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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
Also see this isomorphic question – MJD May 16 '12 at 19:13
Also here it is again – MJD May 16 '12 at 19:15
up vote 2 down vote accepted

If you know that the coins are fair and the tosses are independent, and if the "given at least 1 head" is strictly interpreted (you know that, and just that), your answer is correct. A priori (without the additional "at least 1 head" condition), you had four equiprobable cases. The condition removes one of the four possible cases, but does not say anything more (hence the three remaining cases remain equiprobable). Hence, the probability of the case "2-heads" is 1/3.

If you feel this is incorrect, perhaps you are thinking (wrongly here) along the lines of this related problem. That statement might sound equivalent to this one, but it's not.

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Thanks a lot! In fact I was confident about my answer when drawing out all the same spaces. – Chan Nov 29 '11 at 19:41

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