# Coin flipping, chances for second coin being heads

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?
• 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 $\frac 12$? – lulu Dec 25 '19 at 11:41
• 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. – John Forkosh Dec 25 '19 at 11:48

## 1 Answer

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 $$\frac{\binom42}{2^4}=\frac6{16}, \frac12, \frac{\binom53}{2^5}=\frac{10}{32}$$.

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