Expected Rolls for Two Consecutive Sixes
You roll a fair six-sided die repeatedly. On average, how many rolls does it take until you see two sixes in a row?
Note: you need two consecutive sixes -- not just two sixes at any point during the sequence.
Hints
- Think about what happens after you roll a six for the first time -- you are one step away from winning, but a failed follow-up sends you all the way back to the start.
- Set up a recurrence. Let $E$ be the expected total rolls from the beginning. The expected number of rolls to reach the first six is 6. Once you have that six, work out what happens next in terms of $E$ itself.
- After the first six (which takes 6 rolls on average), write: $E = 6 + \frac{1}{6}(1) + \frac{5}{6}(E + 1)$. Solve for $E$.
Worked Solution
How to Think About It: The key word is *consecutive*. This is not "how long until I roll two sixes" -- it is "how long until I roll a six followed immediately by another six." That distinction matters: after a non-six, you reset your streak to zero; after a six, you are one step from winning. The natural tool is a Markov chain with two states: no six just rolled (call it S0), and exactly one six just rolled (S1). You want the expected time to absorption.
Quick Estimate: How long until the first six? About 6 rolls. Once you have one, there is a 1/6 chance the next roll finishes it and a 5/6 chance you bust and essentially start over. That restart costs another ~6 rolls to get back to a six, and then you try again. Rough intuition: you will need to successfully "reload" the six roughly 6 times before you get the 1/6 follow-up, so the total is in the ballpark of $6 \times 6 = 36$ plus a bit more. The answer should be somewhere in the 36-42 range.
Approach: Set up a recurrence. Let $E$ be the expected number of rolls starting from scratch (state S0). Once we are in S0, we roll until we hit a six -- that takes an expected 6 rolls. At that point we are in state S1 with one six just rolled. From S1 there is a 1/6 chance the next roll is a six (we are done, one more roll) and a 5/6 chance the next roll is not a six (we return to S0 after one roll and must start the whole thing over).
Formal Solution:
This gives us the recurrence:
$E = 6 + \frac{1}{6} \cdot 1 + \frac{5}{6}(E + 1)$
Breaking down the right side: $6$ to get the first six (from S0), then one more roll that is either a success (probability
Simplify:
$E = 6 + \frac{1}{6} + \frac{5}{6}E + \frac{5}{6}$
$E - \frac{5}{6}E = 6 + \frac{1}{6} + \frac{5}{6} = 7$
$\frac{1}{6}E = 7$
$E = 42$
You can verify this with the full two-state Markov chain. Let $E_0$ be the expected rolls from state S0 (no six just rolled) and $E_1$ be the expected rolls from state S1 (last roll was a six):
$E_0 = 1 + \frac{1}{6}E_1 + \frac{5}{6}E_0$ $E_1 = 1 + \frac{1}{6} \cdot 0 + \frac{5}{6}E_0$
From the second equation: $E_1 = 1 + \frac{5}{6}E_0$. Substituting into the first:
$\frac{1}{6}E_0 = 1 + \frac{1}{6}\left(1 + \frac{5}{6}E_0\right) = \frac{7}{6} + \frac{5}{36}E_0$
$\left(\frac{1}{6} - \frac{5}{36}\right)E_0 = \frac{7}{6} \implies \frac{1}{36}E_0 = \frac{7}{6} \implies E_0 = 42$
Both approaches agree.
Answer: The expected number of rolls is $\boxed{42}$.
Intuition
This problem illustrates a general pattern: when you need a streak or a dependent sequence of rare events, the expected wait time compounds multiplicatively. Getting a single six takes 6 rolls on average. Getting a six *followed by* a six does not take $6 + 6 = 12$ rolls -- it takes 42, because each failed attempt at the second six resets you not just one step back but all the way back to zero. The 5/6 failure probability at the final hurdle is what inflates the answer so dramatically.
In quant work, this type of recurrence shows up in streak-based trading signals, run-length statistics, and reliability engineering (how long until a system fails twice in a row). The Markov chain framing -- identifying the minimal set of states that capture "memory" -- is the right mental model. Here, you only need to remember whether the last roll was a six, not the full history. That state compression is the key insight: once you identify the right states, the algebra is straightforward.