{"id":3616,"date":"2023-11-23T08:03:23","date_gmt":"2023-11-23T13:03:23","guid":{"rendered":"https:\/\/sites.williams.edu\/Morgan\/?page_id=3616"},"modified":"2023-11-27T15:59:36","modified_gmt":"2023-11-27T20:59:36","slug":"presidential-election-and-the-length-of-the-shortest-piece","status":"publish","type":"page","link":"https:\/\/sites.williams.edu\/Morgan\/math-chat-archives\/presidential-election-and-the-length-of-the-shortest-piece\/","title":{"rendered":"Presidential Election and the Length of the Shortest Piece"},"content":{"rendered":"

November 16, 2000<\/p>\n

 <\/p>\n

Presidential election with 48% of the US popular vote does not come close to the extreme possibility, even with just two candidates, of winning with less than about 22% of the popular vote. See\u00a0Math Chat\u00a0<\/a>of November 8, 1996.<\/p>\n

Old Challenge<\/b>\u00a0(reappeared on recent “Green Chicken” math competition between Middlebury, Simon’s Rock, and Williams College).<\/p>\n

1. What is the expected (average) length of the shorter piece of a meter stick with a random cut?<\/p>\n

2. What is the expected length of the shortest of 3 pieces from 2 random cuts?<\/p>\n

3. What is the expected length of the shortest of 4 pieces from 3 random cuts?<\/p>\n

4. What is the expected length of the shortest of n pieces from n-1 random cuts?<\/p>\n

Answers<\/b>\u00a0(Joe DeVincentis). We may assume that the first piece is the shortest.<\/p>\n

1. The cut can be anywhere from 0 to 1\/2. The expected length is 1\/4.<\/p>\n

2. The cuts at\u00a0\u00a0must satisfy\u00a0. Averaging x over this region of the unit square yields 1\/9.<\/p>\n

3. The cuts at\u00a0\u00a0must satisfy\u00a0. Averaging x over this region of the unit cube yields 1\/16.<\/p>\n

4. 1\/n2<\/sup>. The n-1 conditions on the unit (n-1)-cube, all meeting at (1\/n,…, 1\/n), yield a pyramid of base b and height 1\/n. Hence the expected value of x is<\/p>\n

\n

<\/p>\n

\n

Substituting u = 1 – nx yields<\/p>\n

\n

<\/p>\n

\n

Math Chat notes that similarly, 1\/n2<\/sup>\u00a0is the expected fraction of a year between the closest of n birthdays. For 20 people, it is less than one day. A small variation on the above proof shows that with probability 1\/2, it is less than (ln 2)\/n(n-1). This is less than 12 hours (so they probably have the same birthday) when (ln 2)\/ n2<\/sup>\u00a01\/730, n2<\/sup>\u00a0510, n\u00a0\u00a023.<\/p>\n

Incredible bridge hands<\/b>\u00a0(Math Chat<\/a>\u00a0of September 21) apparently resulted from dealing out unshuffled new decks, typically boxed as Ace-2-3-…-King of Spades, similarly A-2-3-…-K of Diamonds, then K-Q-…-2-A of Clubs, and similarly K-Q-…-2-A of Hearts, according to analysis by Bart Bramley and Nick Straguzzi in the November 2000 ACBL Bridge Bulletin (p 19).<\/p>\n

New Challenge<\/b>\u00a0(Joe Shipman). On the West Palm Beach ballot of the US Presidential election, it apparently turned out to be a big advantage for the Republican candidate to be listed first. Can you come up with a fair procedure for allocating ballot order from year to year between the two major parties, if you assume that being first is an advantage?<\/p>\n

 <\/p>\n

 <\/p>\n

\n

Send answers, comments, and new questions by email to\u00a0Frank.Morgan@williams.edu,<\/a>\u00a0to be eligible for\u00a0Flatland\u00a0<\/i>and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan’s homepage is at\u00a0www.williams.edu\/Mathematics\/fmorgan.<\/a><\/p>\n

THE MATH CHAT BOOK,<\/a>\u00a0including a $1000 Math Chat Book\u00a0QUEST,\u00a0<\/a>questions and answers, and a list of past challenge winners, is now available from the MAA (800-331-1622).<\/p>\n

Copyright 2000, Frank Morgan.<\/p>\n","protected":false},"excerpt":{"rendered":"

November 16, 2000   Presidential election with 48% of the US popular vote does not come close to the extreme possibility, even with just two candidates, of winning with less than about 22% of the popular vote. See\u00a0Math Chat\u00a0of November 8, 1996. Old Challenge\u00a0(reappeared on recent “Green Chicken” math competition between Middlebury, Simon’s Rock, and […]<\/p>\n","protected":false},"author":2965,"featured_media":0,"parent":3459,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"class_list":["post-3616","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3616","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/users\/2965"}],"replies":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/comments?post=3616"}],"version-history":[{"count":4,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3616\/revisions"}],"predecessor-version":[{"id":3787,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3616\/revisions\/3787"}],"up":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3459"}],"wp:attachment":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/media?parent=3616"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}