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Straw Poll Gives Bush 60% of Vote

October 25, 1987|United Press International

BALTIMORE — Vice President George Bush captured 60% of the vote Saturday in a straw poll of Maryland Republicans, winning by a 4-1 margin over the second-place finisher, Kansas Sen. Bob Dole.

At their annual convention, Maryland Republicans cast 643 votes for Bush, 164 for Dole and 68 for New York Rep. Jack Kemp.

Former television evangelist Pat Robertson received 42 votes, former Secretary of State Alexander M. Haig Jr. had three and former Delaware Gov. Pierre S. (Pete) du Pont IV had one vote.

Former U.N. Ambassador Jeane J. Kirkpatrick received five write-in votes after the Maryland party's Central Committee overwhelmingly passed a resolution urging Kirkpatrick, of Chevy Chase, Md., not to seek the 1988 Republican nomination.

Maryland Republicans, who had talked of Kirkpatrick as a "favorite daughter" candidate, said her stands on economic and social issues would get little support in the party, according to Jim Wright, who offered the resolution to the committee.

Organizers for the Dole and Kemp campaigns said the poll proved only that Bush campaign workers were able to sell more $30 tickets for the convention, alluding to party members who arrived on Bush campaign buses or cast absentee votes.

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