The Probability of only one god = 1 minus probability of no god minus the probability of more than one god.
That said, probability arguments can be telling. However they can also lead to some truly weird situations. For example, consider the Probability of randomly selecting a given digit value from the infinite number of digits in some irrational number's decimal expansion. Let me construct such a number in this way decimal pnt, one, zero, one, zero, zero, one, zero, zero, zero, etc where each time one occurs it is followed by one more zero than the last time it occurred. In no# form we have .10100100010000... It never starts repeating from any location. Thus, it is irrational. It contains a infinite number of one's and zero's, and yet the probability of randomly selecting a 1 across the whole irrational number is 0!
Here's why: take any finite sequence ending in 1 of the irrational number that starts at the decimal and has at least one zero. Count the number of 1's in the sequence and subtract one from that total. So lets say that we count N 1's, our number is thus (N - 1). Because of how the irrational number is constructed, the number of 0's will equal 1+2+3+...+(N-1) in the finite sequence we just studied. Now the sum 1+2+3+...+(N-1) is (N-1)*(N)/2 or (N^2 - N)/2. Adding the number of 1's (N) to the number of 0's (N^2 - N)/2, we get (N^2 + N)/2.
The probability of randomly selecting one of the N 1's in the finite sequence we are studying is equal to the number of 1's in the sequence (N) divided by the numbers in the sequence (N^2 + N)/2. Thus the probability of selecting a 1 is equal to 2N/(N^2 + N) or 2/(N+1). This number decreases as N increases. And using a bit of first year calculus: The limit of 2/(N+1) as N approaches infinity which is equal to 0.