23 kos, each one will be in one of two states: taken by white or taken by black. This effectively means 2^23 - 8,388,608 - right? Well, no. White will never let black get more than two of these kos. So if we assume black having the ko is 1, then there can only be 1 or 2 1's at any given point. This number is then much smaller. However, there are only 23 states of black having 1 ko and this is the key. After 23 moves by white he will not be able to reduce black to 1 ko any longer. Black will take a second ko and white will be unable to take it back. After this, no matter what white does on his next move, black will take a third ko and capture the group on the bottom. This would mean there would be 23 white takes + 23 black takes + 1 white filling a ko or passing + 1 black ko take + 1 white ko take + black ataring the bottom + white taking another ko + black taking the group. 32 moves total.

However, black only has 42 points from his captures and the bottom. White will have 42.5 with his center and komi. So white wins by .5 anyway.

Is this correct?

I found the asymmetry in the number of black states and white states in the ko really interesting, you have Choose(23,2) states where black can approach, but only 23 states where white can hold off the approach, hence, white gets stuck long before black, even though white takes the ko first.

I think black actually wins this one, the same principle should allow black to keep winning the ko situation until the states balance. So black actually should win by around 10, of course that would take so many moves as to be counterproductive

With White taking a ko first, Black can play a triple ko, forcing White to stop taking a ko at move 8. Black can atari at move 9 and take at move 11. Territory result: B +5.5.