Probability question about tossing a coin

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!

\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} 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?
• 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. – Alec B-G Feb 14 at 7:34
• Do you understand the first line? We could put in one more line $$P(H_2|H_1)=\sum_iP(H_2,i|H_1)$$ – Alec B-G Feb 14 at 7:34
• 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. – icecream2727 Feb 14 at 7:54
• 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. – Alec B-G Feb 14 at 7:57