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We have a $501$ coins on the table, and assume that they have all been flipped onto that table (i.e., there is a mix of heads and tails). This also includes a two-headed coing.

Now if we pick up $1$ coin and its heads, what is the probability it is also the two-headed coin?

I've seen questions similar to this, but not the exact same. I did this out but want to hear your thoughts, before I explain how I arrived at the probability.

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What's your explanation? –  Patrick Li May 17 '13 at 4:37
    
You pick a heads-up coin without seeing any of the other coins? –  User58220 May 17 '13 at 6:27

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We will use the powerful Bayes theorem. \begin{align} \mathbb{P}(\text{Two sided coin }\vert \text{ we have a head}) & = \dfrac{\mathbb{P}(\text{we have a head }\vert \text{ Two sided coin}) \mathbb{P}(\text{Two sided coin})}{\mathbb{P}(\text{we have a head})}\\ &= \dfrac{1 \times \dfrac1{501}}{\dfrac12 \cdot \dfrac{500}{501} + \dfrac1{501}} = \dfrac1{251} \end{align}

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Perfect, thank you!! –  chris cole May 17 '13 at 4:49

This seems to be too simple to be rigorous, but it gives the right answer!

There are 502 "heads" on the table. Two of them are on the two-headed coin. You've picked a coin with a head, so the probability that the head you see comes from the two-headed coin is $\frac{2}{502}$ or $\frac{1}{251}$

Apparently, the fact that the coins are spread out on the table in some random heads-tails pattern is immaterial. You could do the experiment by putting the coins in a bag, and then pick one randomly, (with replacement, if the one side you look at is tails) until you see a head...

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