Expected value of a roll of a fair die given that the number rolled is at least 4

Question

I am trying to understand the solution to a probability problem, and I am having trouble understanding where some of the numbers are coming from.

The textbook gives this definition for conditional expectation, which I have no trouble understanding:

The conditional expectation $Ex[R \;|\; A]$ of a random variable $R$ given event $A$ is: $$ Ex[R \;|\; A] ::= \sum_{r\;\in\text{range}(R)} r \cdot Pr[R = r \;|\; A] $$

The textbook gives the expected value of the a die roll given the event that the number rolled is at least 4:

$$ Ex[R \;|\; R \geq 4] = \sum_{i = 1}^6 i \cdot Pr[R = i \;|\; R \geq 4] = 1 \cdot 0 + 2 \cdot 0 + 3 \cdot 0 + 4 \cdot \dfrac{1}{3} + 5 \cdot \dfrac{1}{3} + 6 \cdot \dfrac{1}{3} = 5 $$

I understand why the result is what it is. I do not understand where the multiplier $\dfrac{1}{3}$ in the last three terms comes from.

Answer 1

Because the question asks for the excepted value of the face on a die given that the number rolled is at least $4$, we know for certain that the probability space only includes $4,5,6$. The faces on the die, $1,2,3$, occur with probability $0$ given that the number rolled is at least $4$. Since only $4,5,6$ occur and, in a fair die, each face would occur with equal probability, the probability of rolling $4,5$, or $6$ should be $\frac{1}{3}$.

Another way to see this is that the entire probability space of rolling a die is $\Omega=\{1,2,3,4,5,6\}$ since only these numbers can appear. Now we have a subset $A\subset\Omega$ where $A=\{4,5,6\}$ is the event that the die shows a number at least $4$.

By definition of probability spaces $P(A)=1$. Since any event $\omega\in A$ occur with equal probability, $P(4|A)=P(5|A)=P(6|A)=\frac{1}{3}$.

Thus, $E[R|R\geq4]=5$.

Answer 2

The factor of $ \frac{1}{3} $ applies because that is the probability of rolling a 4, given the result is 4 or greater and the same for 5 and 6. Thus the expected value is 5.

Answer 3

$\Pr[R=i|R\ge 4]$ means "probability that comes out $i$ when only $4,5,6$ are possible outcomes".

You have only $3$ possible outcomes so for each outcome $i$ the probability is $1/3$ if it is possible or $0$ if is impossible.

Another way to see that is decomposing the conditional probability, i.e.

$$\Pr[R=i|R\ge 4]=\frac{\Pr[R=i\text { and }R\ge 4]}{\Pr[R\ge 4]}$$

If $i$ is $4,5$ or $6$ then $\Pr[R=i\text { and }R\ge 4]=1/6$. If $i$ is $1,2$ or $3$ then $\Pr[R=i\text { and }R\ge 4]=0$. And $\Pr[R\ge 4]=1/2$.

Answer 4

The OP asked where the multiplier 1/3 comes from in the last three terms.

Apparently, my answer wasnt clear enough, so I will attempt to clarify. The solution is essentially the equivalent of finding the average of all qualifying results, namely 4, 5, and 6.

So, as with finding any average, you must divide the sum of the terms by the number of the terms you are averaging.

There are three results that qualify for being "at least equal to 4", so you must divide the sum of the qualified results by three to find it's average.