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There is a game in Australia called 2-up in which you toss two coins in the air simultaneously.

Suppose that when you throw the coins, one lands on the table and the other lands on the ground, so for now you can only see the coin on the table, and it is heads.

What is the chance that both coins are heads?

Now, I know the probability of getting at least one heads in this game is 1/3, but is this question not completely different?

user133538
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The probability of two heads before observing any of the coins is $\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$. The probability of another heads after observing a head for one of the coins is $1 \cdot \frac{1}{2}$ (because you now know that one of the coins is definitely a heads).

Yiyuan Lee
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  • Yes, this was my thought as well. Thank you. – user133538 Mar 06 '14 at 14:21
  • @user133538 Glad I could help! Since you seem to be new to StackExchange : Do remember to mark any answer posted for your question if you find it sufficient and helpful, by clicking the tick beside it. – Yiyuan Lee Mar 06 '14 at 14:23
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    @user133538 Yiyuan is right. "Accepting" the answers, provided they are what you need of course, is important. – Guy Mar 06 '14 at 14:34