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Could someone please help me solve the following question:

Say we have 3 coins which have probabilities 1/3, 1/2, 2/3 respectively of getting heads.

We choose a random order in which to flip the coins (all possible orderings being equally likely), once each. Let A_i = the event that the ith flip is heads. Are the events A_1 A_2 A_3 independent?

Attempt at solution: So the possibility of orderings for the coins are 123, 132, 213, 231, 312, 321 (if I label the three coins 1 2 and 3)

Do I then find A1, for example, by finding e.g. P(H first | 123)*1/6 for all the different combinations?

Is there a less tedious approach?

And then for checking independence, I would need P(A1 n A2), P(A2 n A3) etc, would I need to consider the same approach again e.g. P(A1 n A2) = P(HHT or HHH | 123)*1/6 + P(HHT or HHH | 132)*1/6 … ?

Please let me know if I’m on the right track or if I should take a different approach, thank you!

MFFF
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    Just intuitively, If the first two yield $H$, that's evidence (hardly proof) that the third one is the $\frac 13$ coin. To take an extreme case, if the probabilities were $.9999, .9999, .0001$ and the first two came up $H$, you'd be extremely confident that the third was going to come up $T$. – lulu Dec 05 '23 at 18:49
  • @lulu sorry im slow but how does this link to the question – MFFF Dec 05 '23 at 18:54
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    Since the outcomes of the first two rolls can alter your expectation for the third, they can't be independent. – lulu Dec 05 '23 at 19:00
  • @lulu oh i get what you’re saying. but i would like to know what the proper way of showing it will be since this is an 8 mark question on my homework – MFFF Dec 05 '23 at 19:47
  • Sure, so write out exactly what I said properly. Do the Bayesian computation to see what the probability that $A_3$ holds (third toss $H$) conditioned on seeing $H, H$ for the first two tosses. Of course you should first work out the unconditional $p(A_3)$. Were things independent, these computations should give the same value. – lulu Dec 05 '23 at 19:54
  • @lulu sorry i’m slow with maths. To find P(A1) for example do i use the law of total probability so that P(A1) = P(A1 | coin 1 first)P(coin 1 first) + P(A1 | coin 2 first)P(coin 2 first) + P(A1 | coin 3 first)P(coin 3 first)? how would i calculate these probabilities considering the coins can be selected randomly? – MFFF Dec 06 '23 at 14:50
  • Before you have tossed one, each coin has an equal chance of being any of the three, so your application of the law of total probability is correct (with each second factor equal to $\frac 13$). – lulu Dec 06 '23 at 15:25
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    To get $P(A_1\cap A_2\cap A_3)$ is even easier. Regardless of the order, that's $\frac 13\times \frac 12\times \frac 23=\frac 19$. Now, is that equal to $P(A_1)^3? If the events were independent, it would be. – lulu Dec 06 '23 at 15:27
  • @lulu is P(A1) = P(A2) = P(A3) ? I understand why each second factor will be 1/3 (thank you for explaining!) but how do we go on to calculate for example P(A1 | coin 1 first) ? is this just equal to P(H on coin 1) = 1/3 ? and is this the same as P(A2 | coin 1 first) ? – MFFF Dec 06 '23 at 19:18
  • Yes. Before you toss, you have no way to distinguish the coins, they all have the same probability. – lulu Dec 06 '23 at 20:23

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