Wednesday, July 1, 2009

4d6 Less One

This is going to seem simple-Simon for a lot of you out there, but I worked something out a few months ago and I’m fairly proud of it. Since I couldn’t find an example on line, I thought I’d post it here as a sort of public service.

I have always accepted the rolling method for characters of rolling 4d6 and accepting the highest three. Most that I’ve played with over the years found it quite reasonable, the results being high enough without being extraordinary. Now and then a remarkable character occurs.

In designing a character generation sequence on excel, I came across a problem. How do you create an equation which will roll four dice and eliminate the lowest one? I puzzled over that for some time, and there is a solution ... it is long and complicated and requires multiple calculations, and at some point I’ll get into that. For the moment I just want to talk about the short cut.

To begin with, consider the odds on 3d6. The chance of rolling an 18 is, fairly obviously, one in 216 (6x6x6). There are 216 possible combinations and only one will result in an 18. If we break down the die rolls by the chance of obtaining each individual result, we get the following pattern (assume the first number indicates the chance of getting a 3, the second number the chance of getting a 4 and so on):

1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1.

Thus, if you consider the chance of getting a result of 15 or better (the minimum which allows for most modifiers to dexterity, constitution and strength), there is only a 21 in 216 chance. Roughly, 1 in 10. The likeliest results are 10 and 11, the average exactly between, at 10.5.

Suppose, however, we consider 4d6. The total number of combinations are now 1,296 (216x6). I won’t list them all out, it’s unnecessary. Suppose instead that we go through the total number of combinations and determine what the result would be for each combination if we removed the lowest die in each case. This would give us the total number of combinations which would result in statistics from 3 to 18 – in effect, rolling four dice and counting only the highest three, in only one calculation.

In excel, it is possible to throw a die of any number. The formula entered into the box is simply:

=rand()*n

‘n’ being the number you wish to roll.

Thus, if you enter =rand()*1296, you will get a number from 0 to 1296 (not a natural number, as computers don’t understand natural numbers. The number must be rounded off ... and in order not to get a result of ‘0’ whenever the initially created number is less than 0.5, I suggest you add +0.5 to your formula).

But all of this is nonsense if you don’t know excel, so don’t worry about it.

The question is, how do the odds change when comparing 3d6 to 4d6 minus the lowest die?

The following pattern emerges:

1,4,10,21,38,62,91,122,148,167,172,160,131,94,54,21

Keep in mind that those are the odds in 1,216 rather than in 216. The chances of rolling a 15 or better are now 300 in 1,216, or almost 1 in 4. The most likely number to be thrown is 13, and the average is in fact 13.0502.

Interesting, no?

3 comments:

Tim Brannan said...

As a gamer and former stats prof I can dig it.

I'll try it out myself, though I have to admit I rarely use excel for D&D. Other games yes, but D&D demands dice. ;)

Zzarchov said...

Obvious or not its useful to point out to people. Statistics are not always fun to work out, but understanding the math behind a game is very important if you ever want to alter it.

I like numbers though, so I might be biased in my positive views on the post.

tussock said...

The odds are in 1296, as you said at the top.

15+ is 23% (you're 73% likely to have two or more of them, I think), and the average is 12.2446.