Law of total conditional expectation question

Question

I was looking at the accepted answer of this post : Six-sided and four-sided dice question contradiction. It says :

$\mathbb{E}(X \,|\, R_2) = \mathbb{E}(X \,|\, D_6)P(D_6 \,|\, R_2) + \mathbb{E}(X \,|\, D_4)P(D_4 \,|\, R_2)$ where $D_4$ and $D_6$ are disjoint events forming a partition.

I thought however that the right formula was $\mathbb{E}(X \,|\, R_2) = \mathbb{E}(X \,|\, D_6 \cap R_2)P(D_6 \,|\, R_2) + \mathbb{E}(X \,|\, D_4 \cap R_2)P(D_4 \,|\, R_2) $.

Also, I see no reason why $D_6 \cap R_2=D_6$ could hold. Is there something I misunderstand here because the post gets the right answer in the end so it's a bit weird to me.

Answer 1

I thought however that the right formula was $\mathbb{E}(X \,|\, R_2) = \mathbb{E}(X \,|\, D_6 \cap R_2)P(D_6 \,|\, R_2) + \mathbb{E}(X \,|\, D_4 \cap R_2)P(D_4 \,|\, R_2) $.

Lacking formal training, my personal attack on such probability problems relies heavily on having developed intuition. So, I can offer an intuitive explanation (only) why the formula excerpted at the start of this response is wrong.

Dissect, for example, the expression

$$\mathbb{E}(X \,|\, D_6 \cap R_2) \times P(D_6 \,|\, R_2) \tag1 $$

The right hand factor in (1) above, $~\displaystyle P(D_6 \,|\, R_2)~$ captures the fact that $~P(D_6)~$ is being computed under the assumption that $~R_2~$ has occurred. Now, the left hand factor in (1) above should be computing the expected factor of the 2nd throw, under the assumption that the die in question is 6 sided.

Instead, you are attempting to re-use the fact that the first throw was a 2 in computing the expected value of the second throw, when the die is 6 sided.

In effect, you are attempting to use fact that the first throw was a 2 twice, which is inappropriate.

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I emphasize, this is an informal intuitive response, designed to allow you to develop a feel for the right formula, rather than having you memorize the formula for a given problem.

The (intuitive) general formula is that if you have two possibilities, Possibility-1 and Possibility-2, then your expectation is

$$[\text{probability of Possibility-1} \times \text{expectation of Possibility-1}] \\+ [\text{probability of Possibility-2} \times \text{expectation of Possibility-2}]. \tag2 $$

So, examining (2) above, you have to ask yourself, is Possibility-1 represented by the die being 6 sided, or is it represented by the die being 6 sided and having a 2 appear on the first throw?

This is a tricky question. My perspective is that the fact that a 2 occurred on the first roll is already being used to compute the probability of Possibility-1. Therefore, the fact that the 2 occurred on the first roll should not be used a second time.

Answer 2

In this case, $X$ and $R_2$ are conditionally independent given $D_6$ -- both formulas give the same result. Knowing the first die rolled a 2 is evidence whether it was a d4 or d6, but once it is known that the d6 was rolled first, the fact that it rolled a 2 does not affect the subsequent roll of the d4.

If e.g. there was an additional rule that the second die counts as zero if it matches the first die, then $X$ and $R_2$ would not be conditionally independent and only your formula would be correct.