==> probability/lights.s <==
Let E(m,n) be this number, and let (x)C(y) = x!/(y! (x-y)!).  A model
for this problem is the following nxm grid:

     ^         B---+---+---+ ... +---+---+---+ (m,0)
     |         |   |   |   |     |   |   |   |
     N         +---+---+---+ ... +---+---+---+ (m,1)
<--W + E-->    :   :   :   :     :   :   :   :
     S         +---+---+---+ ... +---+---+---+ (m,n-1)
     |         |   |   |   |     |   |   |   |
     v         +---+---+---+ ... +---+---+---E (m,n)

where each + represents a traffic light.  We can consider each
traffic light to be a direction pointer, with an equal chance of
pointing either east or south.

IMHO, the best way to approach this problem is to ask:  what is the
probability that edge-light (x,y) will be the first red edge-light
that the pedestrian encounters?  This is easy to answer; since the
only way to reach (x,y) is by going south x times and east y times,
in any order, we see that there are (x+y)C(x) possible paths from
(0,0) to (x,y).  Since each of these has probability (1/2)^(x+y+1)
of occuring, we see that the the probability we are looking for is
(1/2)^(x+y+1)*(x+y)C(x).  Multiplying this by the expected number
of red lights that will be encountered from that point, (n-k+1)/2,
we see that

            m-1
           -----
           \
E(m,n) =    >    ( 1/2 )^(n+k+1) * (n+k)C(n) * (m-k+1)/2
           /
           -----
            k=0

            n-1
           -----
           \
      +     >    ( 1/2 )^(m+k+1) * (m+k)C(m) * (n-k+1)/2 .
           /
           -----
            k=0


Are we done?  No!  Putting on our Captain Clever Cap, we define

            n-1
           -----
           \
f(m,n) =    >    ( 1/2 )^k * (m+k)C(m) * k 
           /
           -----
            k=0

and

            n-1
           -----
           \
g(m,n) =    >    ( 1/2 )^k * (m+k)C(m) .
           /
           -----
            k=0

Now, we know that

             n
           -----
           \
f(m,n)/2 =  >    ( 1/2 )^k * (m+k-1)C(m) * (k-1) 
           /
           -----
            k=1

and since f(m,n)/2 = f(m,n) - f(m,n)/2, we get that

            n-1
           -----
           \
f(m,n)/2 =  >    ( 1/2 )^k * ( (m+k)C(m) * k - (m+k-1)C(m) * (k-1) )
           /
           -----
            k=1

- (1/2)^n * (m+n-1)C(m) * (n-1)

            n-2
           -----
           \
 =          >    ( 1/2 )^(k+1) * (m+k)C(m) * (m+1)
           /
           -----
            k=0

- (1/2)^n * (m+n-1)C(m) * (n-1)

= (m+1)/2 * (g(m,n) - (1/2)^(n-1)*(m+n-1)C(m)) - (1/2)^n*(m+n-1)C(m)*(n-1)

therefore

f(m,n) = (m+1) * g(m,n) - (n+m) * (1/2)^(n-1) * (m+n-1)C(m) .


Now, E(m,n) = (n+1) * (1/2)^(m+2) * g(m,n) - (1/2)^(m+2) * f(m,n)

+ (m+1) * (1/2)^(n+2) * g(n,m) - (1/2)^(n+2) * f(n,m)

= (m+n) * (1/2)^(n+m+1) * (m+n)C(m) + (m-n) * (1/2)^(n+2) * g(n,m)

+ (n-m) * (1/2)^(m+2) * g(m,n) .


Setting m=n in this formula, we see that

           E(n,n) = n * (1/2)^(2n) * (2n)C(n),

and applying Stirling's theorem we get the beautiful asymptotic formula

                  E(n,n) ~ sqrt(n/pi).