Expected rolls until a 6

Relevant wiki: Conditional Probability - Problem Solving

Here is one solution. It's probably not the most elegant, but it has the advantage of being completely clear.

Let $p = 1/6$ and $q=1/3.$ Then the probability of getting a $6$ in exactly $k$ even rolls is $q^{k-1}p.$ So the probability of having all even rolls until we get a 6 is the sum $\sum_{k=1}^\infty q^{k-1}p = \frac{p}{1-q} = 1/4.$ Then the expected value of our number of rolls, given that we rolled all evens, is just $\sum_{k=1}^\infty kq^{k-1}p$ divided by the probability of rolling all evens, which is $1/4.$ So we get $4p\sum_{k=1}^\infty kq^{k-1} = \frac{4p}{(1-q)^2} = 3/2.$ (That sum is well-known--one way to get it is to notice that $kq^{k-1}$ is the derivative of $q^k,$ and use the geometric series.)

Why is the answer not 3? Well, here is a vague but convincing explanation: 3 is the answer you'd get if we were rolling a 3-sided die. But we're not, and so the fact that we got all even numbers biases our sample toward being shorter, because we're given that all of our rolls were even, and that is harder and harder to do the more we roll. Put another way, suppose the die was 100-sided, and I told you that I rolled until I got a 6 and that all my rolls were either 2, 4, or 6. How many rolls would you expect me to have made? Right, pretty close to 1, because I'm much, much more likely to have rolled a 6 right away than to have rolled $x$ 6 where $x$ is 2 or 4.

When you roll the die repeatedly, you stop when you have an odd number (and throw the rolls away) or a 6 (keep the rolls). As the probability of ending at each roll is $\dfrac{3+1}6=\dfrac23$ , the average number of rolls is $\dfrac1{\frac{2}{3}}=\dfrac32$ .

Now, as the number of rolls is independent from the fact the the last roll is a 6 or an odd number, $\dfrac32$ is also the average number of rolls ending with a 6.