@Martin0: Since the Queen is on f3, etc. in both diagrams, I would call those the same position.

I have no idea what the answer is, but I can tell you with 100% certainty (this number was mathematically calculated) that it is not more than 2.05(10^50).

Flipping the board doesn't change anything. You could look at a chessboard infinitely many ways but who cares?

Ok i was thinking about this problem this morning and came up with my own interpretation regarding the number of possible permutations of the board. I am going to make certain assumptions so I will describe what my assumptions were regarding this math problem.

First I am going to consider the board with no pieces whatsoever as just one position.

Now I am going to begin to put the first piece.

Since there are no pieces on the board yet, there are 64 possible places I can place the first piece.

Now I am going to place the second piece. I am going to assume that pieces will not be allowed to occupy the same square at the same time as though they were legos. 

So the second piece has 63 possible options where it could be placed. So the number of possible unique arrangement locations of the two pieces would be 64 times 63. This is 4,032 unique ways that i can set up the first two pieces.

Now the pieces themselves. I can have both pieces be white, both pieces are black or they can be of opposite colors. Thus 3 potential color schemes.

Thus 4,032 x3 = 12,096

Now i have to decide which pieces themselves i will use. I can use a pawn, a rook, a knight, a bishop, a queen or king. Thus 6 possible piece chooses. However there is nothing to stop me from having 2 of each. So I could have 2 kings, 2 queens, etc. thus 6 x6 =36

12.096 x36 = 435,456 potential permutations.

And remember we are only using 2 pieces right now. What happens when start adding more and more pieces? The numbers as you can see are going to get very big, very fast.

@JoseO: The original question asked how many legal positions are possible. An empty board cannot arise legally, nor can any position with only White pieces, etc., so all these would not count.

This problem seems interesting, and I might have even done it, but if I did do it I would be kicking myself for doing an uneccesary math problem before I did any of the many calculus problems I was assigned! If this is still unsolved on Monday I might give it a whorl, though.

It wasn't the flip that was the point. In the first it was white to move and in the second black to move. Who is on the move can change a position much.

whirlwind2011 wrote:

This is why I was asking for clarification. There would be issues with a lot of positions as a result since i could have a pawn start at row one but that is not legal since a pawn at the beginning starts in row 2.

@JoseO: The original question asked how many legal positions are possible. An empty board cannot arise legally, nor can any position with only White pieces, etc., so all these would not count.

I wish you hadn't.

@Martin0: Oops! I overlooked that detail. Yes, those would count as different positions, I think.

Here's a math problem:

3.45x-56=2.99x+3

Solve for x.

I almost had it, but my pencil ran out of ink.

I would have made the problem longer, but my keyboard ran out of lead.

hah you guys are funny

More than 3, that means approximately infinity.

For more exact answer you can check http://en.wikipedia.org/wiki/Shannon_number

god this is confusing

In Eric Schiller's book: "The Big Book of Chess", it says there are more legal chess positions than there are electrons in the universe. Pretty crazy!

So's reading Schiller.