they are classed as independant events therefore mutiplying the odds of each individual event together will give u the odds

A-level maths...ahhh the joy

Sarah: That sheet is not really correct. I guess this numbers were not cleanly calculated but just made by making 1000 guesses.
You can get the really correct numbers at:

http://kdice.wikispaces.com/

Here you got another table with more digits:
http://kdice.wikispaces.com/How+the+probability+table+is+calculated
+ an explanation how the really correct numbers are calculated

The correct numbers are usually not on the frontpage of the wiki, because nobody cares about the 2nd or more digit after comma, when making in game decisions.
---
if you multiply chances, you cannot multiply the percentages, you have to multiply the factors!

e.g. 2 50% chances give a 25% in total not a 2500% chance!

So its:
(1 - 0.99998013) * (1 - 0.9879401) * (1 - 0.9392361) * (1 - 0.4436728) = 8.1 * 10^-9

This is 0.0000008101%.

While this is the chance only of happening in one sequene! Not the chance of this happening when you are attacked more than 4 times, which a) we miss the according numbers of ryonckc, and b) is far much more difficult to calculate.

Personally I think the 7vs2 is alone already unlikely enough to a) ryonckc is really the biggest unlucky loser all times or b) he has overdrawn a little (especially the 7vs2 part)

fuzzycat

the link u posted is a simpler version of the table i posted...

e.g

2v4 = 3.558 on mine
urs = 3.6?

thats practically the same and cecause mine more decimalsmine = more accurate!

"e.g. 2 50% chances give a 25% in total not a 2500% chance! "

that makes no sense to me. i know my equations work and if u disagree thats fine but u arguing with the brittish government as i passed my Alevel maths...and probability played a part...

and...look at this

http://www.mathgoodies.com/lessons/vol6/independent_events.html

backs up what i previously just said

1-0 sarah

=P

"7v2 = 0.009%
5v2 = 0.587%
4v2 = 3.558%
2v2 = 44.521%

therefore the chances of it happening in one round is...

0.8368567% "

How could the final number be bigger than one of the original probabilities? The product is always equal to or less than the factors, because it can't go higher than one...

it just is right, he asked about in a single round...he asked a ques, i answered it...

has anyone read the webpage i posted???

its just the laws of probability

the question has been answered, thread over!

I believe the correct command is '/thread'.

Incidentally, I now see what sarahxxx did wrong. She forgot what % meant.

Sarah one thing after the other:

a) The table you have has wrong numbers, no matter how many digits there are. The table on the kdice wiki front page, has only 1 digit, but this is correct math, and rounded correctly.

The other link I gave to you http://kdice.wikispaces.com/How+the+probability+table+is+calculated
has on the bottom 5 *correct* digits.
+ a pyhton program, you can calculate as many digit your computer is capable of if you get yourself a python interpreter.

b) The reason why the kdice wiki frontpage has only one digit, is because for normal gameplay decisions... who the fuck cares for more than one digit?

c) You have to multiply the FACTORS of probability not the percent values.

Look you have to events having in series, each having a 1% chance. How high is your total chance of both happing?

1% * 1% = 1% ? -> Wrong!

Remember what '%' stands for, it means "divide by hundret" so its

1 / 100 * 1 / 100 = 1 / 10000

Which is 0,01%

Get it?

d) The math is propably totally wrong.

Did yonckc attack or defend when he said "I lose 7 vs 2
5 vs 2
4 vs 2
2 vs 2"

Was the 7 on his part or the other part?

If I read this by I got attacked by 7 than by 5 than by 4 and then by 2 dice, and lost the defend this is far more likely than he said.

"I attacked with 7 then with 5 then with 4 then with 2 dice" and lost all attacks.

I think the hat is sucking sarah's brains out?

I mean I lose in all the rounds versus 2. Nothing is impossible in this game.