I am sure you can calculate the probability yourself, but anyway here is my approximation:
Consider a circle of radius r at latitude θ and longitude ϕ. In the latitude/longitude coordinate space, if the circle is small (r ≪ graticule dimensions), it will look like an ellipse with semi-major axis ϕr and semi-minor axis θr. The semi-major axis will always be longitudinal due to the non-square shapes of the graticules. The lengths of the axes are:
- ϕr = 1° × r / wgrat,
- θr = 1° × r / hgrat,
where wgrat and hgrat are the width and the height of the graticule respectively at the centre point of the ellipse. Assuming that the Earth is a sphere with radius R = 6371000 m and taking θ = 53.6401767° from 2021-04-10_53_9, the dimensions of the graticule are approximately
- hgrat = π R / 180 = 111195 m,
- wgrat = hgrat cos θ = 65922 m.
Now, if r = 384 m, we get the axes of the ellipse:
- ϕr = 0.00583°,
- θr = 0.00345°.
The probability P1 that a random hashpoint hits this ellipse is simply the area of the ellipse divided by the area of the graticule:
- P1 = (π ϕr θr) / (1° × 1°) = 6.32 × 10−5 = 1 / 15800.
Finally, the probability P2 that a hashpoint hits the ellipse at least once in 13 days is
- P2 = 1 − (1 − P1)13 = 8.21 × 10−4 = 1 / 1220.
An event of this probability will happen on average about once in 3.33 years or three years and four months.
― Kripakko (talk) 13:21, 24 April 2021 (UTC)
- Wow, thanks for your calculations! I just used the area of my graticule, which I once calculated as 7299 km² (for spherical Earth), and then worked with ratios, but you actually put in the work with the ellipsis, and I suppose yours is more accurate, though we both disregarded that graticules are actually not rectangular. Anyway, I get (approximately) once in 7299/(0.384²*π*13) days = 3.32 years on average (3.33 with your calculated graticule area).
- However, I don't visit every hash, so the probability of it happening for one of the geohashes I've been to this far (and could visit again) is only 1-(1-0.384²*π*13/7299)^76 = 6.1%. That's actually more than I expected! So for equal odds of it happening I'd have to do log(base 1-0.384²*π*13/7299)(0.5) = 840 hashes, which at my current pace would take me about 2.87+(840-78)/37 = 23.5 years total! I don't actually know if my method of calculating these is valid, but it makes sense to me. I just made the formulas up on the spot, since I don't really have any stochastical education.
- I love that you're nerdy enough to calculate these as well. I was quite sure you'd notice this as you were the only other geohasher apart from Fippe who went on an expedition on 2021-04-10. Can I put your calculations on the expedition page? I think they'll be more appreciated there. --Π π π (talk) 10:10, 25 April 2021 (UTC)
