Dice Stopping Game (Law of Total Expectation) Question: "You start with £0, and continually roll a d6. If you roll a 1, 2 or 3 then you add £1 to your pot. If you roll a 4 or 5 then you stop, and win your pot. If you roll a 6 then you stop, and win nothing. What are your expected winnings?"
Correct Answer: 2/3
With probability $1/2$, a 4, 5, or 6 will be rolled, thus ending the game. Thus, you expect the game to end after two rolls (as the game ending is a geometric random variable with $p = 1/2$). On the first roll, you therefore expect to make 1 pound. When the game ends, there's a $2/3$ chance you roll a 4 or 5 and a $1/3$ chance you roll a six, thus loosing your money. Therefore the total winnings:
$$(2/3 * 1) + (1/3 * 0) = 2/3$$
My Incorrect Approach:
I want to set up a recursive formula for expressing $E[X]$, the total amount won playing the game. At any step, if either a 1, 2, or 3 is rolled, we go to the next round with 1 more pound. If we roll a 4 or 5 we terminate and get 0 additional value from the round, and if we roll a 6, we loose everything we have gained so far. I thought this could be written as:
$$E[X] = 1/2 * (1 + E[X]) + 1/3 * 0 + 1/6 * (-E[X])$$
However, when I solve this, I get $E[X] = 3/4$, which is incorrect. Can someone tell me where I'm going wrong in the setup? Thanks in advance
 A: The boring way: The probability of winning $k \ge 1$ pounds is $2^{-k} \cdot \frac{1}{3}$ (win one pound in $k$ round, then roll a 4 or 5 in round $k+1$), so the expected winnings is $$\frac{1}{3} \sum_{k \ge 1}k2^{-k} = \frac{2}{3}.$$ (To compute the sum, note that $S:= \sum_{k \ge 1} k 2^{-k}$ satisfies $S-\frac{1}{2} S = \sum_{k \ge 1} 2^{-k} = 1$, so $S=2$.)

The setup of your recursion is incorrect in a subtle way.
Specifically, it seems like you are trying to condition on the result of the first roll, which gives something like
$$E[X] = P(\text{roll 1,2,3}) E[X \mid \text{roll 1,2,3}]
+ P(\text{roll 4,5}) E[X \mid \text{roll 4,5}]
+ P(\text{roll 6}) E[X \mid \text{roll 6}].$$
If you roll 1,2,3, you may think that the game essentially starts over but you have 1 extra pound for your final winnings, so $E[X \mid \text{roll 1,2,3}] \overset{?}{=} 1+ E[X]$. However in your setup where rolling a six means you lose everything you won before, you have one extra pound to lose if you roll a six, so it is not really the same as starting over, and thus that conditional expectation is not $1 + E[X]$. See user2661923's answer for the correct recursion setup.
