Disclaimer: I'm still learning Iota specifications, so take this answer with a grain of salt
We assume that you re-use your key for a transaction that is new and different from the previous (aka
hash(tx1) != hash(tx2))
Best case scenario : only one new bit is revealed, leaving 49.8% of your private key to guess, reducing the computations needed to 2^255.
(note: the very best case would reveal 0 new bit, but that would mean a collision in the transaction hash, which would be disastrous)
Worst case scenario : all the remaining bits are revealed, leaving 0% of your private key to guess, thus completely exposing your private key. This has a 1/2^512 chance of happening.
Average case: assuming on average half of the bits from the hash will match bits from the private key that were used before, that means only 25% of the key will be left to guess. Since
512 * 0.25 = 128 bits will need to be discovered for the private key to be cracked, requiring at most 2^128 computations.