I would expect that player 2 would be able to win almost all of the time for most normal hash functions, as they could just play randomly for the first 39 turns, and then choose one of the 2^8 available moves. It is very unlikely that all of those hashes are zero. (For commonly used hashes, player 2 could just play randomly the whole game and likely win, since the hash of any value is almost never 0.)
Yes, player 2 loses with extremely low probability even for a 1-bit hash (on the order of 2^-256). For a more commonly used hash, or for 2^24 searches on their second-last move, they reduce their probability of loss by a huge factor more.
I would expect that player 2 would be able to win almost all of the time for most normal hash functions, as they could just play randomly for the first 39 turns, and then choose one of the 2^8 available moves. It is very unlikely that all of those hashes are zero. (For commonly used hashes, player 2 could just play randomly the whole game and likely win, since the hash of any value is almost never 0.)
Yes, player 2 loses with extremely low probability even for a 1-bit hash (on the order of 2^-256). For a more commonly used hash, or for 2^24 searches on their second-last move, they reduce their probability of loss by a huge factor more.