yeah i noticed that as well, and started asking about it in games, a bit ago.

Doesn't seem too unlikely to me. Or that big a deal. Now, if an extra pair of sixes showed up on every roll players 3 and 5 make, that may be a problem.

The chance of a doubling is 1/6 or 16.7% I hope this is clear or does it need to be explained. (okay I explain it, 6 possible conditions a die can have after a throw, 1 of it is a double, therefore 1/6 chance.

I counted the number of doublings and non doublings in your post.

Maybe I made one error or two on counting, but in total I got 16 doubles in your number series, and 70 non doubles.

That are 16 doubles of 87 events: meaning it are 18.6%. Well not far of 16.7% if you take in case that you even HANDSELECTED this sample for rolls where you thought they are uneven.

Well, i am not too good with statistics but it seems to me that this is not so weird. I do agree that the dice generator does not seem random(lost 8-4 two times this week)

I applaud pussycat on her clear explanation. You dont happen to have a masters degree in statistics right:p

Otherthan that, the think that ISNT random, is that every so week somebody complaints about the random generator to broken, out of this or the other reason he thinks to haven seen a pattern in random. You can take a look also here:

http://kdice.wikispaces.com/Random+in+Kdice

@Overlast, thx, but no "real" statistics is not really my thing. But there are just highscool basics ;-)

In response to fuzzy cat.

My knowledge of statistics is not thorough, but I believe the correct proabability for doubling in an 8-die roll would be as follows:

(a) Probability of rolling any specific number on the first roll = 1/6
(b) Probability of rolling a the same number on the second roll = 1/6
(c) Probability of rolling two numbers consecutively = 1/6 x 1/6 = 1/36
(d) Probability of rolling two numbers consecutively in any 8 die roll = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 7/36 = .19444

This suggests that 20% of all 8 dice rolls should contain at least one doubling.

In my previous example, 93% of the 8 dice rolls contained doubles. Am I doing the math wrong? I'm not trying to be troublesome, and this is the first time that I've posted to the forum. I am just curious.

Well, chance's of -a- number on the first dice roll is one. Chances of -that same- number on the second is 1/6. Remember, there are six possibilities for doubles, out of 36 total combinations.

This means your next equation adds up to 7/6 probability, so you definitely did something wrong.

Here is another sample taken at random:

Overlast's turn
Grunvagr defended 8v8: 19 to 29 (2,1,1,5,3,1,5,1 to 6,4,3,4,1,4,4,3)

7K's turn
Dirty D defended 8v8: 22 to 35 (2,3,5,3,2,3,3,1 to 6,6,2,2,4,4,5,6)

Grunvagr's turn
Dirty D defended 8v8: 21 to 29 (3,1,2,3,2,5,4,1 to 2,4,2,5,5,6,3,2)
Dirty D defeated 8v8: 26 to 25 (6,2,3,2,5,3,2,3 to 5,5,5,2,2,1,2,3)
Overlast defended 8v8: 29 to 30 (1,4,5,1,6,6,4,2 to 6,6,3,2,6,1,1,5)

Dirty D's turn
Grunvagr defeated 8v8: 32 to 28 (4,5,2,4,1,6,5,5 to 2,3,3,2,5,5,2,6)

Overlast's turn
Grunvagr defended 8v8: 25 to 34 (6,1,2,3,3,2,4,4 to 1,6,4,5,5,3,4,6)

7K's turn

Grunvagr's turn
Dirty D defended 8v8: 21 to 35 (1,3,6,2,2,2,2,3 to 6,6,2,4,6,6,2,3)
Overlast defended 8v8: 29 to 29 (3,2,1,6,6,1,5,5 to 6,5,4,3,2,1,3,5)

Dirty D's turn
Grunvagr defeated 8v8: 30 to 28 (1,6,5,1,5,4,2,6 to 1,2,6,3,6,3,3,4)
Grunvagr defeated 8v8: 30 to 24 (5,5,2,1,1,4,6,6 to 5,1,4,2,4,3,1,4)

Overlast's turn
Grunvagr defended 8v8: 23 to 29 (1,4,1,6,2,1,5,3 to 6,2,2,5,4,3,1,6)

Instances of doubling = 17/24 = 71%

Am I mistaken?