Two sixes in a row
A fair die is rolled repeatedly until two sixes appear on consecutive rolls. Find the expected number of rolls, the last six included.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Keep track of only one thing: whether the last roll was a six. That gives a Markov chain with two transient states.
Let be the expected number of remaining rolls when the last roll was not a six (or at the start), and the same when the last roll was a six. Write one equation for each by conditioning on the next roll.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The past matters only through the last roll. Condition on the next roll and solve the two linear equations that come out.
Step 1: states. Let be the expected number of rolls still needed from the start, or after any roll that is not a six. Let be the expected number still needed right after a six that did not end the game.
Step 2: equations. From state , we roll once. With probability it is a six and we move to state ; otherwise we stay in state : From state , we roll once. With probability it is a six and we stop; otherwise we are back to state :
Step 3: solve. The first equation gives . Substituting into the second, , so and .
The expected number of rolls is .
Check by intuition. A single six takes rolls on average. Each time we get a six, the next roll finishes the game with probability , so we need about sixes, each costing about rolls plus the roll that follows: . In general, for success probability , the answer is .