Mathematics

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Oct 3, 2014 14:43
thanks :)
Oct 3, 2014 14:42
YES
Oct 3, 2014 14:42
Now I want to know from that table, what are my chance to get 12
Oct 3, 2014 14:42
Ok, I will give you the whole thing...

X: (chance to get X) * (chance to get 12-X) = Probability to achieve a total of 12
0:0.24*1.48=00.00
1:1.13*5.05=0.06
2:3.36*13.22=0.44
3:7.04*27.62=1.94
4:11.68*47.35=5.53
5:15.62*68.14=10.64
6:17.38*84.74=14.73
7:16.11*94.5=15.22
8:12.52*98.62=12.35
9:8.08*99.78=8.06
10:4.29*99.98=4.29
11:1.81*100=1.81
12:0.59*100=0.59
Oct 3, 2014 14:39
I gave you the probability to get the desired result... I just want to know if I need to add them all to get the total chance to achieve that result
Oct 3, 2014 14:38
but it doesnt matter
Oct 3, 2014 14:38
and I roll 8 of them
Oct 3, 2014 14:38
6 faces dice with custom numbers on them
Oct 3, 2014 14:37
neither
Oct 3, 2014 14:36
Basically, consider I roll some dice, if X = 3, I need to get 9 on the 2nd roll... the probability to get there is 1.94% in my example
Oct 3, 2014 14:36
@TheGame You just lost me there... I am not so good with mathematique terminology
Oct 3, 2014 14:34
@TheGame and btw... BOO for your name... I just lost the game cause of you :P
Oct 3, 2014 14:34
@TheGame Hmmm.. good question. The % come from the multiplication of the probability to get X (come from the roll of custom dices) and 12-X (come from the roll of custom dices too). (don't know if it is clear)
Oct 3, 2014 14:26
Hello all, I got a simple question about % addition.

I have a probability to achieve a goal when X = a certain number.

0:00.00%
1:0.06%
2:0.44%
3:1.94%
4:5.53%
5:10.64%
6:14.73%
7:15.22%
8:12.35%
9:8.06%
10:4.29%
11:1.81%
12:0.59%


Am I right to say I have 75% total chance to achieve my goal? Or I am totally out of the track?
 
Oct 1, 2014 02:37
You can go check the result on www.121212.ca :)
Sep 30, 2014 20:44
nice idea about the check on 0 for the first array
Sep 30, 2014 20:42
I asked the question on Stackoverflow while we were discussing and someone came up with pretty much all the code (and in a more fashioned way than I would have to):

http://stackoverflow.com/questions/26125981/translate-mathematic-formula-to-javascript/26127810#26127810
Sep 30, 2014 19:09
yeah
Sep 30, 2014 19:09
bye bye and thanks again
Sep 30, 2014 19:08
I will try this and let you know if it is faster than my recursive way
Sep 30, 2014 19:08
i dont know but I thank you A LOT for your help
Sep 30, 2014 19:07
would require 50 dices to get to that (which wont happens)
Sep 30, 2014 19:07
thats why the 101 and 202
Sep 30, 2014 19:07
or could i?
Sep 30, 2014 19:06
but if i roll 6 dices, the "normal" ones could add up to 3 and i could not make the difference between a "special" 1s and 3x "normal" 1s
Sep 30, 2014 19:06
yeah i know i dont need it lol
Sep 30, 2014 19:05
well, will divided by 100 if result is over 100
Sep 30, 2014 19:05
Because on each results, I will minus Z (with a minimum of 0) then divided by 100 (and drop fractions)
Sep 30, 2014 19:04
hehe
Sep 30, 2014 19:03
the 101 and 202 are in fact 1s and 2s (but specials), but to do maths with them i needed them to be way over what I can get when I sum dice values
Sep 30, 2014 19:01
they are d6, but custom face
Sep 30, 2014 19:01
no
Sep 30, 2014 19:01
oki
Sep 30, 2014 19:00
yes i do
Sep 30, 2014 19:00
so i need to make a1=[2,1,1,...,1,...,1] too?
Sep 30, 2014 18:59
i want all dices to be the custom die
Sep 30, 2014 18:59
no no
Sep 30, 2014 18:59
so a4 is like a2?
Sep 30, 2014 18:58
i mean a1 was [0,1,1,1,1,1,1]
and a2 was [2,1,1,...,1,...,1]
Sep 30, 2014 18:58
lol
Sep 30, 2014 18:58
how long is a4?
Sep 30, 2014 18:56
so a1(operate)a2 = a3. then a3(operate)a4 = a5?
Sep 30, 2014 18:55
what do i do?
Sep 30, 2014 18:55
if I add a 3rd die
Sep 30, 2014 18:55
last thing
Sep 30, 2014 18:55
I can work with that stuff to finish up the rest
Sep 30, 2014 18:54
this is something now :D
Sep 30, 2014 18:54
ok
Sep 30, 2014 18:54
so theorically, if we sum all position togheter, we should have 36 (since 2d6 has 36 possibilities)
Sep 30, 2014 18:52
aka, what does the value/position represent