TITLE: Unfair Coin Toss probability, Independence
QUESTION [1 upvotes]: We have two coins. The first one is a fair coin and has $50\%$ chance of landing on head and $50\%$ of landing on tails. for the other one it is $60\%$ for head and $40\%$ for tails. We choose one of them randomly (with equal chance) and toss it three times.The result is Tails-Head-Head. If we toss it again for the 4th time, what is the probability of the coin landing on tails.
At first, it seems like a bayes theorem question, but after some thought I think because the result of every toss is independent of others, the answer is just:
$$\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . \frac{4}{10} = \frac{9}{20} $$
The first $\frac{1}{2}$ is because we choose the coins randomly.
I want to know whether my argument is correct for this probability question or not.

REPLY [3 votes]: The answer, giving two terms corresponding to the actual coin being fair, then the actual coin being biased, will be
$$P(T) = \frac{\left(\frac12\right)^3}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2} \left(\frac12\right)
+ \frac{\frac25\left(\frac35\right)^2}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2}\left(\frac25\right)\\
$$
Doing the arithmetic,
$$
P(T) = \frac{1201}{2690} \approx 44.65\%
$$
The probability that your coin is biased is about 53.5%, which is why the tails on the next throw is less likely than you might think.
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