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Imagine the following scenario: I flip a coin $5$ times. I tell you that I got $3$ times heads and $2$ times tails.

From your point of view: What are the chances, that the second throw was heads and why?

  • $\frac{1}{2}$ because every flip of a coin is $\frac{1}{2}$ chance to be heads or
  • $\frac{3}{5}$ because three out of the five coin flips were heads?
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    $\begingroup$ What are your thoughts? if, say, you were told that you got $5$ tails would you think that the probability that the second toss was $H$ is $\frac 12$? $\endgroup$ – lulu Dec 25 '19 at 11:41
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    $\begingroup$ Your every flip of a coin is 1/2 chance to be heads is only true >>before<< the coin is flipped. Afterwards, the outcome is what it is, regardless of that original probability. $\endgroup$ – John Forkosh Dec 25 '19 at 11:48
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According to Bayes' Theorem:

$Prob(2nd=H|3H2T)=\dfrac{Prob(3H2T|2nd=H)P(2nd=H)}{Prob(3H2T)}$

The respective probabilities on the RHS are $\frac{\binom42}{2^4}=\frac6{16}, \frac12, \frac{\binom53}{2^5}=\frac{10}{32}$.

$Prob(2nd=H|3H2T)=\frac{6\cdot \frac12 \cdot32}{16\cdot10}=\frac{3\cdot32}{16\cdot10}=\frac35$

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