$$\frac{2^{75.51747}}{8*10^9} = 6.759797398 * 10^{12} = 2^{42.62012}$$

So each person would have to generate approximately $$2^{42.6}$$ or 6.7 trillion keys for there to be a 0.1% chance of a Segwit address collision.

_So time to switch to taproot? hmm_

So if we adjust the reply from @shafemtol to calculate how many addresses would have to be generated for a collision:

$$N = 2^{160}$$
$$K \approx \sqrt{−2N*ln(0.999)} \approx 2^{75.5}$$

We get $$2^{75.5}$$ private keys that would have to be generated for there to be a 0.1% chance of a Segwit address collision.

Filiprogrammer

SimpleStacker

Keep in mind that while there are almost 

 valid private keys, there are actually less Bitcoin addresses. For example singlesig native Segwit addresses (bc1q...) are a bech32 representation of a 160 bit hash.

 different singlesig native Segwit addresses. So the probability of guessing a private key for a certain Segwit address is actually 

bitcoin

[Bitcoin Puzzle] Probability of private key collisions, part 1

Keep in mind that while there are almost $$2^{256}$$ valid private keys, there are actually less Bitcoin addresses. For example singlesig native Segwit addresses (bc1q...) are a bech32 representation of a 160 bit hash.

This means that there are only $$2^{160}$$ different singlesig native Segwit addresses. So the probability of guessing a private key for a certain Segwit address is actually $$\frac{1}{2^{160}}$$.