Thank you very much, Tye :)

For the record, a friend who's much better at stats than I am gave me the following equation for the probability that A wins. It's generalized for the case where "probability of rolling heads" is something other than 1/2, and may be different for each player.

the sum over(y=0 to M){
 (M choose y)* probb^y * (1-probb)^(M-y) * 
 the sum over(k=y to N) {
  (N choose k)*proba^k*(1-proba)^(N-k)
 }
}

(Where "proba" is the probability for player a to get a "head", and "probb" is the probability for player b to get a "head")

Thanks again :)

Alan