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If I am not mistaken you could in theory revert the entire blockchain, by producing nonces which probabilistically speaking required more work. So the (hypothetical) question is: if I wanted to steal Satoshi's Bitcoin, should I try to guess the private key or create a new chain with more work? Set aside the economic implications of the value of Bitcoin in either case, purely probabilistically speaking, how do the two compare?

MManke
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By checking the chainwork, which shows the total hashes that are expected to be calculated. As of block 765365, it's 00000000000000000000000000000000000000003a8a576dba35308e77dc8446.

In decimal, that's 18,117,375,202,741,628,848,220,608,424.

Satoshi used P2PK back in 2009, so there is only one valid Bitcoin private key that can sign for the corresponded rewarded public key. Therefore, you have 1 in 2^256 - 2^32 - 977, which is far larger number.

In decimal, it's 115,792,089,237,316,195,423,570,985,008,687,907,853,269,984,665,640,564,039,457,584,007,908,834,671,663.

But, to be fair, calculating SHA256d hashes of block headers, and brute forcing private keys aren't the same computational processes, economically speaking.

Angelo
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    Computing the private key for a given public key needs only ~2^128 group operations, roughly the square root of the number of valid private keys, using algorithms like Pollard's rho. Also, the number of valid private keys is 2^256 - 432420386565659656852420866394968145600, not 2^256 - 2^32 - 977 (which is the size of the field the secp256k1 group is defined over). – Pieter Wuille Nov 30 '22 at 22:29