It's too late to change anything about the count procedure, now that balloting has begun. But I'm going to describe a better Random Ballot procedure. It's for the subsequent poll, and is suggested too for any polls after that. Of course if anyone objects to the Random Ballot procedure that I specified before the balloting, the we could switch to this newer procedure even for the current poll: 1. Each letter of the alphabet can be replaced by a number, so that A=1, B=2,...Z=26. 2. Add up the last letters of all the voters' names. Subtract from that sum the largewt multiple of 26 that is less than that sum. 3. Multiply the resulting number by 10, and subtract from that product the largest multiple of 26 that is less than that product. The resulting number is called "K". 3. Add K to the 2nd to last letter of each voter's name, and subtract from that sum the largest multiple of 26 that is less than that sum. 4. The ballot whose voter's name is the one for which the resulting number is least is the one that is used as the tiebreaking ballot. 5. But if 2 names have the same 2nd to last letter, then, using the same K, add that K to the 3rd to last letter of each voter's name, and proceed as in #3 & #4. If 2 names have the same 2nd to last & 3rd to last letters, then use the 4th to last letters, etc. In general if the nth to last letters are the same, then use the (n+1)th to last letters. [end of random procedure to choose a tiebreaking ballot] This is just a suggestion. Since we've already voted on the voting system, it's probably too late to make this an official recommendation for randomly choosing a tiebreaking ballot. But it could be sent in, with the voting system recommendation, as an informal suggestion for one way of randomly choosing a tiebreaking ballot. Mike _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com
