The sticky to rule them all

[Deleted]

Someone mentioned in the terigon preggo thread that we should get the averages for tervigon spawning in a sticky, which is a great idea but then I thought, why not sticky the other things we all would like readily available? Like 6e charge distance probabilities with different factors? We should come up with a list of things that would be nice to know, have a thread for each where it gets hashed out and then put the best/most relevant tables into a single sticky in the tactics forum. It will be locked so it stays nice and tidy and new info/tables can get tossed in by a mod as necessary.

[wisdomseyes1]

Well Cordump already said the charge factors would be a sticky in a different thread. Might not be a bad idea to just put all the applicable math in one sticky, though.

[killme304]

If your referring to the number of gaunts you spawn, the average is 10 (the straight average of any 3d6 roll).

If you are referring to going dry, I don't remember the odds offhand.

[creep]

Spare me with your math...

a die goes in linear steps from one to 6

so the average roll is lowest+highest divided by 2

(1+6)/2 = 3,5

so we roll 3 D6 3x3,5 = 10,5

not 11
not 10

its 10,5
you dont need to be a genius for that

to be exakt thats a difference of about 5% either way
You think 5% is irrelevant? Next time you play a 1750 game just field 1662 or 1663 points (depends on which way you want to round up as 5% are 87,5 points).
You shouldn't have a disadvantage as its irrelevant right?^^

[coredump]

Winning friends and influencing people...

[demoric]

I like the whole idea of a probabilities thread. Also, creep thanks for a more shorthand way to do linear probabilities. I have written too many lines of code that ruined some of the ways I approach mathematics. I would just do a for each or do until condition adding up the dice and then divide the result.

So when i manually figure the probabilty I just add up each die face value and then divide by the number of sides: (1+2+3+4+5+6)/6=3.5. (1+6)/2 is a lot quicker to figure out.

Still though I don't understand your vehemence in your reply, but thanks again.

[Deleted]

Jul 29, 2012 14:29:28 GMT creep said:

Spare me with your math...

...you dont need to be a genius for that...

Well it's obviously of great interest to the people on TTH because I see threads like this constantly, why not clean up the board and make it easier to find those numbers everyone keeps wanting? You, Mr. Genius, can skip that sticky and work on your spelling!

[Hyrrokin]

Sorry guys but its wrong. the average run on a d6 would be 3,5 not 3 and not 4? Im safe to say I have rolled zero 3,5s. Also I am most likely 3,5 on 1 die thats, im assuming, have equal change on landing on either 6 sides each with a unique number from 1 to 6?

The correct way would to take all the !possible! results and combine them:

1 1 1 = 3 1 2 1 = 4 1 3 1 = 5 1 4 1 = 6 1 5 1 = 7 1 6 1 = 8
1 1 2 = 4 1 2 2 = 5 1 3 2 = 6 1 4 2 = 7 1 5 2 = 8 1 6 2 = 9
1 1 3 = 5 1 2 3 = 6 1 3 3 = 7 1 4 3 = 8 1 5 3 = 9 1 6 3 = 10
1 1 4 = 6 1 2 4 = 7 1 3 4 = 8 1 4 4 = 9 1 5 4 = 10 1 6 4 = 11
1 1 5 = 7 1 2 5 = 8 1 3 5 = 9 1 4 5 = 10 1 5 5 = 11 1 6 5 = 12
1 1 6 = 8 1 2 6 = 9 1 3 6 = 10 1 4 6 = 11 1 5 6 = 12 1 6 6 = 13

3(1) 4(2) 5(3) 6(4) 7(5) 8(6) 9(5) 10(4) 11(3) 12(2) 13(1)

2 1 1 = 4 2 2 1 = 5 2 3 1 = 6 2 4 1 = 7 2 5 1 = 8 2 6 1 = 9
2 1 2 = 5 2 2 2 = 6 2 3 2 = 7 2 4 2 = 8 2 5 2 = 9 2 6 2 = 10
2 1 3 = 6 2 2 3 = 7 2 3 3 = 8 2 4 3 = 9 2 5 3 = 10 2 6 3 = 11
2 1 4 = 7 2 2 4 = 8 2 3 4 = 9 2 4 4 = 10 2 5 4 = 11 2 6 4 = 12
2 1 5 = 8 2 2 5 = 9 2 3 5 = 10 2 4 5 = 11 2 5 5 = 12 2 6 5 = 13
2 1 6 = 9 2 2 6 = 10 2 3 6 = 11 2 4 5 = 12 2 5 6 = 13 2 6 6 = 14

4(1) 5(2) 6(3) 7(4) 8(5) 9(6) 10(5) 11(4) 12(3) 13(2) 14(1)

and so on, making it:

1- 3(1) 4(2) 5(3) 6(4) 7(5) 8(6) 9(5) 10(4) 11(3) 12(2) 13(1)
2- 4(1) 5(2) 6(3) 7(4) 8(5) 9(6) 10(5) 11(4) 12(3) 13(2) 14(1)
3- 5(1) 6(2) 7(3) 8(4) 9(5) 10(6) 11(5) 12(4) 13(3) 14(2) 15(1)
4- 6(1) 7(2) 8(3) 9(4) 10(5) 11(6) 12(5) 13(4) 14(3) 15(2) 16(1)
5- 7(1) 8(2) 9(3) 10(4) 11(5) 12(6) 13(5) 14(4) 15(3) 16(2) 17(1)
6- 8(1) 9(2) 10(3) 11(4) 12(5) 13(6) 14(5) 15(4)16(3) 17(2) 18(1)

out of 216 an x has y possibelities written x(y):
3(1) 4(3) 5(6) 6(10) 7(15) 8(21) 9(25) 10(27) 11(27) 12(25) 13(21) 14(15) 15(10) 16(6) 17(3) 18(1)

its 11
its 10

not 10,5
you dont need to be a genius for that (but takes time)

as a on the go way I take max (18) and divide with 2 (9) and adds the minimum (3) minus 1(2) and adds that to 9 (11)

No harm meant
Pazdzior

Edit: Im pretty sure its ~50% for a double, did math once but not sure and to lazy

[rippertotheface]

^^ His maths, I don't like them^^

I am pretty sure that we have established that the average of 3d6 is 10.5, or 50% 10 and 50% 11.
I think what everyone wants now is the probability of getting doubles, so, can someone get that up?

[WestRider]

There's about a 44% chance per Roll of getting Doubles on 3d6. I don't have the derivation for that saved anywhere, and I don't care enough to work it out again from scratch, but it's around there.

[Hyrrokin]

Jul 29, 2012 21:20:21 GMT rippertotheface said:

^^ His maths, I don't like them^^

I am pretty sure that we have established that the average of 3d6 is 10.5, or 50% 10 and 50% 11.

Then whats not to like? i am indeed saying 10 and 11 tie for most common. Im just saying the 10,5 is wrong as its not a roll and the ,5 comes from the tie between 10 and 11.

Jul 29, 2012 21:20:21 GMT rippertotheface said:

I think what everyone wants now is the probability of getting doubles, so, can someone get that up?

to see double just take all the possibilties for a double, 16 in each column of 36(using my earlier post) add all 6 columns and its 96/216=16/36=44,44 % (endless 4 there as %) Just as WestRider says.

[rippertotheface]

I don't like your math, mainly because it seems very time consuming to get to a result that could be done in one's head. While, obviously it has become useful now with the doubles, I still think that it is far to time consuming and complicated. In my opinion, math should be as simple and elegant as possible in order to mitigate potential errors.

[Hyrrokin]

Jul 30, 2012 1:55:46 GMT rippertotheface said:

I don't like your math, mainly because it seems very time consuming to get to a result that could be done in one's head. While, obviously it has become useful now with the doubles, I still think that it is far to time consuming and complicated. In my opinion, math should be as simple and elegant as possible in order to mitigate potential errors.

true it was to refute creep

[rippertotheface]

Jul 29, 2012 14:29:28 GMT creep said:

Spare me with your math...

a die goes in linear steps from one to 6

so the average roll is lowest+highest divided by 2

(1+6)/2 = 3,5

so we roll 3 D6 3x3,5 = 10,5

not 11
not 10

its 10,5
you dont need to be a genius for that

to be exakt thats a difference of about 5% either way
You think 5% is irrelevant? Next time you play a 1750 game just field 1662 or 1663 points (depends on which way you want to round up as 5% are 87,5 points).
You shouldn't have a disadvantage as its irrelevant right?^^

Creep, I love you man ( ), but when was the last time you rolled a ten and a half on 3d6. There are inherent flaws in using this method for rolls on even-sided die, because you will never roll a decimal on a die. The decimal merely indicates which two numbers will most likely be rolled.

[coredump]

Guys... the average (mean) *is* 10.5. There really is not way around that.

Well, there is one way around that... there are actually 3 'averages', the one most people think of is the Mean. And 10.5 *is* the mean.

The Mode is the number that crops up the most. The Mode for wounds caused by DoM is 0, since that is the most likely result.

The Median, is the number where half of the results are above,and half below that number.

From memory, A spawning tervigon has an average of about 23, a Median of about 17, and a Mode of about 14.


Anyway... the common average (mean) does not have to be a number that is actually possible. So yes, the average is 10.5.