[dice probability] Is this correct?

[TeeSeeJay]

For any n-sided die (1dn), the following is true:

1/n * 100 = % chance of rolling any number from 1 to n

For a roll of 2dn in which the results are summed, the following is true:

The total number of combinations is n^2
The median, mode, and mean of 2dn = n+1
The % chance to roll the mode value is 1/n * 100
The number of possible rolls equal to the mode is n

Therefore,

A roll of 2d8 is 12.5% likely to result in a value of 9, which is the same chance of rolling any given number on 1d8. There are 64 different combinations, 9 of which sum to 9. 9 is also the median value and the average value of these 64 combinations.

I'm posting this because I'm generally not very good with statistics and probability and just want to make sure I understand this correctly.

Is there any similar patterns to easily determine the various probablities of larger rolls, like 3dn? Probably not, since the numbers get very large.

The mode of 3d6 = 10
The mode of 3d4 = 7
The mode of 3d2 = 4

The pattern is mode(3dn)=mode(2dn)+(n/2)

So, that would mean mode(3d8)=9+4=13
mode(3d10)=11+5=16
mode(3d12)=13+6=19
mode(3d20)=21+10=31

I'm not about to verify these the hard way. Can anyone point me to relevant web sites that might have such formulas available? I wasn't able to find anything helpful through my own searches.

My head hurts now.

 

[Tarek]

Alternate way

Let's take an example of 47d12.

This is actually fairly easy to calculate.

You figure that the mode on 2d12 is 13, which is absolutely correct, *but* you can then break it down to being 6.5 per die (2 being the minimum).

Thus, where n = the maximum on a die, and where x equals the number of dice (minimum of 2), the formula is x(n/2). Since the fraction is always .5, you round it down at the end of the calculation.

47d12 averages to 47(6.5) = 305.5 = 305

For a tricky problem, figure out the probability that an "exploding" die meets or exceeds a particular number. (an exploding die being one that you re-roll if you roll the maximum, and add the results together. Thus 4+4+3=11 if you roll a d4 and "explode" twice.)

Tarek

 

[Kahuna Burger]

Re: Alternate way

Let's take an example of 47d12.

This is actually fairly easy to calculate.

You figure that the mode on 2d12 is 13, which is absolutely correct, *but* you can then break it down to being 6.5 per die (2 being the minimum).

This is for the mean (arithmetic average) not mode (number most often obtained.) There is no mode for one die because each number will be rolled equally often (on a fair die.)

I'm thinking that there will be a split mode on odd numbers of dice rolled, but its too late for me to be sure of this...

Kahuna Burger

 

[Meds]

Yep, I agree with Kahuna Burger.

As I understand 'mode', there are 2 modes when x is odd (and greater than 1).
e.g. For 3d6 the modes are 10 and 11, whereas the mean/median is 10.5

The formula for mean/median of xdn is:
x(n+1)/2

The mode is the same when x is even.
The modes are (mean + 0.5) and (mean - 0.5) when x is odd.

 

[ichabod]

It is correct that any odd number of same, even sided dice will have two modes. The distribution will be symetric, and the the mean will be a fractional value. That will make the two values on either side of the mean the two modes. So the to get the mode of 5d8, figure the mean (5 x 4.5 = 22.5). That means 22 and 23 will be the modes (that is, they both have the maximum probability of occuring).

 

[Mercule]

Re: Re: Alternate way

This is for the mean (arithmetic average) not mode (number most often obtained.) There is no mode for one die because each number will be rolled equally often (on a fair die.)

I'm thinking that there will be a split mode on odd numbers of dice rolled, but its too late for me to be sure of this...

The nice thing about die rolls is that mean = mode for even numbers of dice. For odd numbers of dice, the mode will be the integer values immediately above and below the fractional mean.

I'm pretty sure that medium = mean for die rolls, too.

Exploding dice or drop-the-lowest are ugly and beyond my ability without either a spreadsheet (quick way) or grabbing my statistics book from college.