Why is a low (<1) chi-square measure unlikely when degress of freedom are high? If the chi-square measures variability, where 0 is no variability and a very small measure is no variability, shouldn't all chi-square distributions be shaped like the dof-1 distribution, where the most likely values are the smallest?
 A: As I understand it, the chi-square statistic is generated by adding the values of normal random variables together.
Let $ X \sim N(0,1) $. For one degree of freedom, the random variable $ X $ value is squared and then plotted. The probability of getting a large value of $ X $ is small, since the normal curve becomes asymptotic to 0 moving farther from the mean. This explains why the chi-square distribution is right skewed. The hump happens at low values of $ X $, because the probability of choosing a value of $ X $ that is small is high. Think of the normal distribution: the "bell" part happens close to 0.
For two degrees of freedom, the process above is done twice. Let $ X \sim N(0,1) $ and $ Y \sim N(0,1) $. Choose values of $ X $ and $ Y $, square them, and add them together. The "hump" of the $ \chi^2 $ distribution moves right. Why? Because the possibility of getting larger numbers increased.
This will continue for 3, 4, 5, etc. degrees of freedom. Therefore, the more degrees of freedom, the farther right the distribution will be.
Note: I had a source for this answer that I can no longer find, but this Wikipedia article has information that supports what is said in this answer.
