Is there a gametheoretic interpretation of nimber multiplication? There is such for addition (a single move in a+b is either a move in a or a move in b).

As Alex says, there's no good one. One way to see the problem is to compare nimbers with (surreal) numbers. Addition in both is just game addition, so you can consider nimbers and numbers together and add them consistently. But there is no consistent multiplication. (What would the unit be?) A good game theoretic multiplication would explain how to multiply arbitrary games. Instead, we have nimber multiplication, which is constructed ad hoc and based on the algebraic structure of the nimbers: nimbers include all additive groups of characteristic 2, and the multiplication makes them into the universal field of characteristic 2 (in the sense that any field of characteristic 2 embeds in the nimbers); any totally ordered field embeds in the surreal numbers. 


Not a pleasingly natural one, I think. Somewhere or other I've seen a discussion of a game played with rectangular cards in which the move is to replace an a*b card by three cards of sizes a'*b, a*b', and a'*b' for some 0≤a'<a and 0≤b'<b, but it was clear that this game had been concocted to match nimmultiplication. 


There is a recursive definition of the product of two games here: http://en.wikipedia.org/wiki/Nimber Surreal numbers have a multiplicative identity see the section on multiplication here: http://en.wikipedia.org/wiki/Surreal_number In terms of ease of computation there are the following rules which may make computing the product of finite nimbers. There are two rules any number of the form 2^(2^i) multiplies normally with any number of the form 2^(2^j) j and i different. However 2^(2^i) times itself is equal to 3/2(2^(2^i). Using these rules we can break up two integers into the sums of powers of two and multiply then for each of the powers of two we break up the exponent into powers of two then we can use the two rules to evaluate the products of the exponents and that should make the computation easier. 


check out this comment. this is also mentioned towards the end of chapter 6 in conway, but i don't have that on hand at the moment. 

