The runoff
In the town of Miniville, Tom and Dick are in a runoff for alderman. As it happens, 3 votes are cast for Tom and 1 for Dick. (Tom has two family members living in town, and Dick has no remaining family.)
Harry takes the ballot box and counts the votes, removing one vote at a time (they still use paper ballots in Miniville.) What are the chances that Harry's "running count" has Tom ahead of Dick at all times, starting with the first vote? Express as a percentage, to two decimal places.
The answer is 50.00.
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1 solution
In order for Tom to be ahead of Dick at all times, the first two ballots removed must be votes for Tom. The chance of the first ballot being removed being a vote for Tom is 3 / 4 . Since there are then two ballots for Tom and one for Dick remaining, the chance of the second ballot being removed being a vote for Tom is 2 / 3 . Multiplying, you get a total chance of 2 / 4 = 50% exactly.