On Sat, 5 Apr 2008 18:56:10 -0700 (PDT),
TheTravellingSalesman
saad.zaman@gmail.com> wrote:

Can anyone help me with this question?

At the end of an election, the eventual winner,
candidate B, receives
m votes, and
H receives n < m votes. Assuming that as the votes
are counted they
come as
independent random variables with probability half
that they are for
either candidate,
show that the probability B always stayed ahead of
H through the
counting process
is given by
(m - n)/(m + n)

I've never seen that problem, and I'm surprised that
the answer is apparently that simple.

Well, it is true for n+m = 3. Maybe you can show it
by induction?

--
Rich Ulrich

http://www.pitt.edu/~wpilib/index.html

Can anyone help me with this question?

At the end of an election, the eventual winner, candidate B, receives
m votes, and
H receives n < m votes. Assuming that as the votes are counted they
come as
independent random variables with probability half that they are for
either candidate,
show that the probability B always stayed ahead of H through the
counting process
is given by
(m - n)/(m + n)

Can anyone help me with this question?

At the end of an election, the eventual winner, candidate B, receives
m votes, and
H receives n < m votes. Assuming that as the votes are counted they
come as
independent random variables with probability half that they are for
either candidate,
show that the probability B always stayed ahead of H through the
counting process
is given by
(m - n)/(m + n)

On Sat, 5 Apr 2008 18:56:10 -0700 (PDT),
TheTravellingSalesman
saad.zaman@gmail.com> wrote:

Can anyone help me with this question?

At the end of an election, the eventual winner,
candidate B, receives
m votes, and
H receives n < m votes. Assuming that as the votes
are counted they
come as
independent random variables with probability half
that they are for
either candidate,
show that the probability B always stayed ahead of
H through the
counting process
is given by
(m - n)/(m + n)

I've never seen that problem, and I'm surprised that
the answer is apparently that simple.

Well, it is true for n+m = 3. Maybe you can show it
by induction?

--
Rich Ulrich

http://www.pitt.edu/~wpilib/index.html



Rich, this is known as the ballot problem. It counts the number of paths from (0,0) to (m+n,m-n) for which the y coordinate is positive from x=1 on, divided by the total number of paths. This gives the probability that B is always ahead of H during the entire counting process. You can find the derivation in Feller and similar books.