Multiple cheap IMUs vs one expensive IMUs
That really will depend on the variance (standard deviation) of the measurements you get from the five vs the one.

Suppose your expensive IMU gives a measurement distributed as:
$$
x_\mbox{expensive}  \sim  N\left(x_\mbox{truth}, \sigma^2_\mbox{expensive}\right)
$$
and suppose your cheaper ones give measurements distributed as:
$$
x_\mbox{cheap}^i  \sim  N\left(x_\mbox{truth}, \sigma^2_\mbox{cheap}\right)
$$
for $i=1\ldots 5$ and that you form the average of these cheaper measurements to get the "real" measurement:
$$
\hat{x} = \frac{1}{5} \sum_{i=1}^5 x_\mbox{cheap}^i
$$
then $\hat{x}$ is distributed as:
$$
\hat{x} \sim  N\left(x_\mbox{truth}, \frac{\sigma^2_\mbox{cheap}}{5}\right)
$$
So if 
$$
\frac{\sigma^2_\mbox{cheap}}{5} \le \sigma^2_\mbox{expensive}
$$
you should be OK with the cheaper sensors.

Of course, there may be other problems with your cheaper sensors: they may be biassed (so that the mean of their distributions will all be different from $x_\mbox{truth}$ ).

And I'm not quite sure what you mean by "voting"? Do you vote off the one furthest from the mean of the 5?  What if the mean changes for the remaining four?  I'll do a bit more digging to see if there's a viable algorithm there, but the direct mean of all five might be the easiest.