4 ms·
I do not see how that’s a good estimate. For example, take a game length of, on average, 4 and a branching factor of 10. That gives an estimate of 10,000. Chan
by Someone 9mo ago
I do not see how that’s a good estimate. For example, take a game length of, on average, 4 and a branching factor of 10. That gives an estimate of 10,000.
Chances are there are games of lengths 3 and 5, too. With that branching factor, there are 1,000, respectively 100,000 of those, for a total of 111,000. That’s over ten times as many games as estimated.
The larger the spread in game length towards games that are larger than average, the more the proposed estimate underestimates the actual number.
- tromp 9mo ago> That’s over ten times as many games as estimated. That's still a pretty good estimate of an exponentially large quantity; the exponent being off by only 1. With these estimates you cannot hope to do better than estimating the exponent.
- Someone 9mo agoBut for chess, the spread in number of moves is a lot larger, and the branching factor is higher. 20 more half moves and a branching factor of 35 isn’t unreasonable, and gives you an underestimation of over 10³⁰. And these tweaks do not complicate the math.