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One bag contains two coins. One is fair, the other is biased with Heads probability = $0.6$. One coin is randomly picked and it is tossed. It lands heads up. What is the probability that the same coin will land heads up if tossed again?

Now, the probability that for a coin randomly picked up the toss will result in a head is given by the formula for total probability. Writing $B_1, B_2$ the events that the fair coin and the biased coin is selected, respectively and $E_1$ the event that the coin will land heads up in the first toss, we obtain $$ P(E_1) = P(E_1|B_1)P(B_1) + P(E_1|B_2)P(B_2) = 0.5\cdot0.5 + 0.6\cdot0.5 = 11/20. $$ However, it is not clear to me how to calculate now the probability $P(E_2)$. I would like to use again the total probability formula for $E_2$, i.e., $$ P(E_2) = P(E_2|E_1)P(E_1) + P(E_2|E_1^c)P(E_1^c) = P(E_2|E_1)\cdot 11/20 + P(E_2|E_1^c)\cdot 9/20 $$ But I don't see how to calculate $P(E_2|E_1), \ P(E_2|E_1^c)$. So in the end I suspect that my method doesn't work and a better solution should be given.

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    $\begingroup$ If you know the probability that the fair coin was selected, the rest is easy. Could you calculate this ? $\endgroup$
    – Peter
    Jun 20 '20 at 15:49
  • $\begingroup$ @Peter Sorry but I don't find your comment particularly clear. Could you elaborate? $\endgroup$
    – RandomGuy
    Jun 20 '20 at 16:06
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Next you need to use bayes theorem to figure out the posterior probability $P(B_1|E_1)$ that the coin picked is fair $$P(B_1|E_1)=\frac {P(E_1|B_1)P(B_1)}{P(E_1)}=\frac {0.5 \cdot 0.5}{11/20} = 5/11$$

To determine the probability of $E_2$ given $E_1$ $$P(E_2|E_1) = P(E_2 B_1|E_1) + P(E_2 B_1^c|E_1) = P(E_2|B_1 E_1)P(B_1|E_1) + P(E_2|B_1^c E_1)P(B_1^c|E_1) $$ Using the fact that flips of the same coin are independent, $P(E_2|B_1 E_1)=P(E_2|B_1)=0.5$ and $P(E_2|B_1^c E_1)=P(E_2|B_1^c)=0.6$, and substituting these probabilities in gives $$P(E_2|E_1) = P(E_2|B_1)P(B_1|E_1) + P(E_2|B_1^c)P(B_1^c|E_1) = .5 \cdot 5/11 + .6 \cdot 6/11$$

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  • $\begingroup$ Thank you, very good explanation. However, I don't see why the last relation, $P(E_2|E_1) = P(E_2|B_1)P(B_1|E_1) + P(E_2|B_1^c)P(B_1^c|E_1)$ is true? $\endgroup$
    – RandomGuy
    Jun 20 '20 at 16:57
  • $\begingroup$ If you didn't know that $E_1$ was true, you would have $P(E_2) = P(E_2|B_1)P(B_1) + P(E_2|B_1^c)P(B_1^c)$, which is like the equation you used to determine the prior probability of $É_1$.Then you just add "given $E_1$" to all the terms. $\endgroup$
    – dovalojd
    Jun 20 '20 at 17:03
  • $\begingroup$ Or approached from a different direction, it's breaking $P(E_2|E_1)$ into the cases $P(E_2 \& B_1|E_1)$ and $P(E_2 \& B_1^c|E_1)$ $\endgroup$
    – dovalojd
    Jun 20 '20 at 17:08
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    $\begingroup$ Wait a minute: we certainly have $P(E_2B_1 |E_1) = P(E_2 |B_1E_1)P(B_1|E_1)$, but why this should be equal to $P(E_2 |B_1)P(B_1|E_1)$, i.e. why $P(E_2 |B_1E_1) = P(E_2 |B_1)$ as you claim, is not clear at all. I even doubt it's true. Can you write a proof? $\endgroup$
    – RandomGuy
    Jun 20 '20 at 17:31
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    $\begingroup$ It's not a mathematical invariant, it's part of the problem description. We're told a value of $P(E_2|B_1)=0.5$, and that value is independent of $E_1$, because that's how coins work; two flips are independent trials, so $P(E_2|B_1)=P(E_2|B_1 E_1)=P(E_2|B_1 E_1^c)$. If you have a fair coin, and it previously landed heads, you have a 50% chance it'll land heads again. If instead you have a fair coin, and it previously landed tails, you still have a 50% chance it'll land heads on the next flip. $\endgroup$
    – dovalojd
    Jun 20 '20 at 17:48

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