I'm having trouble working out the conditional probabilities in this problem.
I tell you that I'm about to toss a coin, which is either fair, or will land heads every time.
You are told there is a 0.5 chance I pick either coin. Given this information, your subjective probability of the toss landing heads should be 0.75.
If I toss the coin 3 times for you, and it lands heads each time, how should you adjust your subjective probability, and to what value?
Thanks!
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Let us call $R$ the result event (3 times head), $H$ the event "the coin falls always on heads" and $F$ the event "the coin is fair".
The probability that we get 3 heads is $$ P(R)=P(R|H)P(H)+P(R|F)P(F)=1\times\frac12+\frac18\times\frac12=\frac9{16}.$$
So the probability that we have a fair coin is, after Bayes formula, $$P(F|R)=\frac{P(R|F)P(F)}{P(R)}=\frac{\frac18\times\frac12}{\frac9{16}}=\frac19.$$ We should therefore adjust our subjective probabilities to $P_{\rm new}(F)=\frac19$ and $P_{\rm new}(H)=\frac89$.
Let $X$ be the event that the coin being flipped is fair. Let $Y$ be the event that you have $3$ starting flips all landing on heads.
We can calculate $P(X|Y)$, the probability that the coin being flipped is the fair coin given you have $3$ flips all landing with heads. $$P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$$
$P(X\cap Y)=\dfrac{1}{2^3}\bigg(\dfrac{1}{2}\bigg)=\dfrac{1}{16}$. This is the probability that the coin is fair and you flipped heads thrice.
$P(Y)=\bigg(\dfrac{1}{2}\bigg)1+\bigg(\dfrac{1}{2}\bigg)\dfrac{1}{8}=\dfrac{9}{16}$. This is the total probability that we got heads thrice (either from the all-heads coin or the fair coin).
Hence $P(X|Y)=\dfrac{1}{9}$.
Since there is a $\dfrac{1}{9}$ chance that the coin is fair, and a $\dfrac{8}{9}$chance that the coins is all-heads, the probability of the next flip also being heads is $\bigg(\dfrac{1}{2}\bigg)\dfrac{1}{9}+\dfrac{8}{9}=\dfrac{17}{18}$.