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?
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

According to Bayes' Theorem:


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.