Jasper County Democrat, Volume 3, Number 32, Rensselaer, Jasper County, 17 November 1900 — What Might Have Been. [ARTICLE]

What Might Have Been.

Mr. McKinley’s plurality in the popular vote will probably reach 750,000. It would seem at first thought that the Republican ticket might have suffered a loss of one teftth of its popular plurality without serious consequences. Seventy-five thousand votes are only about one half of 1 per cent of the total number cast last Tuesday in the Presidential election. Yet the resalt depended upon that apparently insignificant fTaction of the immense total. If 75,000 citizens who voted for McKinley in certain states of the Union had voted instead for Bryan, McKinley would have been defeated and Bryan elected. The table below mentions 12 states with the electoral yotes belonging to them and their several, pluralities for McKinley and Roosevelt, as unofficially ascertained or as estimated yesterday. Changes in the figures by later returns will not greatly affect the general proposition: Electoral Repub. Vote. Plural'y. Delaware ..... 3 5,000 Indiana..... 15 27,400 Kansas io 25,000 Maryland 8 14,360 Nebraska 8 5,000 North Dakota 3 8,000 Oregon.... ...I 4 14,000 South Dakota 4 10,000 Utah 3 4.000 Washing.on 4 5,000 West Virginia 6 15,000 Wyoming y 3,0-0 Total 71 135,760 Thus, leaving a margin of 14,000 for possible increase in the aggregate Republican plurality in these states, it is evident that the change of 75,000 votes would have reversed the result in all 12 of them. That is to say, if 75,000 citizens who voted for McKinley had voted for Bryan, 71 electoral votes nowin the McKinley column would have gone into the Bryan column, thus: McKinley’s electoral vote as it i 5.... 2c>2 Electoral vote of these 12 states 71 McKinley’s electoral vote as it would have been. 221 Necessary to a choice 224 On the other band: Bryan s electoral vote as it is 155 Electoral vote ot these 12 slates.... 71 Bryan’s electoral vote as it would have been 226 Necessary to a choice 224 The change of 75,000 votes, therefore, properly distributed in the 12 states in the list, would have given to Bryan two more than the necessary number of electoral votes; his electoral majority over McKinley would have been 5. —N. Y. Sun.