u/Noise-Theorem

To reduce the bias of standard dice I throw four dice (D6) and use a random Latin square, set-up so that rx/cy produces a different number than ry/cx. These produce a number 1-6.

I can also use 3 different Latin squares using the same principle so that:

d1/d2 =n1 -> Latin square 1

d3/d4 = n2 -> Latin square 2

For Latin square 3: n1 = row; n2 = column.

I am not able to work out if three squares would produce a more random result than just one square.

I do this out of interest, I could of course use precision dice. Here are the Latin squares:

4 2 6 1 5 3

6 4 2 5 3 1

2 6 4 3 1 5

3 1 5 2 6 4

1 5 3 4 2 6

5 3 1 6 4 2

------------

6 5 1 4 2 3

1 6 5 2 3 4

5 1 6 3 4 2

3 4 2 5 1 6

4 2 3 6 5 1

2 3 4 1 6 5

-------------

5 1 2 4 3 6

2 5 1 3 6 4

1 2 5 6 4 3

6 4 3 1 2 5

4 3 6 5 1 2

3 6 4 2 5 1

A professional website I am on has asked everyone to change their password and it can be up to 255 characters long. Why allow such a long password?

Also they insist a character cannot be repeated, and upper and lowercase letters, numbers and special characters must be used. If a password is long enough it doesn't matter. Is their IT department out of touch with current NIST recommendations?

Of course I had to take advantage of this and used a password this length.

A simple Latin Square is:

1 2 3

3 1 2

2 3 1

If there are two more Latin Squares such as:

3 1 2

2 3 1

1 2 3

2 3 1

1 2 3

3 1 2

Does this constitute a 3 dimensional Latin Square providing the following conditions are met?

sq1 r1 c1, sq2 r1 c1, sq3 r1 c1 does not contain any repeats.

sq1 r2 c1, sq2 r2 c1, sq3 r2 c1 does not contain any repeats.

through to . . .

sq1 r3 c3, sq2 r3 c3, sq3 r3 c3 does not contain any repeats.

If so, by following this method and adding a 4th Latin Square and using a 4*4 grid for each square, could a 4 dimensional Latin Square be constructed?

It is suggested that the bias of a basic standard die is less that 1%. Assuming this is true, by using two dice and a Latin Square, what would the expected probability be? My thinking is that the chances of the bias of both dice occurring at the same time is greatly reduced.

I throw two dice: D1 = row; D2 = column. This is the Latin Square I have been using, set-up so that D1 = x and D2 = y does not produce the same result as D1 = y and D2 = x:

4 5 3 6 1 2

3 4 5 1 2 6

5 3 4 2 6 1

2 6 1 5 3 4

6 1 2 4 5 3

1 2 6 3 4 5

(This is an interest of mine, I know I could buy precision dice instead.)

I have read that using one hand has more chance of producing a random throw than a shaker. I have also read the opposite. I would be interested in the views of dice game players. It seems that most studies suggest two hands is the worst method.

I need the dice throws to be as random as possible.

To get random numbers from 1 - 12 I throw three dice and use the number on dice 1, adjusted as described below.

If dice 2 is odd, 0 is added to the number on dice 1.

If dice 2 is even, 6 is added to the number on dice 1.

If dice 3 is odd I use the result, if dice 3 is even I ignore the result and throw three dice again.

I believe this should remove the bias of standard dice - is this correct.

In response to the excellent comments:

I am assuming all standard dice have a slight bias and was wondering if by using three dice it would help cancel out any bias. The biases of each dice would appear at random.

NEW METHOD:

I know this is overkill but I have 8 dice:

Throw 8 dice.

Dice 7 decides which of the dice 1- 6 will be the basic number.

Dice 8 decides which of the dice 1- 6 will be the offset.

If dice 7 = dice 8, re-throw dice 7 and 8.

If dice 7 and 8 are 4 and 2 respectively, choose dice 4 for the basic number and dice 2 for the offset. If dice 2 is even, add 6 to dice 4, if odd add 0.

My theory is that this should mitigate any bias. A random number from one dice is deciding which random number from the other dice to use.