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One about flipping coins. Coin A has 0.6 probability of being heads, Coin B has a 0.3 probability of being heads and Coin C has a 0.1 probability of heads. Given that a coin is flipped and lands on heads, what is the probability that when it is flipped again it will also be heads. I'm not sure how I can apply Bayes' rule here.

I did a Monte Carole simulation the answer come out to be around 0.46, I am not sure how to give a analytical answer.

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2 Answers 2

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As a way to get intuition for such problems: suppose we toss each coin $10$ times. Then we expect to get $10$ Heads... $6$ Heads from $A$, $3$ from $B$, and $1$ from $C$. Accordingly if our prior was that our mystery coin could be any of the three equally, then seeing Heads now makes us rethink. Now Bayes' tells us that there is a $.6$ probability the coin was $A$, $.3$ that it was $B$, and $.1$ that it was $C$.

Note: the fact that these numbers coincide with the weightings is a consequence of the fact that the weightings add to $1$, it is not generally true.

Thus the probability that tossing the coin again yields another $H$ is: $$.6^2+.3^2+.1^2=.46$$

Which, unsurprisingly, lines up with the result of your Monte Carlo simulation.

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  • $\begingroup$ I'm not fully convinced, each coin should have 1/3 of chance being picked however, given the fact that the first coin is more biased to flip a head then it should carry more weight that the first flip was the first coin. Maybe you have skipped a few steps in the calculation. $\endgroup$
    – szd116
    Feb 19, 2016 at 20:44
  • $\begingroup$ No...I'm saying that your prior was $\frac 13$. Meaning before you toss the coin for the first time. After the first toss, your estimates of the probability change. $\endgroup$
    – lulu
    Feb 19, 2016 at 20:45
  • $\begingroup$ Thank you, this is a very elgant answer, I would really need something like this during interviews. I think i can impress them by giving them something above, thanks again. $\endgroup$
    – szd116
    Feb 19, 2016 at 20:52
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Making use of tree diagram,

\begin{align*} P(H|H) &= \frac{P(HH|A)+P(HH|B)+P(HH|C)}{P(H|A)+P(H|B)+P(H|C)} \\ &= \frac{\frac{1}{3} \times 0.6^{2}+ \frac{1}{3} \times 0.3^{2}+ \frac{1}{3} \times 0.1^{2}} {\frac{1}{3} \times 0.6+ \frac{1}{3} \times 0.3+ \frac{1}{3} \times 0.1} \\ &= 0.46 \end{align*}

enter image description here

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  • $\begingroup$ Very nice and very clear thank you. $\endgroup$
    – szd116
    Feb 19, 2016 at 20:51
  • $\begingroup$ And thanks for the tree diagram $\endgroup$
    – szd116
    Feb 19, 2016 at 20:53

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