Ah yes, 1 in 2¹⁶⁰ needs to be compared.

Well since we allready have 2.142 * 10⁴³ water molecules in the pacific and ~ double that (approx 4.28 * 10⁴³) in all oceans, we just need 10 ⁵ times of all of earths oceans

But thats not a nice comparisson. How about "Earth's volume is approximately 1,083,210,000,000 km³" https://en.wikipedia.org/wiki/Volume_of_the_Earth

Thats approx. 1,083 * 10¹² km³, lets just see how low we need to go with the unit till we hit 10⁴⁸

1,083 * 10¹²km³ = 1,083 * 10¹²*10¹⁸ mm³ ... hmm 10¹⁸ more to go the particles/molecules might fit, but lets take something else.

According to this site http://htwins.net/scale2/

the known universe is approx 10²⁷ m (diameter) so we just need to find something has an diameter of ~ 10^-21 m...

While zooming in @ 10^-16,3 i noticed "length shorter than this are not confirmed", well lets keep going...

10^-21 has nothing, but @ 10^-22 there is a Top Quark. So lets just take 10 of those


So getting the correct private key to a given public address is like hitting a specific wall of 10*10 top quarks from the edge of the observable universe.

This assumes that the observeable universe is flat, which makes it easier to imagine.


Is it still? If your answer is yes, please visit the page I linked above and play with it a bit. I allready though 2¹⁶⁰ is big, but this...

Edit: After thinking about this for a while, I think I made a big mistake. The problem is that the Information given is a diameter which has 1 dimension. So instead of thinking about the observeable universe as a big disc youd have to think about it as a line, which I dont like tbh. I will edit this again later with a better 3d comparission (as the fraction of a drop of water in the ocean was).


Edit2:

Lets do this proper now

We need to find something with approx 1.5 * 10⁴⁸m³ in order for 1m³ to be equivalent to 1 in 2¹⁶⁰

http://htwins.net/scale2/
Gives us diameters of pretty much anything. E.g. the observeable universe (OU) has a diameter of approx 9.3 * 10²⁸m. This gives the OU a Volume of:

V_OU = 1/6 * pi * (9.3 * 10²⁸)³ ~ 4.2*10⁸⁶

Thats way bigger than we need. So in order to find something with a volume of approx 1.5 * 10⁴⁸ we can take the same formula but this time we let the diameter be unknown. Since this forum is not supporting latex commands Ill spare you the details.

1.5 * 10⁴⁸ = 1/6 * pi * d³ is equiv. to d = 1.4 * 10¹⁶

which gives us the "Rotten Egg Nebula", "Raw Egg Nebula" or "Calabash Nebula" https://en.wikipedia.org/wiki/Calabash_Nebula

In this nebular I have hidden a ball with diameter ~ 1.24 meter and you have to find it.

Im not sure I like this comparisson better, but at least the numbers add up.


Last Edit (promise):

Earth has a diameter of 1.27 * 10⁷m which gives it a volume of 1.07 * 10²¹m³
So in order for the comparisson to hold up we need something with a volume of 7.13*10^-28m³ or a diameter of approx 1.11 * 10^-9m.
According to http://htwins.net/scale2/ a single water molecule has a diameter of 2.8*10^-10m, thats close but a little to small. But a buckyball* has a diamter of 1*10^-9m. The water molecule might be a better comparisson though.

Thats the last one. I like it. 1 in 2¹⁶⁰ is like finding a single buckyball in the whole earth.

* https://en.wikipedia.org/wiki/Buckminsterfullerene