Thomas Andrews already answered the question, but I'd like to present a more analytical solution to the problem.

The average of averages is only equal to the average of all values in two cases:

- if the number of elements of all groups is the same; or
- the trivial case when all the group averages are zero

Here's why this is so.

Consider two sets $X = \{x_1, x_2, ..., x_n\}$ and $Y = \{y_1, y_2, ..., y_m\}$ and their averages:

$$ \bar{x} = \frac{\sum_{i=1}^{n}{x_i}}{n} \,,\,
\bar{y} = \frac{\sum_{i=1}^{m}{y_i}}{m}
$$

The average of the averages is:

$$ average(\bar{x}, \bar{y})
= \frac{\frac{\sum_{i=1}^{n}{x_i}}{n} + \frac{\sum_{i=1}^{m}{y_i}}{m}}{2}
= \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2m}
$$

Now consider the whole group
$Z = \{x_1, x_2, ..., x_n, y_1, y_2, ..., y_m\}$ and its average:

$$ \bar{z} = \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{m}{y_i}}{n + m}$$

For the general case, we can see that these averages are different:

$$ \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2m}
\ne \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{m}{y_i}}{n + m}
$$

This answers the first OP question, as to **why the average of averages usually gives the ***wrong* answer.

However, if we make $n = m$, we have:

$$ \frac{\sum_{i=1}^{n}{x_i}}{2n} + \frac{\sum_{i=1}^{m}{y_i}}{2n}
= \frac{\sum_{i=1}^{n}{x_i} + \sum_{i=1}^{n}{y_i}}{2n}
$$

This is **why the average of averages is equal to the average of the whole group when the groups have the same size**.

The second case is trivial: $\bar{x} = \bar{y} = average(\bar{x}, \bar{y}) = 0$.

Note that the above reasoning can be extended for any number of groups.