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Quick basic question here to make sure I understand conditional probability properly.

You flip two coins, and at least one of them is heads. What is the probability that they are both heads?

Now, I think the answer to this is $\frac{1}{3}$, for the following explanation. If $A$ is the event that the first coin lands heads, and $B$ is the result that the second coin lands heads, then what we're looking for is $P(A \cap B | A \cup B)$. The probability of both $A$ and $B$ is $\frac{1}{2}$, so $A \cup B$ is $\frac{3}{4}$. The probability of $A \cap B$ is $P(A|B)P(B)$, which is $\frac{1}{2}(\frac{1}{2}) = \frac{1}{4}$. Therefore, the probability is $\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$.

Now, besides the obvious question of whether or not this is actually right, I'm wondering: why is it that $P(A|B) = P(A)$? Is it because these are independent events? If so, can I use this fact as proof that these are independent events?

Thanks for the help!

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    It is correct and yes it is proof that they are independent events. Definition of independence is $P(A\cap B) = P(A)P(B)$ which you find that $P(A\cap B) = P(A)P(B) \Leftrightarrow P(A) = \frac{P(A\cap B)}{P(B)} \Leftrightarrow P(A) = P(A|B)$ – JMoravitz Oct 26 '14 at 00:20
  • If one is heads the probability for the other being head is 50%. – Masacroso Oct 26 '14 at 00:21
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    You may be interested to read this: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox which discusses at length the difference between the question of knowing "at least one of the coins is heads" versus knowing "the first of the coins was heads." The boy-girl paradox is simply a reformulation of the problem you just looked at worded as genders of children instead. – JMoravitz Oct 26 '14 at 00:25
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    @Masacroso: If the first is heads, then the probability that the second is heads is 50%. But if all we know is that one of the two is heads, the probability that the other is also heads is 1/3. – Charles Oct 26 '14 at 00:26
  • I’m not sure what your question is. Your calculation is correct, if you assume that the flips are independent, and the first statement you make that depends on the independence of $A$ and $B$ is “The probability of both [meaning each one of] $A$ and $B$ is $\frac{1}{2}$, so [the probability of] $A\cup B$ is $\frac{3}{4}$.” – Steve Kass Oct 26 '14 at 00:27
  • Sorry it wasn't more clearly defined. My question was mostly regarding what constituted proof of independent events. –  Oct 26 '14 at 00:31
  • I've asked another question here which relates to this concept but perhaps provides a better approach to solving my concerns. –  Oct 26 '14 at 01:07

2 Answers2

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Very easy solution to this not even requiring any formulas. There are only $4$ possible outcomes of $2$ fair coin flips: (HH, HT, TH, TT). If we know one of them is a H, then we can concentrate on just (HH, HT, and TH) since TT has no heads. Both HT and TH will result in the other coin being a tail and only the HH will result in the other coin being a head so it is $1$ good outcome out of $3$ possible and they are all equally likely so the correct answer is $1/3$ probability. Done!

David
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It depends on how you came by the information that "at least one of them was heads".

I'm not sure why no one brings this up, it's an incredibly important piece of information that's missing. Try and apply this theory to reality and prepare to be confused.

For example, if you're blind, and you flip two coins, a machine sees them and says "at least one of the coins is heads" what are the odds the other coin is heads?

In that case you flipped two coins and are given that "at least one of them is heads".

It's not 33%! It depends on the rule your machine is following when he talks. If his rule is "If at least one coin shows heads, I will always inform them of such and never anything else" then yes, it'll be 33%. But if it's rule is almost anything else, it won't be!

For example if it's rule is "I will always inform him regarding one of the coins that I see" Then half the time He'll say heads, half the time he'll say tails.

The sample space in that scenario is (assuming each flip combination happens twice for full sample space)

HH says Heads 2 times TH says Heads 1 time and Tails 1 time HT says Heads 1 time and Tails 1 time TT says Tails 2 times

In that more likely case where the machine randomly chose one coin to tell you of, If he tells you "one of the coins is heads" the 2/4 times the other coin will be heads too. 50%, not 33%

Whenever you see the word "given that..." always wonder, "but how is it given?". (I mean maybe not useful in a math test, but 100% needed in real life probabilities)

All I'm saying is knowing that "at least one of the coins is heads" is a given, is not nearly enough information if you don't know how it was given.

Jacob Weigensberg
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