Tromp does indeed estimate 4.8 x 10^44 legal positions with promotion, and your wish to ignore all promotions is ridiculous. The reasoning that underpromotions to bishop and rook by the proponent only of a drawing strategy can be ignored is valid. This means that we can manage with positions where one side only promotes to queens and knights. [Note you can't assume the opponent of a strategy will not underpromote because that risks reaching a position where the opponent can win only by underpromotion. Evaluating this as a draw would be fatal.
Where is the estimate of the number of these positions where one side has unlimited promotions and the other is only permitted to promote to queen or knight? On a log scale it is nearer the number for all legal positions than the number for all legal positions without promotion (because one side has all the possibilities and the other has a lot of them). Somewhere around 10^42.
@12789
"I only inspected and found none can result from optimal play by both sides."
++ You can verify yourself. Here is the random sample file of 10,000 positions without promotions to pieces not previously captured. The file is raw, before legality check, i.e. contains illegal positions. The file is labelled 'noproms', meaning no promotions to pieces not previously captured, i.e. it contains only positions from 1 box of 32 chess men, but some positions have e.g. 2 dark-square bishops on one side. I inspected the positions and found none that could result from a reasonable, let alone perfect game with optimal play from both sides.
If you think one of these could, then try to come up with a reasonable game that leads to it.
You do not have to prove optimal play, only present reasonable play.
"but not why you should be starting with 10^38 in the first place"
++ The number of legal positions is (4.82 +- 0.03) * 10^44.
However, the 3 random samples shown cannot result from optimal play by both sides,
as both sides have multiple underpromotions to rooks and/or bishops,
and such underpromotions only make sense to avoid a draw by stalemate,
and it cannot be optimal play for both sides to avoid the draw.
Per Tromp the breakdown of raw position count before legality check is:
promotions: 0 positions: 19201527561695835455154058755564594798074
promotions: 1 positions: 382355871178268365234183218244670372695068
promotions: 2 positions: 3666683498600457464891752992187014354136188
promotions: 3 positions: 22267499667290257736558400874926183060238400
promotions: 4 positions: 95095065373967146179514528215894174339720228
promotions: 5 positions: 300571414300527313744528888013946849776424304
promotions: 6 positions: 721668497316402902485416452421325823057710432
promotions: 7 positions: 1329934072135692805837128923570048899100334756
promotions: 8 positions: 1874962044164806332602085236357597905810647344
promotions: 9 positions: 1980800128935921108339671872170042183548439128
promotions: 10 positions: 1492529839915108301878747832838229979840571492
promotions: 11 positions: 722080907452760073481816196266539169729817880
promotions: 12 positions: 175351843526979273665005184194531833618491680
promotions: 13 positions: 7338473695924787177946719990630518998574920
promotions: 14 positions: 45087168602668580254351850721788483191140
promotions: 15 positions: 55323182237139471340692375109727946960
promotions: 16 positions: 11716401834002951530424702440978260
Total: 8726713169886222032347729969256422370854716254
Promotions to pieces not previously captured occur in master games and in ICCF WC Finals draws, but positions with 9 promotions to pieces not previously captured make up the lion's share of the (4.82 +- 0.03) * 10^44 legal positions.
No, they do not, according to your quoted numbers. The positions with 9 or more positions are similar in number to those with 8 or fewer.
That is why 10^44 is no good starting point.
Faulty premises give faulty conclusions.
A better starting point is 3.8521 * 10^37 from An upper bound for the number of chess diagrams without promotion.
That's great for chess variants where pawns cannot advance passed the seventh rank. Not for chess, though.
The 10^37 is too restrictive, as positions with 3 or 4 queens do occur in master games,
and in perfect games with optimal play from both sides as we know from ICCF WC Finals draws.
And the number of such positions is greater than those without (because the game starts with fewer queens than pawns)
I arbitrarily multiply by 10 to include such positions, leading to 10^38.
That is why starting from 10^38.
You're such a comedian! You didn't even do a dodgy calculation for that did you?
"what optimal play has to do with anything"
++ Weakly solving chess is about optimal play by both sides: positions that cannot result from optimal play by both sides are not relevant to weakly solving chess.
WRONG!
Positions that cannot result from ONE side playing moves THAT ACHIEVE THE OPTIMAL RESULT are not relevant to weakly solving chess. These moves may be suboptimal (draw instead of win) and the other players moves are restricted only by legality. THAT'S THE DEFINITION OF WEAK SOLUTION, regardless of your long held error.
The 110 draws out of 110 games of the ongoing ICCF WC Finals are examples of optimal play by both sides and constitute at least part of a weak solution of Chess.
What's that? About 5000 positions? Job nearly done then.
Moreover it is redundant, as it shows several strategies to achieve the game-theoretic value instead of the required one.
You correctly point out that not only is 5000 positions a bit short of a full weak solution, only a fraction of the positions can be in one, because they diverge. While a weak solution has to include the 20 positions after white's first move in the white strategy, it only has one move in the white strategy. To be more precise, if there are games where both players diverge between them, at least one of them is worthless to a weak solution. (eg no weak solution contains 1. d4 Nf6, 1. d4 d5, 1. e4 c5 and 1. e4 e5).