What is the probability that a randomly written n x n binary matrix (all elements 1 or 0) is singular?

A similar problem showed up in this newsgroup 4 months ago.  Here is my
old article, without the usual quote symbols (after all, I'm just
quoting myself :-)  )

***BEGIN OLD ARTICLE***

Subject: Re: A Probability Problem on Determinants.
Date: 20 Apr 1995 17:40:12 -0500

In <3n3h7n$jln@zip.eecs.umich.edu>,
kanad@colorado.eecs.umich.edu (Kanad Chakraborty) wrote:

>Determine the probability that a randomly chosen {0,1}-matrix
>of order n has odd determinant.
>
>[Hint: Permute !)

Reinterpret this problem as follows: if you are working over the
2-element field, what is the probability that a random matrix of order
n is non-singular?    Answer:

   n-1
--------
  |  |      n    r     n
  |  |  ( (2  - 2 ) / 2  )
  |  |
   r=0

Proof: In a nonsingular matrix, the first r rows must form a set of
independent vectors, for all r from 1 to n.  So the first row can be
any one of the 2^n possible vectors, except the vector of all zeroes.
The probability of randomly getting such a first row is (2^n -1)/(2^n).

Fix r<n, and assume the first r rows have been chosen so that they form
a set of independent vectors.  Look at the subspace of n-dimensional
space spanned by these r vectors.  That subspace contains exactly 2^r
vectors.  So the probability that a random vector is not in this
subspace is (2^n - 2^r) / (2^n).  This gives the probability that the
random vector chosen for the next row is "successful".

Continue choosing each row in this manner.  Then the final probability
of getting a nonsingular matrix is given by the product of these
factors, which is the formula I gave.

Now please explain your        >[Hint: Permute !)

***END OLD ARTICLE***

The above formula can be rewritten as follows, by reading the factors
from right to left:

 (1-1/2) (1-1/4) (1-1/8) (1-1/16) ... (1-1/2^n)

That is the probability of getting a non-singular matrix.  However, the
latest question asked for the probability of getting a singular
matrix.  So the final answer is:     1 - (that last product)

Note that this answer assumes that all calculations are done in base 2.
If you are working over the reals, then this answer is wrong, and I do
not know the correct answer to that version of the problem.  The
following example will show the distinction I am making:

  /  1  1  0  \
 |   0  1  1   |
  \  1  0  1  /

This matrix is singular in base 2, but not over the reals.  Your
statement of the problem was unclear, and I'm guessing that my answer
is what you are really looking for.

-- Barry Wolk <wolk@ccm.UManitoba.CA>  (August 1995)