Suppose we have a coin with bias $\theta$ within [0, 1]. If we define the outcome of flipping a coin as a random variable X with value in $\{0, 1\}$ and flip it twice (given $\theta$ known), would the following hold true?

$P(x_{1:2}=00 | \theta)=P(x_1=0|\theta)P(x_2=0|\theta)$, where $x_{1:2}$ as shorthand for the sequence $x_1, x_2$.

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    $\begingroup$ The events are conditionally independent but not unconditionally independent. $\endgroup$
    – Xi'an
    Oct 10, 2020 at 12:50
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    $\begingroup$ The question in the title is different from the question in the text. $\endgroup$ Oct 10, 2020 at 15:09
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    $\begingroup$ @Xi'an - that looks like a Bayesian comment. I agree with it, but it seems to make assumptions about the nature of $\theta$ $\endgroup$
    – Henry
    Oct 10, 2020 at 15:36

1 Answer 1


As also mentioned in the comments, the two tosses are independent only if we know the probability of head of the coin itself. Otherwise, the probability of head can be inferred from past outcomes (in this case there is only one) and the events are dependent, i.e. $P_{X_1,X_2}(x_1,x_2)\neq P_{X_1}(x_1)P_{X_2}(x_2)$.


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