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I am stuck on a probability question:

We have 10 (biased) coins. When the $i$-th coin is tossed, the probability of heads is $i/10$ $(i = 1, ..., 10)$. We randomly select a coin, toss it, and get heads. Toss again the same coin. What is the probability that it lands heads up again?

I started with doing this:

Let $ H_1 = $ " We get heads on the first toss of a coin", $ H_2 = $ " We get heads on the second toss of a coin" , $ C_i = $ " The coin selected is the $i$-th coin"

Now I want to find $P(H_2$ $\cap$ $ H_1$ $ / C_i)$ for all $i \in $ {1,2,...,10}, that is, $\sum _{i=1}^{10} {P(H_2 /C_i) \cdot P(H_1/C_i)}$

But I'm getting a feeling that I'm doing something wrong. Any help would be much appreciated.

Thanks!

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\begin{equation}\begin{aligned}P(H_2|H_1) &= \sum_i P(H_2|H_1,i)P(i|H_1)\\ &= \sum_i \frac i{10}P(i|H_1),\end{aligned}\end{equation} The second line uses the fact that the coins are memoryless. So now we need the probability that given a coin said heads, it was coin $i$. For this we can use Bayes' theorem. Can you finish it from here?

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  • $\begingroup$ What do you mean by memoryless? I don't understand how you came up with the equation. $\endgroup$ – icecream2727 2 days ago
  • $\begingroup$ Memoryless means that $P(H_2|i,H_1)=P(H_2|i)$. This is because the coin $i$ does not remember what it previously threw. Every time you throw coin $i$ (knowing that it is coin $i$) it is $i/10$ chance of giving heads regardless of what happened last time. $\endgroup$ – Alec B-G 2 days ago
  • $\begingroup$ Do you understand the first line? We could put in one more line $$ P(H_2|H_1)=\sum_iP(H_2,i|H_1) $$ $\endgroup$ – Alec B-G 2 days ago
  • $\begingroup$ But shouldn't it be like $P(H_2|H_1)=\sum_iP(H_2|i,H_1)$ because this way we are saying: given that we selected the coin $i$ and got heads on the first toss, what's the probability that we get heads on the second toss? I mean this sound more intuitive to me. $\endgroup$ – icecream2727 2 days ago
  • $\begingroup$ You need to multiply the summand by the probability that you have chosen coin $i$ given that you got $H_1$. Otherwise you overcount. This is what the first line of my answer is saying. $\endgroup$ – Alec B-G 2 days ago

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