Problem Two coins are tossed. What's the probability of 2-head given at least 1 head?

I got the answer to be: $$P = \dfrac{(1/2)(1/2)}{1 - 1/4} = \dfrac{1}{3}$$ but I have a feeling it should be greater than this, could anyone help me out?

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This is correct. Stipulating at least one head leaves three equally probable cases (HT, TH, and HH), one of which is the 2-head case. –  mjqxxxx Nov 29 '11 at 19:32
–  MJD May 16 '12 at 19:13
Also here it is again –  MJD May 16 '12 at 19:15