My solution to the betting problem: (defun bet (total maxgames score1 score2 &optional err-check-off) (let ((thresh (floor (/ maxgames 2)))) (unless err-check-off (cond ((evenp maxgames) (error "Only odd numbers of total games allowed")) ((or (> score1 thresh) (> score2 thresh)) (error "Scores must be below half the total number of games")) ((or (< maxgames 0) (< score1 0) (< score2 0)) (error "maxgame, score1 and score2 must not be negative")))) (cond ((or (= score1 (1+ thresh)) (= score2 (1+ thresh))) 0) ((and (= score1 thresh) (= score2 thresh)) total) (t (/ (+ (bet total maxgames (1+ score1) score2 t) (bet total maxgames score1 (1+ score2) t)) 2))))) Given the you bet $a this round, and $b / $c next round, then a win-loss or a loss-win must put you back at the same value, meaning that a - b = -a +c and therefore a = (b+c)/2 That's the recursion that took me some time to realize. Otherwise the coding is straightforward. -- ******************************************* Chiao-Lun Cheng Van Voorhis Group Department of Chemistry, Room 6-234 Massachusetts Institute of Technology 77 Massachusetts Avenue Cambridge, MA 02139-4307 Email : chiao@mit.edu Tel : +1 (617) 253-1539 Website : http://zeno.unixboxen.net *******************************************