2

checksum = 1st4byte(sha256(sha256(PubKeyHash)))

checksum1 = 1st4byte(sha256(sha256(PubKeyHash1)))

What is the P(checksum == checksum1)? where PubKeyHash != PubKeyHash1

Is it 1/(2^^32)?

8

Yes, exactly 1/2^32.

There isn't much more to say about it. Hashes of distinct values are uniformly randomly distributed. The chance for two independently uniformly randomly distributed hash to match in 32 bits is 1/2^32.

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