- 23 Sep '07 19:37I have two standard 6-sided dice; a red one and a blue one. What is the average (mean, mode & median) difference between them.

What about with two red dice and two blue. What is the average difference between the total of the red dice and the total of the blue dice?

And finally for mathematicians ... N dice. - 23 Sep '07 22:35 / 1 editI have only an answer when the number of dice tends to infinity.

Let there be n red dice and n blue dice. The number on one dice has mean 3.5 and variance 35/12. So the total of the red dice has approximately normal distribution, with expected value 3.5n and variance (35/12)n. The same is true for the blue dice, and also for minus 1 times the total of the blue dice.

The difference of two independent random variables ditributed normally has also normal distribution, so Total of red minus total of blue = total of red plus (minus 1 times total of blue) is approximately distributed normally with mean 0 and variance (35/6)n, or standard deviation sqrt(35/6)*sqrt(n).

If a random variable x is distributed normally with mean 0 and standard deviation s, then the expected value of |x| is sqrt(2/pi)*s. So:

for large n, the expected diffrence between the total of the n red

dice and the total of the n blue dice is approximately sqrt(35/3pi)*sqrt(n), or about 1.927*sqrt(n). - 24 Sep '07 01:26 / 1 edit

36 possibilities.*Originally posted by wolfgang59***I have two standard 6-sided dice; a red one and a blue one. What is the average (mean, mode & median) difference between them.**

What about with two red dice and two blue. What is the average difference between the total of the red dice and the total of the blue dice?

And finally for mathematicians ... N dice.

In 6 of them, the difference is 0.

In 10 of them, the difference is 1.

In 8 of them, the difference is 2.

In 6 of them, the difference is 3.

In 4 of them, the difference is 4.

In 2 of them, the difference is 5.

(6*0 + 10*1 + 8*2 + 6*3 + 4*4 + 2*5)/36 = 70/36

Yep, 1.9444...

N dice? Too much work for my brain right now. - 24 Sep '07 20:05 / 1 edit
*Originally posted by AThousandYoung***36 possibilities.**

In 6 of them, the difference is 0.

In 10 of them, the difference is 1.

In 8 of them, the difference is 2.

In 6 of them, the difference is 3.

In 4 of them, the difference is 4.

In 2 of them, the difference is 5.

(6*0 + 10*1 + 8*2 + 6*3 + 4*4 + 2*5)/36 = 70/36

Yep, 1.9444...

N dice? Too much work for my brain right now.**slight flaw in that.**

In some cases the red is greater in some the blue is greater; so you have two questions:

What is the absolute difference?*"ATthousandYoung" answered the mean part of this question*

In 6 of them, the difference is 0.

In 10 of them, the difference is 1.

In 8 of them, the difference is 2.

In 6 of them, the difference is 3.

In 4 of them, the difference is 4.

In 2 of them, the difference is 5.

Mean = 70/36 = 1.94...*(last digit repeats)*

Mode = 1

Median = 2

**What is the difference of red with respect to blue?**

In 1 of them, the difference is -5

In 2 of them, the difference is -4

In 3 of them, the difference is -3

In 4 of them, the difference is -2

In 5 of them, the difference is -1

In 6 of them, the difference is 0.

In 5 of them, the difference is 1.

In 4 of them, the difference is 2.

In 5 of them, the difference is 3.

In 2 of them, the difference is 4.

In 1 of them, the difference is 5.

Mean = 0

Median = 0

M0de = 0

**As for two dice;**

relative difference gives the same results

**Absolut diference gives:**

In 146 cases, the diffeence is 0

In 280 cases, the diffeence is 1

In 250 cases, the diffeence is 2

In 208 cases, the diffeence is 3

In 160 cases, the diffeence is 4

In 112 cases, the diffeence is 5

In 70 cases, the diffeence is 6

In 40 cases, the diffeence is 7

In 20 cases, the diffeence is 8

In 8 cases, the diffeence is 9

In 2 cases, the diffeence is 10

Mean = 3556/1296 = 889/324 = 2.74382716049...*(last 9 digits repeat)*

Median = 2

Mode = 1 - 25 Sep '07 11:59

Difference is always positive.*Originally posted by preachingforjesus***[b]slight flaw in that.**

In some cases the red is greater in some the blue is greater; so you have two questions:

What is the absolute difference?*"ATthousandYoung" answered the mean part of this question*

In 6 of them, the difference is 0.

In 10 of them, the difference is 1.

In 8 of them, the difference is 2.

In 6 of them, the di ...[text shortened]... an = 3556/1296 = 889/324 = 2.74382716049...*(last 9 digits repeat)*

Median = 2

Mode = 1[/b]

The difference between a and b is |a - b|. - 25 Sep '07 16:21

How do you define the difference with N dice? Do you mean for us to first take the difference between every possible pair of dice, then average them out?*Originally posted by wolfgang59***Thank you .. that is what I meant as the difference. Now anyone going to tackle N dice? (Other than the excellent approximation for large values of N)??** - 25 Sep '07 20:51

yes*Originally posted by PBE6***How do you define the difference with N dice? Do you mean for us to first take the difference between every possible pair of dice, then average them out?**

mean, mode & median

I guess we all realise the mode is 1 for any N

And probably (but how to prove?) that the median is 2

But whats the mean? - 26 Sep '07 20:39

Oh, I thought we were supposed to total the result of N red dice, same for the blue dice and take the diference between the two totals. My bad :p*Originally posted by wolfgang59***yes**

mean, mode & median

I guess we all realise the mode is 1 for any N

And probably (but how to prove?) that the median is 2

But whats the mean? - 28 Sep '07 15:13 / 2 edits

This is false, because the variance of the sum goes to infinity, as well. So you can't use the Central Limit Theorem and say it approaches normality.*Originally posted by David113***I have only an answer when the number of dice tends to infinity.**

Let there be n red dice and n blue dice. The number on one dice has mean 3.5 and variance 35/12. So the total of the red dice has approximately normal distribution, with expected value 3.5n and variance (35/12)n. The same is true for the blue dice, and also for minus 1 times the total of th ...[text shortened]... and the total of the n blue dice is approximately sqrt(35/3pi)*sqrt(n), or about 1.927*sqrt(n).