Independency and flipping a coin that might be a biased coin

Forums: Math, Probability, Biased Coin
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Setup:
There are two coins in an urn: A fair coin and a biased coin that lands Heads with probability q. At time 0, we choose a coin at random, and then write down unconditional probabilities P(H1), P(H2), ...P(H100). Here, P(Hi) = probability (estimated at time 0) of getting Heads on flip i. So event Hi is defined as the event that the coin land Heads on flip i. Event Ti = Tails on flip i. Then at time 1 we flip the coin the first time, at time 2 we flip the coin a second time, etc. We make a total of 100 coin flips.

It is the case that
P(H1) = P(H1|fair)P(fair) +P(H1|Biased)P(Biased) = (0.5)(0.5) + 0.5q = 0.25 +0.5q

P(H2) = P(H2|fair and H1)P(H1|fair)P(fair) + P(H2|fair and T1)P(T1|fair)P(fair) + P(H2|biased and H1)P(H1|biased)P(biased) + P(H2|biased and T1)P(T1|biased)P(biased) =

= (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5) + (q)(q)(0.5) + (q)(1-q)(0.5) =
=0.25 + 0.5q.

Thus unconditional probabilities P(H1) = P(H2).

Question:
Do I need to provide a proof that P(H1) = P(Hj) = 0.25 + 0.5q, where j is an integer 1 < j < 101? Or is this a trivial result? I guess it all boils down to independency. The flips are independent. However, events Hi and Hj are not independent, as P(Hj|Hi) is not equal to P(Hj), assuming j>i.

Is there some property of the setup above that I could cite to justify a statement
"P(H1) = P(Hj), where j is an integer 1 < j < 101" as being self-evident, without having to prove it?

Thank you for your kind help!

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@CanadianBill,

Doesn't the fact that you're to compute unconditional probabilities imply that P(H2) is not to be conditional on the outcome of H1? If so, doesn't it boil down to the fact that once you have a coin in hand, P(Hi) is constant?

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@markr,

Does this mean that events H1 and Hi are independent events? The thing is that if q is, say, 0, then the more Tails I see, the more likely it is that I have a biased coin. In other words, when q=0, the unconditional probability P(H2) = 0.25, but P(H2|H1)=0.5. Doesn't this prove that H1 and H2 are not independent? I am not sure what to make of this, and what technical reasons one could give to back up the fact that P(Hi) is constant...

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@CanadianBill,

In other words, my question is:
Is it the case that P(H1|information set at time 0) = P(Hj|information set at time 0) because H1 and Hj are assumed to be independent for all j > 1? Or is there another reason why those probabilities are equal?

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@CanadianBill,

Two things:
1) You say you are to compute unconditional probabilities, but it seems you are computing conditional probabilities.

2) What does P(H2|fair and H1) mean? It's the "and H1" that has me puzzled.

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@markr,

> P(H2|fair and H1)

This is the probability the second flip results in Heads given the coin is fair and the first flip had resulted in Heads.

In other words, I am describing a decision tree. At the first level of the tree, the coin can be either fair or biased. Given it is fair, the coin can land Heads or Tails. Given "fair and H1", the coin can land either Heads or Tails. To find the probability P(Fair and H1 and H2), we multiply the probabilities along the branches of the decision tree that we have to cross to get to that outcome...

> you are to compute unconditional probabilities, but it seems you are computing conditional probabilities.

P(Hi) are unconditional probabilities for all 0<i<101. But to test for independency between Hi and H1, I thought we needed to compare P(Hj) to P(Hj|H1).

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@CanadianBill,

Once you have a coin in hand, aren't all of the flips independent of one another?

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@CanadianBill,

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@markr,

> Seems to me that your computation assumes independence:

I used the result on the third page of
http://web.as.uky.edu/statistics/users/viele/sta320u04/condprob.pdf

The result is that in general (when we DO NOT assume independence), P(A intersect B intersect C) = P(C|A intersect B)*P(B|A)*P(A).

For the our purposes, no matter how many Heads we observe on the past flips, a fair coin will land heads with probability 0.5, and a biased coin will land Heads with probability q.

But it is still the case that P(Hj|H1) is not equal to P(Hj), so I am hesitant to call events Hj and H1 independent.

1 Reply

@CanadianBill,

I'm not saying I know this to be absolutely correct, but this is how I look at it. Once you have a coin in hand, the flips are independent. Unconditional probabilities calculated beforehand will be identical for each flip. After you start flipping, historical outcomes affect the conditional probability of the coin being fair or biased which affects the computed probability of the upcoming flip.

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@markr,

I am writing an academic paper (in a field related to social sciences). I am wondering how one can use technical language to explain why P(H1) = P(Hj)...

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@CanadianBill,

How could we improve using Bayes rules to compute probability backwards?

We might establish the chance to get a sample in case we picked the biased or fair coin

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Independency and flipping a coin that might be a biased coin

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