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I'm having trouble gaining intuition as to why the answer is 2/3 as opposed to 1/2.

Here is the question:

Alice has five coins in a bag: two coins are normal, two are double-headed, and the last one is double-tailed. She reaches into the bag and randomly pulls out a coin. The coin lands and shows heads face-up. What is the probability that the face-down side is heads?

So our sample space starts off as {H,T H,T H,H H,H T,T} Given that face-up is {H} our probability will be conditioned on the event

{H,T H,T H,H H,H }occurs, which leaves these possibilities -> {H, H, T, T}. Intuitively speaking, shouldn't the answer be 2/4?

The answer uses law of total probability to get the answer, which makes sense mathematically, but I don't get what's wrong with my logic in the intuitive answer.

Thank you.

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    $\begingroup$ See here. $\endgroup$ – C. Oliveira Feb 18 '18 at 18:42
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You are of course correct that it can only be one of the four coins as indicated, but when you say that the sample space $HT, HT, HH, HH$ leaves as options $T,T,H,H$ you are effectivly assuming that of the double-headed coins, you could only have gotten the 'first' head being face-up. But since the 'second' one could be the one that's face-up as well, your possible outcomes for the face-down side are $T,T,H,H,H,H$

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  • $\begingroup$ Thank you. This makes sense. I wrote this in another comment, but essentially, "I was assuming that since {H} has already occurred our sample space gets reduced to {T, T, H, H}. But since there are more heads in the original sample space, we have a higher chance of a double-headed coin being chosen, which is something I didn't look into. $\endgroup$ – moondra Feb 18 '18 at 19:02
  • $\begingroup$ @moondra Yes, that's it! :) $\endgroup$ – Bram28 Feb 18 '18 at 19:12
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"Intuitive thinking doesn't always give the right answer." is the short answer.

{H,T H,T H,H H,H }occurs, which leaves these possibilities -> {H, H, T, T}

First off I'm not sure what you mean here, but I think something like "Head means we can ignore the double-tailed coin. Then we could have head in two cases and tails in two cases." Your intuitive is wrong simply because in {H,T H,T H,H H,H } the H occurs six times but the T occurs only twice. There is no reduction to {H, H, T, T} I could see.

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  • $\begingroup$ I was assuming that since {H} has already occurred our sample space gets reduced to {T, T, H, H}. But since there are more heads in the original sample space, we have a higher chance of a double-headed coin being chosen. I think that's what I didn't look into. $\endgroup$ – moondra Feb 18 '18 at 19:00
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The sample space is: $$S=\{HT,HT,H_1H_2,H_2H_1,H_1H_2,H_2H_1\},$$ where the first (second) is top (bottom) and the index numbers are sides.

Hence: $$P=\frac{n(\text{bottom is head})}{total}=\frac46.$$

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Our configuration space is \begin{eqnarray*} \{H,T \,\,\, H,T \,\,\, H_1,H_2 \,\,\, H_1,H_2 \,\,\, T_1,T_2 \}. \end{eqnarray*} There are $6$ possible ways to get a head \begin{eqnarray*} \{\color{red}{H},T \,\,\, \color{red}{H},T \,\,\, \color{red}{H_1},H_2 \,\,\, \color{red}{H_1},H_2 \,\,\, \color{red}{H_2},H_1 \,\,\, \color{red}{H_2},H_1 \}. \end{eqnarray*} So ...

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