A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly…

Concept. Waiting for a run of two consecutive successes is a different problem from waiting for a single success, so it is solved with states rather than with one reciprocal.

In independent trials where a success has probability p, track how far the current run has advanced. State A means the run is empty, because nothing has been tried yet or the last trial failed; state B means the previous trial was a success. Let EA and EB be the expected numbers of further trials from these two states.

Every trial costs one attempt, after which the run advances, completes, or resets, so EA = 1 + p·EB + (1 − p)·EA and EB = 1 + p·0 + (1 − p)·EA. Solving that pair gives the general identity EB = 1/p2 and EA = 1/p + 1/p2 = (1 + p)/p2.

Application to this die.

1. A throw counts as a success when the face is even. The even faces are 2, 4 and 6, that is three of the six equally likely faces, so p = 3/6 = 1/2.

2. Name the two states: EA is the expected number of further throws when nothing has been thrown yet or the previous throw was odd, and EB is the expected number of further throws when the previous throw was even.

3. Equation from state A: EA = 1 + p·EB + (1 − p)·EA. Subtracting (1 − p)·EA from both sides gives p·EA = 1 + p·EB, hence EA = 1/p + EB.

4. Equation from state B: the next throw is even with probability p and the process stops with nothing further to count, and is odd with probability 1 − p, which sends the run back to state A. Hence EB = 1 + (1 − p)·EA.

5. Substitute the expression for EA from step 3 into step 4: EB = 1 + (1 − p)(1/p + EB) = 1 + (1 − p)/p + (1 − p)·EB.

6. Collect the EB terms: p·EB = 1 + (1 − p)/p = 1/p, so EB = 1/p2 and therefore EA = 1/p + 1/p2.

7. Put p = 1/2: EB = 1/(1/2)2 = 4 and EA = 1/(1/2) + 4 = 2 + 4 = 6. The die starts in state A, so the expected number of throws is 6.

Cross-check and contrast.

• The standard run-length identity for k consecutive successes, 1/p + 1/p2 + … + 1/pk, gives 2 + 4 = 6 at k = 2 and p = 1/2, reaching the same value by an independent route.

• Cutting the sequence into disjoint pairs of throws and taking 1/(1/4) = 4 answers a different question: it counts pairs as units instead of throws, and it misses runs that straddle a pair boundary, such as odd, even, even.

• A sanity bound agrees: the count must exceed 4, the wait measured from a throw that has already come up even, and it must exceed 2, the shortest length the experiment can have.

Result: the expected number of throws until two consecutive even faces appear is 6.