I'm working on a maths exercise and came across this question.

The probability of a "heads" when throwing a coin twice is 2 / 3. This could be explained by the following:

• The first time is "heads". The second throw is unnecessary. The result is H;

• The first time is "tails" and twice "heads". The result is TH;

• The first time is "tails" and twice "tails". The result is TT;

The outcome: {H, TH, TT}. two of the three results include a "heads", it follows that the probability of a "heads" is 2/3

What's wrong with this reasoning?

I think the answer is 1/2, is that right?

Ps. my first language isn't english,

Thanks Jef

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You are correct that the probability of exactly one head from two tosses is $1/2$. However the probability of at least one head from two tosses is $3/4$. – Henry Dec 16 '11 at 1:07

The three events in the sample space are not equally probable. P(H?)=1/2 P(TH)=1/4 P(TT)=1/4 Probability of getting at least one head is then 1/2+1/4=3/4.

In probability problems, you need to be careful about choosing your sample space.

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Sorry there was an error in my question, The probability of a "heads" when throwing a coin twice is 2 / 3 instead of getting heads twice – Jef Dec 15 '11 at 22:25
Thanks for your help! – Jef Dec 16 '11 at 12:00

The reason your result, as Shitikanth has already pointed out, is wrong, is that you've applied the principle of indifference where it doesn't apply. You can only assume that events will all be equally likely if they're all qualitatively the same and there's nothing (other than names and labels) to distinguish them from each other. The prototypical examples are the two sides of a coin and the six sides of a six-sided die.

In your case, on the other hand, the event H is qualitatively different from the two events TH and TT, so there's no reason to expect that these three will be equiprobable, and the principle of indifference doesn't apply. To apply it, you need to look at qualitatively similar events. In this case, that would be HH, HT, TH and TT. Of these four, three contain a heads, so, as Shitikanth has already stated, the probability is $3/4$.

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Thanks for your explanation and help! – Jef Dec 16 '11 at 12:00