Talk:2021-04-23 53 9

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 ϕ_r and semi-minor axis θ_r. The semi-major axis will always be longitudinal due to the non-square shapes of the graticules. The lengths of the axes are:

ϕ_r = 1° × r / w_grat,

θ_r = 1° × r / h_grat,

where w_grat and h_grat 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

h_grat = π R / 180 = 111195 m,

w_grat = h_grat cos θ = 65922 m.

Now, if r = 384 m, we get the axes of the ellipse:

ϕ_r = 0.00583°,

θ_r = 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 ϕ_r θ_r) / (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 for every given hash (3.33 with your graticule area).

However, 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)