No no no, I had that wrong. Silly of me. Sorry. It's been too long since I did that calc. It's not the product of the p's that is a chi-square distribution, it's the following, given p1, p2..., pn: 2*((ln p1) + (ln p2) + ... + (ln pn)) That expression has a chi-square distribution with 2n degrees of freedom. So you feed THAT into the inverse-chi square function to get a p-value. Let invchi(x, f), where s is the random variable and f is the degrees of freedom, be the inverse chi square function. Let S be a number near 1 when the email looks spammy and H be a number near 1 when the email looks hammy. Then you want S = 1 - invchi(2*((ln (1-p1)) + (ln (1-p2)) + ... + (ln (1-pn)), 2*n) and H = 1 - invchi(2*((ln p1) + (ln p2) + ... + (ln pn)), 2*n) I am a little out-of-practice but I am about 99.9% sure that the above is right. I looked up some of my notes from a few years ago to get the calc. --Gary -- Gary Robinson CEO Transpose, LLC grobinson@transpose.com 207-942-3463 http://www.emergentmusic.com http://radio.weblogs.com/0101454