A fair six-sided die is rolled repeatedly until three sixes are rolled consecutively. What is the expected number of rolls?

Question

I am asking this question because I don't know what is wrong with my method and for someone to explain where I went wrong. However, other methods are welcome as long as they don't use calculus(I'm in precalculus.)

I got this idea from André Nicholas's answer and Brilliant. I tried to use his method, however I didn't get the answer. Brilliant's solution just confused me.

My solution:

Let a be the expected number of rolls if the previous roll was not a 6. Let b be the expected number of rolls if the previous roll was a 6. Let c be the expected number of rolls if the previous 2 rolls were 6. Therefore:

The reasoning for this is that each roll has a probability of either not getting a six or getting one. c adds 2 because there were 2 rolls before it. If anyone needs clarification on what I mean please comment.

The correct answer is 258, but I got 54. I'm confused why my solution doesn't work.

Sidenote- From this post, I found out that for numbers on a die less than 6, the answer is 43(258/6). I'm unsure why it's 43 instead of 258. This isn't part of my question, however, if somebody could please explain it to me in the comments that would be great.

Answer 1

• Note that $b$ should be "expected rolls needed if the previous two rolls were "not 6, then 6." Otherwise there is overlap between "previous roll is 6" and "previous two rolls were both 6"
• Expanding on lulu's correction: the "$+1$" in each equation is counting the current roll, not the previous ones. For instance, the logic for the first equation is really "if the current roll is not a six, then the expected remaining rolls needed is $a$; otherwise the expected remaining rolls needed is $b$, so it is $a = \frac{5}{6}(1 + a) + \frac{1}{6}(1+b) = 1 + \frac{5}{6} a + \frac{1}{6} b$. For the last equation, if the current roll is not a six, then you are back to "$a$," otherwise you are done and you just need to count the current roll, so it is $\frac{5}{6}(1+a) + \frac{1}{6} \cdot 1 = 1 + \frac{5}{6} a$.