The recipe gives R^2/(8Q)>=16(beta+3)^2*nu. Thus the outside probability is at most 2*exp(-16(beta+3)^2*nu), which is no more than 2^{-((beta+2)*nu)} for beta>=1 and nu>=1. Since lambda<=2^nu, this is at most lambda^{-(beta+2)}. Finally b<=lambda^beta and r^2<=lambda imply lambda^{-(beta+2)}<=1/(b*r^2*lambda)=epsilon=tau/(r*b). Therefore the exceptional fraction satisfies the assumption needed for at most 2*tau*n split measurement rows. All parameter and balancing checks remain required; a small tail alone is not a complete layout.