From: John Conover <john@email.johncon.com>

Subject: Re: Bush/Kerry optimal wagers

Date: 28 Oct 2004 10:10:44 -0000

There is a simple C program at the bottom; it wades through all possible combinations-all 2048 of them-of either Bush or Kerry taking each of the 11 States in question; so the output can be worked on with the Unix sort(1) program to calculate the statistics of certain things happening. There are 28 possibilities where Bush and Kerry would tie, and who would be president would be decided by Congress-the last time that happened was 1824. The probability of that happening is about 28 / 2048 = 1.3671875%. Calculating the chances of the Presidency being decided by the courts, again, (Bush can not have a commanding lead in the state-remember the last president that won by about 60% was Lyndon Johnson in the 60's-which lets FL, OH, WI, IA, NM, and CO, off the hook): State Electoral Votes Bush's Chance of Winning State MN 10 0.65 MI 17 0.50 HI 4 0.65 NH 4 0.17 NJ 15 0.50 meaning the electoral spread would have to be within 17 votes to make it reasonable to challenge a state's count. There are 812 possibilities, so the probability of at least one count challenge is 812 / 2048 = 39.6484375%, with the highest probability in MI and NJ, then MN, follwed by NH and HI. John -- John Conover, john@email.johncon.com, http://www.johncon.com/ /* Set up a binary count of 2^11 = 2048 possibilites of the 11 states in question, where a zero in the count means the state goes to Bush, and a 1 means it goes to Kerry. */ #include <stdio.h> struct state { char name[2]; int e_votes; }; struct state electoral[11] = { {"FL",27}, {"OH",20}, {"WI",10}, {"IA",7}, {"MN",10}, {"MI",17}, {"HI",4}, {"NM",5}, {"CO",9}, {"NH",4}, {"NJ",15} }; void main (void) { int i, bush, kerry, i0, i1, i2, i3, i4, i5, i6, i7, i8, i9, i10, x[11] = {1,1,1,1,1,1,1,1,1,1,1}; for (i0 = 0; i0 < 2; i0 ++) { x[0] = (x[0] == 1) ? 0 : 1; for (i1 = 0; i1 < 2; i1 ++) { x[1] = (x[1] == 1) ? 0 : 1; for (i2 = 0; i2 < 2; i2 ++) { x[2] = (x[2] == 1) ? 0 : 1; for (i3 = 0; i3 < 2; i3 ++) { x[3] = (x[3] == 1) ? 0 : 1; for (i4 = 0; i4 < 2; i4 ++) { x[4] = (x[4] == 1) ? 0 : 1; for (i5 = 0; i5 < 2; i5 ++) { x[5] = (x[5] == 1) ? 0 : 1; for (i6 = 0; i6 < 2; i6 ++) { x[6] = (x[6] == 1) ? 0 : 1; for (i7 = 0; i7 < 2; i7 ++) { x[7] = (x[7] == 1) ? 0 : 1; for (i8 = 0; i8 < 2; i8 ++) { x[8] = (x[8] == 1) ? 0 : 1; for (i9 = 0; i9 < 2; i9 ++) { x[9] = (x[9] == 1) ? 0 : 1; for (i10 = 0; i10 < 2; i10 ++) { x[10] = (x[10] == 1) ? 0 : 1; bush = 213; kerry = 171; for (i = 0; i < 11; i ++) { if (x[i] == 0) { bush = bush + electoral[i].e_votes; } else { kerry = kerry + electoral[i].e_votes; } } (void) printf ("%d, %d,", bush, kerry); for (i = 0; i < 11; i ++) { if (x[i] == 0) { (void) printf (" %s", electoral[i].name); } } (void) printf (","); for (i = 0; i < 11; i ++) { if (x[i] == 1) { (void) printf (" %s", electoral[i].name); } } (void) printf ("\n"); } } } } } } } } } } } }

Copyright © 2004 John Conover, john@email.johncon.com. All Rights Reserved. Last modified: Thu Oct 28 03:14:08 PDT 2004 $Id: 041028031100.21687.html,v 1.1 2004/10/28 10:18:06 conover Exp $