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I'll skip the first 90% of Mike's message due to its condescending tone,
and go
<br>straight for the last paragraph.
<p>MIKE OSSIPOFF wrote:
<blockquote TYPE=CITE>I'm not asking for more symbolic argument. I'm asking
for an
<br>example consisting of actual probabilities, whatever ones that you
<br>consider relevant, like the probability that A & B will be frontrunners,
the
<br>probability that if there's a 2-way tie it will
<br>be betweent them, and the probability that there will be a 2-way tie.
<br>I ask that in your example all the probabilities that you mention
<br>be explicitly numerically stated.
<p>Mike Ossipoff</blockquote>
Granted. For a three-way race,
<p><tt>probability that: AB
AC BC Sum</tt>
<br><tt>_________________________________________________</tt>
<br><tt>both are front</tt>
<br><tt>runners
0.5 0.3 0.2 1.0</tt>
<br><tt>_________________________________________________</tt>
<br><tt>a tie occurs</tt>
<br><tt>involving
0.1 0.03 0.01 0.14</tt>
<br><tt>both</tt>
<br><tt>_________________________________________________</tt>
<br><tt>if a tie occurs,</tt>
<br><tt>both are front 0.71
0.21 0.07 1.00</tt>
<br><tt>runners</tt><tt></tt>
<p>In the first row after the header, the probabilites for each pair of
<br>candidates of being the two front-runners are given; they must add
up
<br>to one (I'm ignoring the remote possiblity of a three-way tie).
<p>The next row gives the probability of a tie involving each pair. The
<br>sum of these probabilities is the probability of any tie occurring.
<p>The last row is the probability, given a tie, that it is between the
<br>specified pair. These numbers are derived by dividing the probability
<br>of a tie between the two candidates (from the second row) by
<br>the probability of any tie (0.14). The sum in the third row is one
<br>(ignoring the rounding error after the division).
<p>Note that the numbers on the first row do not equal those in the
<br>last row.
<p> -- Richard
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