I copy pasted you're question into Claude 3.5, but none of the formulas were copied, nonetheless I think it got it right.

Response:

I notice some mathematical expressions might have been cut off in your message. From context, this appears to be about Bitcoin private key collisions and the birthday problem. Let me help fill in the details:

Bitcoin has 2^256 possible private keys (this is the total number of possible keys in the keyspace). Let's call the number of randomly generated keys n.

Let's solve this step by step:

P(collision) = 1 - P(no collision)

P(no collision) = (2^256)/(2^256) × (2^256-1)/(2^256) × ... × (2^256-(n-1))/(2^256)

This can be rewritten as:

P(no collision) = ∏(i=0 to n-1) (1 - i/2^256)

P(collision) ≈ n^2 / (2 × 2^256)

0.001 = n^2 / (2 × 2^256)

Solving for n: n = √(0.001 × 2 × 2^256) n ≈ 2^128 × √0.002 n ≈ 4.8 × 10^38 keys

These calculations assume we're using the approximation formula, which is valid when the probability is small relative to the keyspace (which it is in this case).

This demonstrates why Bitcoin private key collisions are practically impossible - even with billions of people generating enormous numbers of keys, the probability of a collision remains infinitesimally small due to the massive keyspace of 2^256.

Note: I performed these calculations directly. While they should be accurate given the approximation formula used, for absolute precision with such large numbers, you might want to verify with specialized cryptographic software or mathematical tools.