How does a hole's size affect the distance that water will squirt I took a bucket, drilled 2 diffe

I believe the physically relevant parameters here are: 1. internal pressure P at the bottom of the bucket but not too close to the hole, units [$kg\ast m^{-1}{s}^{-2}$], the fluid density $\rho$ [$kg\ast m^{-3}$], hole area $A$ units $m^2$, and viscosity $\eta$$kg{m}^{-1}{s}^{-1}$. First, let's ignore the viscosity, and find which combinations can give us a velocity, units [$m/s$]. The unique combination is $\sqrt{(}P/\rho )$, and so in the limit of vanishing viscosity the velocity will be independent of the hole size. Now if we add in viscosity, there is a dimensionless combination available, $\rho {\eta }^2{P}^{-1}{A}^{-1}$. We could have any function of this dimensionless ratio, but common sense tells us that at least in the limit of small $\eta$ it should be a decreasing function, since viscosity should take kinetic energy away from the fluid. From this ratio we see that decreasing the size of the hole is equivalent to increasing the viscosity, and thus the smaller hole should have a smaller exit velocity than the bigger one. I guess this is a long-winded way of saying the big hole will have higher Reynolds number and be less effected by the viscosity.

I like the notation from Ben in his answer. So we will compare two situations with equal water column height but different areas of the hole, $A_1$ and $A_2$. We find that the distance the water squirts is different due to friction. I propose that the friction is from three different sources.
1.Fluid internal friction as it approaches the hole
2.Boundary layer friction around the perimeter of the hole
3.Air friction after it leaves the hole
I would actually imagine that the 3rd one would be the most significant. Some meaningful relations for this could be deduced exclusively from the hole size. If we assume the first two sources of friction are small, then the initial velocity leaving the hole is the same for both cases, thus we have a certain flow rate and velocity for each. If we assume the holes to be circular and extend this assumption to the stream itself, we can find the perimeter interacting with air and the linear density of the stream of water (close to the hole).
Then the problem is similar, but not exactly the same, as that of a projectile with air resistance. The metric that matters is the ratio of the drag coefficient to the mass, but the linear mass density in the case here. I'll try to come back and put some equations up here later so we can maybe even get a general figure of merit for where deviation from a parabolic path of the water jet will be observed.
Local forms loss of the hole
I realized that the parts of 1&2 is really a pretty standard engineering problem.
K = Local forms loss coefficient
H = fluid head, used in place of pressure
$\mathrm{\Delta }H=\frac{v^2}{2g}+K\frac{v^2}{2g}$
$v=\sqrt{\frac{2gh}{1+K}}$
Now, we won't be able to avoid the inevitability of finding Reynolds number, so here we are.
$\rho =1000kg/m^3$
$\mu =0.001Pas$
If we're talking about in the neighborhood of a hole $1c{m}^2$ and a velocity $1m/s$ then we find Reynolds number to be around 1,000. Now, I found no choice to use an actual academic paper to get the loss coefficient.
$K=4.183\times {10}^{-5}Re+0.152$
So K will be in the neighborhood of 0.2. This is the major difference for the two cases regarding 1&2 types of losses. I tried some numbers, like 1 m for the head, and found without the forms loss you get about 4.4 m/s exit velocity, with 1 cm^2 hole, you get 2.8 m/s. Increasing the size of the hole 10x I get 2.1 m/s and decreasing it 10x I get 3.4 m/s. This isn't what I expected. I thought that a small hole would make the fluid velocity smaller, but maybe I've messed something up with the signs although I can't find it right now. I know how I would approach #3 as well, but I don't have the time. I certainly don't think anyone has sufficient evidence to say one type of loss is greater until some more work is done.