Anyway, being a rather contentious
lot, the AEIOU had a bitterly fought
and drawn-out election campaign for
its first-ever chairperson. The details of
the campaign don’t concern us, just the
outcome. The final vote was a landslide.
Amy Anderson won by a two to one margin over Bob Brady, 512 votes to 256.
The final vote count was conducted
in accordance with the AEIOU’s bylaws,
each vote being read aloud in a public
setting and held up for everyone to see,
to insure that there was no possible
manner in which the count could be off.
After each vote was read, the running
tally of the vote was marked on a board,
clearly visible to everyone present. You
can just imagine the crushed spirits of
the Bob Brady bunch as the long painfully drawn-out process dragged on into
the night and they watched Amy Anderson’s lead grow and grow.
Now, here’s the puzzle. Given the
lopsided result, what’s the probability
that at some point in the vote tally Bob
was tied with Amy? Put slightly differently, what’s the probability that Amy
and Bob had the same number of votes
after counting the second, or fourth,
or sixth vote, on up to the 768th vote?
Once you’ve settled that, please also
answer the question for a general two-way vote—that is for the final tally being
a:b, what is the probability that at some
point the ultimate loser was tied with
the ultimate winner?
And every sequence that starts with B
starts with B taking a lead. So, if you take
twice the number of sequences that start
with B, divided by the total number of
sequences, you have the probability of B
being in a tie at some point.
In the puzzle’s specific case, 256/
(256+512) = 1/3 of all the possible sequences start with B and so in two-thirds
of possible sequences, B at some point
ties with A. For the puzzle’s general case,
a>b, 2b/(a+b) of the time, B will be tied
at some point with B.
solvers
Robert Bartholomew, Benjamin Bock,
Bob Byrne, Andrew Dean, David
Engelmayer, Mark Evans, Rui Guo,
Lee Michelson, Geoff Moak, Chris
Norman, David Oakden, David
Promislow, Sally Jane Smith, John
Snyder, Al Spooner, Joel Winters
Honorable mention:
Thomas Dahl, Megan Farris, Adam
Hoelscher, Robert Johnson, Bob Rietz,
Jeff Schwartzmann, Stephen Vincent
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solution
There are clearly more possible sequences of votes than you would want to list
out. However, it’s helpful to list out a few
of the possible beginning sequences, because they let you observe that in those
sequences that create a tie, there’s a certain symmetry. That is, if after six votes
you have a tie (say, AABABB), there’s another sequence with the votes reversed
that gets the same tie but with the initial
lead reversed (BBABAA, in this case).
Now, every sequence that starts with
B taking a lead must eventually come to
a tie, because A ultimately overtakes B.
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Insureware ..................... 44–45
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Jacobson ......................... 71
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JPMorgan ......................... 34
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MaxBermuda ..................... 17
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Mid America Search ................ 72
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Milliman ..................... 3, 36–37
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Pinnacle Group ..................... 73
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PolySystems, Inc. .................. C3
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Pryor Associates/Pauline Reimer . . . . . 64
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tillinghast
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Valani Consulting .................. 20
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Watson Wyatt...................... 9