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You have a bag with two coins. One will come up heads $40\%$ of the time, and the other will come up heads $60\%$. You pick a coin randomly, flip it and get a head. What is the probability it will be heads on the next flip?

My approach to the problem is the following one.

We want to compute the probability $\mathbb{P}(hh\mid h)$. By Bayes, this is equivalent to $$\frac{\mathbb{P}(h\mid hh)\cdot\mathbb{P}(hh)}{\mathbb{P}(h)}$$ It is immediate that $\mathbb{P}(h\mid hh)=1$. On the other hand $$\mathbb{P}(hh)=1/2\cdot (0.6)^2 + 1/2 \cdot (0.4)^2=0.26$$ and $$\mathbb{P}(h)=1/2\cdot 0.6+1/2\cdot 0.4=0.5$$ Therefore, $$\mathbb{P}(hh\mid h)=0.52$$ Is my approach correct?

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    $\begingroup$ Its $P(h|hh)=1$, and $.26/.5=.52$, but otherwise correct. $\endgroup$
    – JMP
    Jun 6 at 8:46
  • $\begingroup$ @JMP Thanks, I have just edited the post with the right solution. $\endgroup$ Jun 6 at 12:13

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Imagine doing this experiment 1000 times. 500 times we pick coin A that has 40% probability of coming up heads so the first time we flip the coins we get 200 heads and 300 tails. 500 times we pick coin B that has 60% probability of coming up heads so the first time we flip the coins we get 300 heads and 200 tails.

So we got a total of 500 heads. 200 of those times it was with coin A so we will get .4(200)= 80 heads the second time we flip the coins. 300 or those times it was with coin B so we will get .6(300)= 180 heads the second time we flip the coins. That is, we got a total of 80+ 180= 260 heads in those 500 times. Given that we got heads on the first flip the probability we also get heads on the second flip is 260/500= 0.52.

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