Conditional expectation of a dice game. Conflicting answers. Problem Statement:
You will roll a fair die until the game stops. The game stops when you get a 4, 5, or 6. For every number 1, 2, or 3 you have thrown your score increases by + 1. If the game stops with a 4 or 5, you get paid the accumulated score. If the game stops with a 6 you get nothing. What is the expected payoff of this game?
Approach 1 (Conditional Expectation):
Let $Y$ be the random variable denoting our payoff. Let $E_6$ denote the event that our game ends with throwing a 6 and let $E_{4, 5}$ denote that event that our game ends with throwing a 4 or 5. Then, by the law of total expectation, we have
$$
\mathbb{E}(Y) = \mathbb{E}(Y \mid E_6) \cdot \mathbb{P}(E_6) + \mathbb{E}(Y \mid E_{4, 5}) \cdot \mathbb{P}(E_{4, 5}) = 0 + \mathbb{E}(Y \mid E_{4, 5}) \cdot \frac{2}{3}
$$
So, it remains to find $\mathbb{E}(Y \mid E_{4,5})$, which we can calculate using recursion:
\begin{align*}
\mathbb{E}(Y \mid E_{4, 5}) &= \mathbb{P}(\text{throwing 1, 2, or 3} \mid E_{4, 5}) \cdot (1 + \mathbb{E}(Y \mid E_{4, 5})) + \mathbb{P}(\text{throwing 4 or 5} \mid E_{4, 5}) \cdot 0 \\
&= \frac{3}{5} \cdot (1 + \mathbb{E}(Y \mid E_{4, 5})) + \frac{2}{5} \cdot 0 \\
\implies \mathbb{E}(Y \mid E_{4, 5}) = \frac{3}{2}
\end{align*}
Therefore, we conclude $\mathbb{E}(Y) = \frac{2}{3} \cdot \mathbb{E}(Y \mid E_{4, 5}) = 1$
Approach 2 (Definition of Expectation):
Let $y$ denote the payoff we get. Then
$$
\mathbb{E}(Y) = \sum_{y = 0}^\infty y \cdot \underbrace{(\frac{3}{6})^y}_{\text{probability of throwing 1, 2, or 3}} \cdot \underbrace{\frac{2}{6}}_{\text{probability of throwing 4 or 5}} = \frac{2}{3}
$$
In summary, Approach 1 gives $\mathbb{E}(Y) = 1$ while Approach 2 gives $\mathbb{E}(Y) = \frac{2}{3}$. After searching online for this question, it seems the answer from Approach 2 is correct. However, I just can't see what's wrong with Approach 1. Any ideas? What went wrong in Approach 1?
 A: Explaining the mistake
Your mistake lies in the computation of $\mathbb{E}[Y \mid E_{4,5}]$, and more specifically in computation of $P[\text{throw }1,2,3 \text{ on turn 1} \mid E_{4,5}]$. Let $T$ be the event "we throw 1,2, or 3 on turn 1" and write
$$\begin{align}
P[T \mid E_{4,5}]
&= \frac{P[T \cap E_{4,5}]}{P[E_{4,5}]} \\
&= \frac{\left(\frac 1 2\right) \left(\frac 2 3 \right)}{\frac 2 3} \\
&= \frac 1 2.
\end{align}$$
That's different from the $\frac 3 5$ result you used in your calculation, and if you use $\frac 1 2$ instead of $\frac 3 5$ in your recurrence then you'll get the right answer.
Side note: You also have a wrong value $P[\text{throw }4,5 \text{ on turn 1} \mid E_{4,5}] = \frac{2}{5}$ when the correct value is $\frac 1 2$ again. You could fix that using the same formula method above, but also it doesn't directly impact your answer because it gets multiplied by 0 in the recurrence.
What went wrong in terms of intuition?
I think you got $\frac 3 5$ by intuitively reasoning like: "Once we condition on $E_{4,5}$, there are only 5 possible outcomes and 3 of them are 1,2,3".
My intuitive explanation of why your approach doesn't work is: Imagine playing the game 1000 times and tracking how many times $k$ shows up as the first roll. We want to condition on $E_{4,5}$, so in our scorekeeping, we discard any games that eventually end with $E_6$. If we roll 1, 2, or 3 on the first round, this may or may not get counted, since there's still a $\frac 1 3$ chance that this game will be omitted from tracking later (because we end with a 6). If we roll 4 or 5 on the first round, that will always get counted in our tracker (since this game definitely does not end with a 6). That explains why e.g. $$P[\text{roll 1 on first round} \mid E_{4,5}] < P[\text{roll 4 on first round} \mid E_{4,5}].$$
A: You need to define with precision what the payoff $Y$ means to see if your calculations are right or not. Let $X:=\{X_n\}_{n\in\mathbb{N}}$  an i.i.d. sequence of r.v. that represents the throw of dice and let $T:=\min\{k\in \mathbb{N}: X_k\in\{4,5,6\}\}$ the time when the game stops. Then $Y=(T-1) \mathbf{1}_{\{X_T\neq 6\}}$, therefore
$$
\begin{align*}
\operatorname{E}[Y]&=\operatorname{E}[(T-1) \mathbf{1}_{\{X_T\neq 6\}}]\\
&=\sum_{k\geqslant 1}\operatorname{E}[(T-1)\mathbf{1}_{\{X_T\neq 6\}}|T=k]\Pr [T=k]\\
&=\sum_{k\geqslant 1}\operatorname{E}[(k-1)\mathbf{1}_{\{X_k\neq 6\}}|T=k]\Pr [T=k]\\
&=\sum_{k\geqslant 1}(k-1)\Pr [X_k\neq 6,T=k]\\
&=\sum_{k\geqslant 1}(k-1)\left(\frac1{2}\right)^{k-1}\frac1{3}\\
&=\frac1{3}\left[\sum_{k\geqslant 0}k x^k\right]_{x=1/2}=\frac1{3}\left[x\frac{d}{d x}\sum_{k\geqslant 0}x^k\right]_{x=1/2}\\
&=\frac1{3}\left[x\frac{d}{d x}\frac1{1-x}\right]_{x=1/2}=\frac1{3}\left[\frac{x}{(1-x)^2}\right]_{x=1/2}\\
&=\frac{2}{3}
\end{align*}
$$
∎
I don't understand why you'd written the recursion of your first approach, i.e., where it comes from and why you think that it could be correct.
