Forum adverts like this one are shown to any user who is not logged in. Join us by filling out a tiny 3 field form and you will get your own, free, dakka user account which gives a good range of benefits to you: No adverts like this in the forums anymore. Times and dates in your local timezone. Full tracking of what you have read so you can skip to your first unread post, easily see what has changed since you last logged in, and easily see what is new at a glance. Email notifications for threads you want to watch closely. Being a part of the oldest wargaming community on the net. If you are already a member then feel free to login now.

Hello,

I was wondering if there is a way or a formula to calculate '2D6 discard the lowest or take the highest'.

I would be really useful for Mathammering, as there is a lot of '2D6 discard the lowest' rules in this current edition.

Thanks in advance!

Friend of mine said he finally found the math on this and it came to 4.5 average, but I can't personally vouch for it. Sounds good though

you mean a probability curve for 2d6 pick the highest? Those are easy.

Re-roll lowest is kinda tricky

It's 4.47222 average.

Easy way to do it is to put it into excel as 2 columns, D1 and D2, using the 'max()' formula into a third column for them all, and using 'average()'.

Hey, I'm not a mathematician, but my solution for this is following:

Add all the outcomes as sum and divide by number of outcomes:

Highest 6 can be achieved 6 ways 6&6-1
Highest 5 can be achieved 5 ways 5&5-1 (6 is included above)
Highest 4 can be achieved 4 ways 4&4-1 (as above)
and so on.

So then I calculated the average:
6X6+5X5+4X4+3X3+2X2+1
divided by
6+5+4+3+2+1
Resulting in average of
4.33

My result is below the one mentioned before (4.47), but still quite close. I however am interested how this is actually calculated correctly if someone knows?

Cheers!

Ghorgul wrote:
Hey, I'm not a mathematician, but my solution for this is following:

Add all the outcomes as sum and divide by number of outcomes:

Highest 6 can be achieved 6 ways 6&6-1
Highest 5 can be achieved 5 ways 5&5-1 (6 is included above)
Highest 4 can be achieved 4 ways 4&4-1 (as above)
and so on.

So then I calculated the average:
6X6+5X5+4X4+3X3+2X2+1
divided by
6+5+4+3+2+1
Resulting in average of
4.33

My result is below the one mentioned before (4.47), but still quite close. I however am interested how this is actually calculated correctly if someone knows?

Cheers!

You idea was correct, but your values are wrong
Let's say you roll
1s
11 = 1

2s
12 = 2
21 = 2
22 = 2

3s
13 = 3
23 = 3
33 = 3
32 = 3
31 = 3

so it should be 1*1 + 2*3 + 3*5 + etc
Also divide it by 36, since 6*6 = 36

For an easy tool google rumkin die roll stats.

Be aware for mathammering results , the curve isn't normal the chances ramp linearely not in bell curve.

 Farseer Anath'lan wrote:
It's 4.47222 average.

Easy way to do it is to put it into excel as 2 columns, D1 and D2, using the 'max()' formula into a third column for them all, and using 'average()'.

https://docs.google.com/spreadsheets/d/1eYW-VySR-KgUWazWPka7eWkOKNNL2vD6Vk1XQCPf-qw/edit?usp=sharing


Automatically Appended Next Post:

Ghorgul wrote:
Hey, I'm not a mathematician, but my solution for this is following:

Add all the outcomes as sum and divide by number of outcomes:

Highest 6 can be achieved 6 ways 6&6-1
Highest 5 can be achieved 5 ways 5&5-1 (6 is included above)
Highest 4 can be achieved 4 ways 4&4-1 (as above)
and so on.

So then I calculated the average:
6X6+5X5+4X4+3X3+2X2+1
divided by
6+5+4+3+2+1
Resulting in average of
4.33

My result is below the one mentioned before (4.47), but still quite close. I however am interested how this is actually calculated correctly if someone knows?

Cheers!

Using your method to speed it up would work, but just needs a correction
Highest of 6 can be achieved in 11 ways, 6 * 2 - 1
Highest of 5 in 9 ways, 5 * 2 - 1

Which means...
(6 * (6 * 2 - 1)) +
(5 * (5 * 2 - 1)) +
(4 * (4 * 2 - 1)) +
(3 * (3 * 2 - 1)) +
(2 * (2 * 2 - 1)) +
(1 * (1 * 2 - 1))

Then divide that all by total number of permutations or 6 * 6

((6 * (6 * 2 - 1)) + (5 * (5 * 2 - 1)) + (4 * (4 * 2 - 1)) + (3 * (3 * 2 - 1)) + (2 * (2 * 2 - 1)) + (1 * (1 * 2 - 1))) / (6 * 6) = 4.472222
You can literally copy paste this into Windows Calculator (Make sure it's on Scientific)




Automatically Appended Next Post:
and you can expand on it if your game uses dice values higher than d6s

Example using 2d8
(8 * (8 * 2 - 1)) +
(7 * (7 * 2 - 1)) +
(6 * (6 * 2 - 1)) +
(5 * (5 * 2 - 1)) +
(4 * (4 * 2 - 1)) +
(3 * (3 * 2 - 1)) +
(2 * (2 * 2 - 1)) +
(1 * (1 * 2 - 1))

Then divide that all by total number of permutations or how high each dice goes, so 8 on dice A, and 8 on dice B. 8 * 8

((8 * (8 * 2 - 1)) + (7 * (7 * 2 - 1)) + (6 * (6 * 2 - 1)) + (5 * (5 * 2 - 1)) + (4 * (4 * 2 - 1)) + (3 * (3 * 2 - 1)) + (2 * (2 * 2 - 1)) + (1 * (1 * 2 - 1)) ) / (8 * 8) = 5.8125

Windows Calculator once again! (Make sure it's on Scientific)

Thanks Talamare! I see what you did there and it makes sense, I did only one "side" of each possible pair.

Hmmm, but as the lowest is discarded, it doesnt matter if it is 6-1 or 1-6, or does it? Now I am confused again.

Ghorgul wrote:
Thanks Talamare! I see what you did there and it makes sense, I did only one "side" of each possible pair.

Hmmm, but as the lowest is discarded, it doesnt matter if it is 6-1 or 1-6, or does it? Now I am confused again.

it does matter hecause those are two Seperate results that just happen to have the same out come

Ghorgul wrote:
Thanks Talamare! I see what you did there and it makes sense, I did only one "side" of each possible pair.

Hmmm, but as the lowest is discarded, it doesnt matter if it is 6-1 or 1-6, or does it? Now I am confused again.

I honestly didn't know the correct way to do it either until I saw your post and read Farseer's suggestion and did the entire thing.
After doing it on the spreadsheet on google, I realized the pattern and setup the equation.

I don't really know what you mean by 1 side of a possible pair
but I know that
Dice A = 1
Dice B = 6

is different than

Dice A = 6
Dice B = 1

of the 36 possible cases 1/6 end up showing identical numbers on both dice, which averages out to 3,5

That leaves 30 more cases in which each number appears 10 times.

As soon as one of those dice shows a 6, we're done. Doesn't matter what the other is. I.e. 10 cases with a result of 6.

For the number 5 we have to deduct the two cases where it is paired with a 6. I.e. 8 cases with a result of 5.

6 cases with a result of 4, 4 with a result of 3, 2 with a result of 2

1/6x7/2+5/6x(6x10+5x8+4x6+3x4+2x2)/30
=
7/12+5/6x140/30
=
7/12+70/18
=
21/36+140/36
=
161/36
~
4,4722...

Doctoralex wrote:
Hello,

I was wondering if there is a way or a formula to calculate '2D6 discard the lowest or take the highest'.

I would be really useful for Mathammering, as there is a lot of '2D6 discard the lowest' rules in this current edition.

Thanks in advance!

Roll thousands upon thousands of times?

I've found this to be an invaluable tool when trying to figure out the probabilities on all sorts of nonsense;

http://anydice.com/

The summination of n(2n-1) divided by n^2 where n = 6

So you end up with a calculation looking like:
(1(2(1)-1)+2(2(2)-1)+3(2(3)-1)+4(2(4)-1)+5(2(5)-1)+6(2(6)-1))/6^2

Ever feel like people are putting too much thought into it?

I mean, top marks for your maths. But....blimey!

 Mad Doc Grotsnik wrote:
Ever feel like people are putting too much thought into it?

I mean, top marks for your maths. But....blimey!

It's really a pretty basic thing to work out if you have the background for it. There are going to be lots of people who can work out the probabilities of each value and the average in under a minute. I can picture the 6 by 6 grid of 2d6 outcomes and just "see" that there are going to be 11 outcomes with a 6 as the highest value, 9 with a 5, etc. I'd use a calculator or my browser's search bar to get the actual average (doing "1/36*(11*6 + 9*5 + ...)" ) but it's immediately apparent that the probability distribution is linear.

You don't have to put that much work into fairly basic stats if you do it a lot, or you know, teach it. Plenty of mathematicians, scientists and engineers in the hobby as well as others who just like maths.

Here's the grid Dionysodorus mentioned if you want to count things up yourself!

Alternatively, this is what I was referring to doing, although this is rows rather then columns (makes it fit nicely on the screen).
Can be altered to work for almost any probability work, although some are more involved then others. This was quick and easy to put together-more complicated scenarios can take a while.

As I said, impressive application of maths.

Me, I'm more a 'just chuck the dice and hope for the best' type!