Question

Two fair dice are rolled and the absolute value of the difference of the outcome is denoted by \(\displaystyle X\). What are the possible values of \(\displaystyle X\), and the probabilities associated with them. Finally a problem that does not lead to confusion. \(\displaystyle X = |6-0| = |0-6| = 6\) \(\displaystyle X= |6-1| = |1-6| = 5\) \(\displaystyle X = |6-2| = |2-6| = 4\) \(\displaystyle X = |6-3| = |3-6| = 3\) \(\displaystyle X = |6-4| = |4-6| = 2\) \(\displaystyle X = |6-5| = |5-6| = 1\) \(\displaystyle X = |6-6| = |6-6| = 0\) Is it correct to say that the probability to get any \(\displaystyle X\) is \(\displaystyle \frac{1}{7}?\)

Answer 1

Your dice are different from any one I have ever seen since your dice have 7 sides and each contain the numbers 0,1,2,3,4,5,6. You should make a Cayley table similar to what Plato did but instead of putting (2,6) for example, I would put 4. Label the left hand side and the top 1 2 3 4 5 6 and STOP using 0 as a face on a die.

Answer 2

SupremeCookie said: A lot of the outcomes are missing, what about e.g. (5,0), (5,1),...(4,0),(4,1), ect...? Click to expand... ahhh mouse slip. I did only the 6. CaptainBlack said: You can't throw a "0" on a D6. Click to expand... The zero is problematic. Plato said: Here are all thirty-six of the pairs. Now just count. \(\displaystyle \begin{array}{*{20}{c}} {(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)} \\ {(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)} \\ {(3,1)}&{(3,2)}&{(3,3)}&{(3,4)}&{(3,5)}&{(3,6)} \\ {(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)} \\ {(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)} \\ {(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)} \end{array}\) Click to expand... I have already done the 6. Remove |6-0|, so we have X =,0,1,2,3,4,5. There is a clear pattern what will happen for the rest. 5 4,4 3,3,3 2,2,2,2 1,1,1,1,1 0,0,0,0,0,0

Answer 3

Here are all thirty-six of the pairs. Now just count. \(\displaystyle \begin{array}{*{20}{c}} {(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)} \\ {(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)} \\ {(3,1)}&{(3,2)}&{(3,3)}&{(3,4)}&{(3,5)}&{(3,6)} \\ {(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)} \\ {(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)} \\ {(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)} \end{array}\)

Answer 4

george357 said: Two fair dice are rolled and the absolute value of the difference of the outcome is denoted by \(\displaystyle X\). What are the possible values of \(\displaystyle X\), and the probabilities associated with them. Finally a problem that does not lead to confusion. \(\displaystyle X = |6-0| = |0-6| = 6\) \(\displaystyle X= |6-1| = |1-6| = 5\) \(\displaystyle X = |6-2| = |2-6| = 4\) \(\displaystyle X = |6-3| = |3-6| = 3\) \(\displaystyle X = |6-4| = |4-6| = 2\) \(\displaystyle X = |6-5| = |5-6| = 1\) \(\displaystyle X = |6-6| = |6-6| = 0\) Is it correct to say that the probability to get any \(\displaystyle X\) is \(\displaystyle \frac{1}{7}?\) Click to expand... You can't throw a "0" on a D6.

Answer 5

george357 said: Two fair dice are rolled and the absolute value of the difference of the outcome is denoted by \(\displaystyle X\). What are the possible values of \(\displaystyle X\), and the probabilities associated with them. Finally a problem that does not lead to confusion. \(\displaystyle X = |6-0| = |0-6| = 6\) \(\displaystyle X= |6-1| = |1-6| = 5\) \(\displaystyle X = |6-2| = |2-6| = 4\) \(\displaystyle X = |6-3| = |3-6| = 3\) \(\displaystyle X = |6-4| = |4-6| = 2\) \(\displaystyle X = |6-5| = |5-6| = 1\) \(\displaystyle X = |6-6| = |6-6| = 0\) Is it correct to say that the probability to get any \(\displaystyle X\) is \(\displaystyle \frac{1}{7}?\) Click to expand... A lot of the outcomes are missing, what about e.g. (5,0), (5,1),...(4,0),(4,1), ect...?