sorry folks, there were many errors in the first versionand I ran out of time to edit it.
Interesting approach PBE6, I got 132 moments, and 66 possible hand configurations, with the following proof:
It simplifies things to work with the position of a hand represented by a number between 0 and 12, where 0 and 12 are midnight.
The following equation will produce such a number:
Pos_1 = H + X/12 + F/12
where H and X are Integers between 0 and 11 (inclusive), and 0<=F<1
In such a position, H will correspond to hours, and x/12 + f/12 to "fractions of an hour", the definition ensures that the fractional part cannot exceed 1.
So let us assume Hand_A of the clock is at such a position, and first of all let us assume that it is the Hour Hand.
We know that Hour = H, and, working in fractional minutes we can state:
Minutes = 5(X + F)
(note that Minutes cannot exceed 60)
If Hand_A is the Hour Hand, then Hand_B must be the minute hand, and in that case, to show 5(X + F) minutes, it must be at:
Pos_2 = X + F
But, assuming this is an ambiguous time, Hand B could alternatively be an Hour Hand
Noting that X ranges from 0 to 11 and F from O to 1, so from the mapping of position to time, if this is the case the time will be X hours, and F*60 minutes.
So if Hand B showed the hour, the Hand_A would be the minute hand, to show F*60 minutes, it would point to:
Pos_3 = 12F
Because they refer to the same hand, Pos_1 and Pos_3 must be equivalent, so:
12F = H + X/12 + F/12
which rearranges to:
F = (H + X/12)*12/143
We can therefore cancel F out of the first equation and rearrange to give:
Pos_1 = (144H + 12X)/143
Similarly we can rearrange the second equation to give the pleasingly symmetric:
Pos_2 = (12H + 144X)/143
So, every combination of H and X is an ambiguous time, giving 12^2 = 144 ambiguous moments.
Of these 144, for some Pos_1 and Pos_2 will be identical, meaning that while we don't know which hand is the hour and which is the minute we do know the time. Noting that in Pos 1, H is the hour, and in Pos 2 X is the hour, and for the times to be identical the hours must be equivalent, therefore for these times X=H
So there are 12 moments where the two hands are in the same place
This leaves 144 - 12 = 132 moments
But, if we assume the hands are identical these, there are only half this number of unique hand patterns so:
Unique, ambiguous hand patterns= 132/2 = 66
ambiguous moments = 132