadArcanist]

Awesome! Thanks for the help:smallsmile:

[Aeson]

Because no one has actually told you how to compute the answer analytically:

The probability that you will get k desired results in a sequence of n trials, where the probability of getting a desired result in any given trial is p and is independent of the results of any other trial in the sequence, is


P(k) = (n! / (k! * (n - k)!)) * p^k * (1 - p)^(n - k)

The probability of getting at least k desired results is the sum of the probabilities of getting k' desired results for k <= k' <= n. n! is the factorial of n, or the product of all integers m such that 1 <= m <= n for any integer n > 0, and 0! is defined to equal 1.

To get an 18 in 4d6 take best 3, you need to get 3 or 4 sixes. Therefore, the "desired result" is rolling a six on the die, you want to get 3 or 4 sixes, and the probability of rolling a 6 on an unbiased six-sided die is 1 in 6. The probability of getting an 18 using this stat-rolling method can thus be computed as


P(k = 3) + P(k = 4) = (4! / (3! * (4 - 3)!)) * (1/6)^3 * (1 - 1/6)^(4 - 3) +


(4! / (4! * (4 - 4)!)) * (1/6)^4 * (1 - 1/6)^(4 - 4)

This works out to about a 1.6% chance of getting an 18 using 4d6 take best 3 stat rolling.

[Tyrrell]

Here's a description of how to calculate it that, to me, is more straightforward than Aeson's.

To find the probability of getting a result we just ask how many total options are there and how many of them give us our result.

The probability of getting an 18 is:

the number of possibilities that give us an 18 / total number of possibilities.

So how many different ways can the dice turn up.

there are 6 options for the first die, for each of these there are 6 options for the second die (giving us 36 total), for each of these 36 there are 6 ways for the third die to roll (216) , and with the fourth die we get to 1296 different arrangements of the dice.

How many of these results get us 18?

There is the combination where all 6 of the dice come up 6, there are 5 combinations where the first die isn't a 6 but the other three are, 5 where the second die isn't a 6 but the other three are, 5 for the third die and 5 for the fourth die. So the total number of combinations where we get an 18 is 21.

This means the probability of getting an 18 on 4 dice take the best three is 21/1296 or about 1.6% (0.016203704).

[Democratus]

Great explanation, Tyrrell!

[Knaight]

It's worth observing that there are two possible questions being asked. One is the probability of getting an 18 in just one roll, the other is the probability of getting at least 1 18 in a 6 roll set (yet another is getting exactly one 18 in a six roll set). To get the probability of at least one 18, we want one minus the probability of no 18 results. In other words:

1 - (1275/1296)^6 = 0.09336788352

So you have about a 9.34% chance of getting at least one 18.

[Tanarii]

For those who are interested, you can also fairly easily calculate the probability of getting: At least K results of a number or higher in N rolls.

You just use the binomial probability mass function. That boils down to:
(n! / k!*(n-k)!) * pk * (1-p)n-k

P is the probability of getting the number or higher, which you can easily pull from anydice. (For 18 on 4d6b3, it's ~1.6%)

So of you want to know the chance of getting exactly 2 18s in 6 attributes, each one rolled using 4d6b3, the odds are:
(6! / 2!*(4)!) * (.016)2 * (.984)4
Or
15 * .000256 * .49787136
Or approximate 0.2%

(The odds of getting exactly 2 17 or higher using 4d6b3 (~5.8%) is ~4%. )