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The following problem was posed to me:

A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, i.e. the probability that the fair coin is selected is 0.5. When the gambler flips the chosen coin, it shows heads.

(A) What is the probability that it is the fair coin?

(B) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin?

(C) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

I am concerned with (C).

The following solution was provided:

Let $F$ be the event that the coin is fair, $F^c$ is the complement of $F$. Let also $H$ be the event that it shows a head.

$$P(F|HHH) = \dfrac{P(HHH|F)P(F)}{P(HHH)} = \dfrac{P(HHH|F)P(F)}{P(HHH|F)P(F) + P(HHH|F^c)P(F^c)} = \dfrac{1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2}{9/6} = 1/9$$

But isn't this a solution to the problem of probability that it is the fair coin when flipping the coin a third time and it showing heads? Shouldn't we instead be calculating $P(F|HHT)$ ?

But if we should be calculating $P(F|HHT)$, since only one of the coins (the fair coin) has a tails side, wouldn't $P(F|HHT)$ (the probability that the coin is fair instead of the two-headed coin) equal to $1$? In that case, we wouldn't even need to calculate anything.

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ The coin shows heads the first two times? $\endgroup$ – Magma Aug 25 at 22:27
  • $\begingroup$ @Magma Edited. Sorry, I got ahead of myself and forgot the rest. $\endgroup$ – The Pointer Aug 25 at 22:29
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You are correct, $P(F|HHT) = 1$, since $P(F^c|HHT) = 0$.

Is this question from a textbook? My guess is that there must have been a mistake while editing the source material, maybe the question was "Calculate $P(F|HHH)$" in a previous edition and the answer wasn't updated to match.

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  • $\begingroup$ Thanks for the answer. It's a practice question given in a lecture, but it likely was from the course textbook. $\endgroup$ – The Pointer Aug 25 at 22:39
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    $\begingroup$ @ThePointer - in that case, the false solution was surely because the Lecturer made a mistake in reading the question. This is particularly likely when they speak extemporaneously in response to an audience question. $\endgroup$ – Paul Sinclair Aug 26 at 16:39
  • $\begingroup$ It should be noted that until the first T shows up, there is a difference between the perceived probability and the actual probability. E.g. Once we know it's the fair coin (having seen a T), we can say that the actual probability of it being the fair coin on all the prior flips is and was always 1, but the perceived probability was lower since we did not yet know. $\endgroup$ – Darrel Hoffman Aug 26 at 20:02
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The given solution to C is not only wrong, it is also absurdly complicated.

A double headed coin (one where both sides show heads) never shows tails. That's it. If you see tails, then it is absolutely impossible that you have the double headed coin, so the probability for having the double headed coin is zero.

Even if you have a million heads and one tail, the probability that you picked the double-headed coin is zero. Obviously it is very very (repeat a few tenthousand times) very unlikely that this happens with the fair coin, but it is impossible to happen with the double headed coin.

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Problem C is ridiculously simple.

Dont lose time with F, H, P, T...

If you have a fake 2-headed coin, and a fair coin with heads and tails, and the third flip shows tails... Dude, the probability it is the fair coin is 100%.

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    $\begingroup$ This answer doesn't add anything new that isn't already in the other answers. $\endgroup$ – JBentley Aug 26 at 19:16

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