1

I have heard of Shannon's number (10^120) which is the supposed number of possible chess games computed in the 1950s. But Shannon's estimate is criticised for including illegal moves. What is a truer estimate under the given parameter

  1. no illegal positions
  2. FIDE rules of 3 fold rep and 50 moves are in force
  3. we start from the regular starting position
Rewan Demontay
  • 17,514
  • 4
  • 67
  • 113
anti - Marshall
  • 625
  • 3
  • 11

1 Answers1

3

Peruse this link. Also see link for possible duplicate.

Sum-up of first link:

  • Longest game under FIDE rules: 8848.5 (Fabel)
  • Maximum mobility number: 218 moves
  • Shannon's number, assuming 40 (full) moves, 30 moves per ply, ignoring FIDE rules and duplicates: 10^120
  • Upper bound: 10^34082
  • Realistic lower bound (Monte Carlo): 10^29241
Hauke Reddmann
  • 16,242
  • 3
  • 27
  • 70
  • Noted, with thanks. Your attached link is very helpful. – anti - Marshall Feb 29 '24 at 08:04
  • Also Shannon's number is an estimate of the number of positions possible after 40 ply or 40 moves? – anti - Marshall Feb 29 '24 at 08:21
  • I doubt that there are that many positions taking longer than either (although the retro genre knows many exceptions - but I bet they are rare). – Hauke Reddmann Feb 29 '24 at 19:27
  • The answer might be better if it at least summarized the link's estimate on the answer to the question. Links can disappear. – D M Mar 01 '24 at 05:20
  • The gap between the upper bound and lower bound is HUGE. – Zuriel Mar 02 '24 at 03:25
  • @Zuriel: Not for a professional mathematician - the gap between Ramseys upper bound and the actual value (whatever exact problem my remark references to - I am no professional mathematician) - THAT is huge :-) – Hauke Reddmann Mar 02 '24 at 18:22