Gambler coin problem: fair coin and two-headed coin

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.


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.


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%.

Tags:

Probability

Bayes Theorem

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