Variance of a die roll I am a little unclear if this question makes sense. Say I have a fair die with sides 1 to 6. Can I ask what is the variance of a single roll of the die? The calculation I was thinking was the following. $\mu = 3.5$
$$\frac{1}{6}\times\left[2.5^{2} + 1.5^{2} + .5^{2}\right]\times 2 = 2.91$$
So then the standard deviation is 1.70. Does this further mean that within 3.5 $\pm$ 1.7 is 68% of all the outcomes? (Not sure if this makes sense in this example where prob are same for each outcome)
 A: I am not sure your question makes sense. Wanting that, here are some ideas about variance:
Variance is a measure of how spread out the sample data are about the mean, or, alternately, how spread out the population values are about the population mean.
If by "a single value" you mean "a single (sample) observation," then the variance must be zero, since the sample mean is just the value of the one observation, and there is no spread of observations about it.
If by "a single value" you mean a single value from the distribution of a fair six-sided die, then, while there is no spread of data around (i.e. a distribution of values above and below) the population mean ($\mu=3.5$), there is a deviation of the single observation from that population mean: $x-\mu$.
A: This is a classic example of 'experts' deliberately misinterpreting the intent of the OP's question and refusing to provide a clear answer on the basis that the original question lacked technical precision.  I suspect this is the actual answer you are looking for:
Yes, your calculation is correct.  However, your 68% proposal implicitly assumes a normal distribution.  For a single die roll, this doesn't make much sense.  However, were you to roll a die say 100 times, then a normal distribution would provide a very close approximation to the actual distribution of results.  The more times you roll the die, the better the approximation becomes.
However, now you have to scale the Variance according to the number of die rolls.  The variance of the total scales according to n (100), while the variance of the average scales according to 1/n.  Therefore, if you roll a die 100 times:
Total sum : Expected value 350, Variance roughly 17 (101.7)
Average : Expected value 3.5, Variance roughly .17 (1/101.7)
And your 68% rule applies.  Notice how the expected value and variance of the total sum is 100 * those of the average.
