A coin is flipped twice. $A$ is the result of the first coin flip and $B$ the result of the second one. I know that $A$ and $B$ are independent. Following $$P(A\cap B)=P(A)P(B),$$ when it comes to write down the event of the intersection of $A$ and $B$ I am stuck because the event $A$ excludes the event $B$, and yet $$P(A\cap B)$$ is not empty. Can somebody help me understand the right modeling of flipping $2$ coins. Thanks.

  • $\begingroup$ Do you mean $B$ is the result of the second coin flip? $\endgroup$ – applyb Jun 5 '17 at 9:16

It doesn't make sense to talk about the intersection of $A$ and $B$ as you have defined them, because the result of a coin flip is a variable, not an event.

If you define them as events, say $A$ is the event that the first flip is a head, and $B$ is the event that the second flip is a head, then the formula $P(A\cap B)=P(A)P(B)$ makes sense and is correct. Here $A\cap B$ is the event that both $A$ and $B$ happen, i.e. both flips show heads. This can happen, so the events do not exclude each other, and it has probability $1/4$ if the coin is fair.


You call $A$ a "result".

That is okay but then it can take several values and e.g. you can look at it as function $A:\Omega\to \{H,T\}$.

Likewise $B$ is such a function.

Behind all this we have a probability space $\langle\Omega,\mathcal A,P\rangle$.

The independence tells us things like: $$P(A=H,B=H)=P(A=H)P(B=H)$$

Actually this is an abbreviation of:$$P(\{\omega\in\Omega\mid A(\omega)=H,B(\omega)=H\})=P(\{\omega\in\Omega\mid A(\omega)=H\})P(\{\omega\in\Omega\mid B(\omega)=H\})$$

The events $\{A=H\}$ and $\{B=H\}$ do not exclude eachother.

You might wonder: what is the look of $\langle\Omega,\mathcal A,P\rangle$?

There are lots of possibilities for that.

One of them is $\Omega:=\{\langle H,H\rangle,\langle H,T\rangle\langle T,H\rangle\langle T,T\rangle\}$ together with $\mathcal A=\wp(\Omega)$ and where $P$ is determined by $P(\{\omega\})=\frac14$ for every $\omega\in\Omega$.

Let $A$ be the function $\omega=\langle\omega_1,\omega_2\rangle\mapsto\omega_1$.

Let $B$ be the function $\omega=\langle\omega_1,\omega_2\rangle\mapsto\omega_2$.

This models two flips with an unbiased coin.

  • $\begingroup$ Ok. Thanks. But in the case i take $$ A=H,B=T $$, one should again verify the same law, which i don't see how. $\endgroup$ – user249018 Jun 5 '17 at 9:57
  • $\begingroup$ $$P(A=H,B=T)=P(\{H,T\})=\frac14=\frac12\frac12=$$$$P(\{HH,HT\})P(\{HT,TT\})=P(A=H)P(B=T)$$ in the model described in my answer. Btw, also $P(A=H,B=T)=P(A=H)-P(A=H,B=H)=\frac12-\frac14=\frac12$ $\endgroup$ – drhab Jun 5 '17 at 10:10
  • $\begingroup$ Thanks for the great clarity. I guess the key was the right modeling of the problem. $\endgroup$ – user249018 Jun 5 '17 at 21:12

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