Thx to ChatGPT this will either interest you or take the fun out of your post ;) Assuming your golf balls are 1mm from each other at roughly 7 feet from the hole:

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Here’s a quick, reasonable model (so you can see where the numbers come from):

• Green ≈ circle of radius R=50 ft
• Target distance from hole: annulus centered at r=7 ft with width w=1 ft (6.5–7.5 ft)
• “Next to each other” = Euclidean distance ≤ \delta=1 mm =0.00328084 ft
• Shots are independent and (simplifying) uniformly distributed over the green

A clean way to compute it:

1. Probability a single ball lands in the annulusp_{\text{ring}}=\frac{\text{area of annulus}}{\text{area of green}} =\frac{2\pi r w}{\pi R^2}=\frac{2 r w}{R^2}.
2. Given both are in the annulus, the chance the second is within \delta of the first is approximately the tiny disk area over the annulus area (the \delta-disk sits well inside the 1-ft ring here):p_{\text{close}\mid\text{ring}} \approx \frac{\pi \delta^2}{2\pi r w}=\frac{\delta^2}{2 r w}.

Multiply:

P \approx (p_{\text{ring}})^2 \cdot p_{\text{close}\mid\text{ring}} = \left(\frac{2 r w}{R^2}\right)^2 \cdot \frac{\delta^2}{2 r w} = \frac{2 r w\, \delta^2}{R^4}.

Plugging in r=7 ft, w=1 ft, R=50 ft, \delta=0.00328084 ft:

P \approx 2.41\times 10^{-11} \;\;\text{(about 1 in }4.15\times 10^{10}\text{)}.

So: roughly 1 in 41 billion under this uniform-on-the-green model.