Individual Postings 1st appeared(& were copied in html form) on the Email List Mail Jewish

From: Russell Hendel <rhendel@saber.towson.edu> Date: Sun, 9 Jul 2000 23:33:24 -0400 (EDT) Subject: The MJ Answer to the "ending in 50" problem Chaim Manaster enlarges upon my grandfathers question. It seems to me that the same answer (that was given in MJ several years ago) would apply to his enlarged question. Here is a brief summary Assume the following rule: Round all Biblical numbers to the nearest 100 UNLESS the number ends in exactly 50. So eg 1234 1289 and 1250 would appear as 1200 1300 and 1250 respectively We can now ask the probability that among 16 tribes exactly 1 should have an exact count of 50. This is a routine exercise in introductory probability courses and is equal to 13.8%, not an unreasonable number. (If we only deal with 12 tribes the probability of exactly one census ending in 50 is 10.7%) An intuitive defense of this number is as follows: The probability of a census ending in 50 is 1% (since only 1 2-digit number ends in 50) However there are 16 tribes which can have this census ending in 50. Hence the probability that one of them ends in 50 is roughly 16%. We however must multiply this 16% by the probability that NO OTHER tribe ends in 50. There are 99 out of 100 ways that a tribe will not end in 50. Combining all the numbers we get 13.8% Hope this helps Russell Jay Hendel; Phd ASA; Math Towson; RHendel@Towson.edu Moderator Rashi is Simple http://www.RashiYomi.Com/---------NEW IMPROVED