Mathematics

billprovince, it looks like you calculated all the positions with 32 pieces. You have to add to that all the positions with 31 pieces, 30 pieces, 29, etc.

 Yeah like The Joy of Sex that will do it

Why not talk about a truly large number...like Graham's number?

http://en.wikipedia.org/wiki/Graham%27s_number

 too big for me...i prefer Graham's cracker.

Yes.  I acknowledge that fact at the beginning of the rather long post. :)

Sorry I was a bit confused. You acknowledged that your calculation was for all 32 pieces. But you didn't note that there were unique positions for fewer pieces as well.

Anyway, it's not a simple thing to calculate. Someone pointed out that with all 32 pieces on the number of legal positions is severely limited compared to the number of ways to arrange the pieces on the board --  no pawn captures is a huge restriction.

I wonder how bad chess960 throws a monkey wrench into the value of the number of games.

Indeed, even getting good estimates is a bit difficult.

I can write a program to calculate the number of random positions with 32 pieces or less down to just 2 pieces (the two kings) and not considering pawn promotions (and situations like having 6 bishops or two white-squared bishops), but explaining the program in this forum would be tedious.

The point of my post was to suggest that the number *felt like* an overestimate.  (It was clearly not sufficient to be a proof of this fact).  And true, I did not sufficiently acknowledge the fact that I ignored 31-piece positions and 30-pieces positions, etc., I felt it valid to consider it an over-estimate based on the method that I used to generate the positions allowing for vastly more illegal positions than legal ones.  I felt that if I were to consider the positions with fewer pieces that the added number of positions there would not make up for the fact that almost all of the random positions are not legal.

Just for fun, here's an argument that indicates there are more positions with 31 pieces than with 32. Not taking into account position legality, just completely random positions.

Consider each position with 32 pieces. For all these positions there are 32 ways to generate a position with 31 pieces because there are 32 ways to remove a piece from the board. So a naive estimate is that there are 32 times as many positions with 31 pieces as there are with 32 pieces.

Why is this naive estimate incorrect?

Is the answer because there would be so many duplicate positions?

For example take any random position with 32 pieces as you've said.  Lets say the only change you'll make for a certain set involve moving the white king to each empty square.  For each unique position of this set, removing the white king will give identical 31 piece positions while the naive argument assumes a unique position will occur each time a piece is removed.

I think it is calculatable with the nCr formula. The problem probably just comes in when you have to weed out the ILLEGAL positions

Actually, the 31 piece case is very close to the 32 piece case, but is just slightly less (by a factor of 32/33).

To do this calculation, consider in turn each of the cases of removing a white or black king or queen compared to the case of removing a white or black rook, knight, or bishop and then consider the case of removing a white or black pawn.  We add up each of the results since they are mutually exclusive.

  Considering white or black king or queen removal, we get 4 * 64!/(33! * 8!^2 * 64).

Removing a white or black rook, knight or bishop, we get 6 * 64!/(33! * 8!^2 * 32).

Removing a white or black pawn, we get 2 * 64!/(33! * 8!*7! * 64).

Now, we get to add up these results:

4 * 64!/(33! * 8!^2 * 64) +

6 * 64!/(33! * 8!^2 * 32) +

2 * 64!/(33! * 8! * 7! * 64)

Finding a greatest common denominator and doing a bit of factoring, and we get:

(64! / (33! * 8!^2 * 64)) * (4 + 12 + 16) = 64! * 32 / (33! * 8!^2 * 64)

Compare this with Shannon's number (which covers the case with 32 pieces): 64!/(32! * 8!^2 * 64), and we get our ratio of 32/33.

As to why the naive estimate of a factor of 32 in favor of the 31-piece solution having more was incorrect, I would say that it did not account for two things: first, the ratio 64!/31!*33! being slightly less than 64!/(32!^2) (the number of ways to select 32 squares compared to 31 squares out of 64), and second, it did not account for the fact that many selections of 31 pieces are identical: picking the first white pawn compared to picking the second white pawn does not generate distinct positions.

If you are a genius people, give your reasonable answer for this test. "When Cheryl's birthday is? Explain your answer with logic argument, please!

       You've been here 7 days and you've bumped 3 threads that are 5 yrs old. Why?  Show your answer on a separate paper.