Take two dice and choose between these patterns: double, double, non-double, or non-double, double, double. You will keep rolling until one of them appears in three consecutive rolls. Which would you rather back?

Each contains two doubles and one non-double. In any three rolls selected in advance, each pattern has a probability of 1/6 × 1/6 × 5/6, or about 2.3%. Changing the order of those factors does not change the answer.

But in the race to appear first, the second pattern wins with probability 35/36, about 97.2%. The first wins only 1/36 of the time.

This is a version of Penney’s game, the established pattern-matching game credited to Walter Penney. It is usually explained with coin tosses. Here we use actual two-dice rolls, where a double has probability 1/6 rather than 1/2, and choose a pair of patterns whose result we can explain without a long calculation.

Each letter represents a whole two-dice roll

We will write D for a double, such as 4–4, and N for a non-double, such as 2–5. We do not care which particular double appears. Among the 36 equally likely ordered dice pairs, six are D and 30 are N.

The two patterns are therefore DDN and NDD. After every roll from the third onwards, look at the most recent three letters. Stop as soon as one of the patterns appears. The three-roll windows overlap; you do not restart in separate blocks of three.

Start with a fresh stream of ordinary independent two-dice rolls. This is a dice experiment, not a normal backgammon opening: in backgammon, each player rolls one die and opening ties are rolled again. That opening procedure would give different starting conditions.

DDN needs its chance right at the beginning

Suppose the first two rolls are doubles. The sequence starts DD. The next non-double will complete DDN, however many extra doubles come first. NDD cannot appear before that non-double, because there has not yet been an N.

So DDN wins whenever the experiment starts DD. The chance of those first two doubles is 1/6 × 1/6 = 1/36.

Now suppose the first two rolls are anything else: NN, ND or DN. At least one N has appeared before the first pair of consecutive doubles. When that first pair finally arrives, the roll immediately before it must be N; otherwise an earlier pair of consecutive doubles would already have existed. NDD has just completed.

DDN cannot have beaten it. To finish DDN later, the sequence would first need DD, and that first DD would already have completed NDD. DDN can therefore win only if the first two rolls of the experiment are already doubles.

That covers every possible start. DDN gets 1/36; NDD gets the other 35/36. An endless sequence that never completes either has probability zero under the stated fair, independent-roll assumptions.

Why equal three-roll chances do not mean an equal race

Look at the four-letter sequence NDDN. NDD appears at roll three, then DDN appears at roll four. Counting occurrences in a long record would count both. A race stops at the first one, so only NDD gets the win.

This is where the order matters. The patterns can arrive close together, sharing some of the same rolls. They do not behave like two independent lottery tickets drawn anew after each turn.

Neither calculation was wrong. A fixed window asks which pattern occupies those particular three places. A first-arrival race asks what happens throughout the sequence leading up to a pattern, including everything that could have ended the race sooner.

It is related to the confusion about consecutive doubles: changing when you start observing, or what you count as success, changes the probability problem.

The two dice really do change the number

With a fair coin, if D and N were simply the two equally likely faces, DDN would win with probability 1/4 and NDD with 3/4. That is already an unequal race, despite the equal fixed-window probabilities.

For backgammon dice, D is much less common. The special opening DD occurs only 1/36 of the time, rather than 1/4, making NDD’s advantage much larger. We cannot take a coin-toss percentage and apply it unchanged to doubles.

The dice themselves remain fair throughout. After any history, the next roll still has probability 1/6 of being a double. What changes is which pattern can finish first, given the rolls already seen.

Try the race with two dice

Roll one pair at a time, or sample 100 fresh races. The exact chances above are calculated from the rules. The sample below is newly generated and will vary; a short run need not match 35 wins to one exactly.

The DDN versus NDD race

D = double; N = non-double. Each letter is one two-dice roll.

DDN1/362.8%

NDD35/3697.2%

Exact chances of winning a fresh race, not simulation results.

An example without simulation

2–5, 4–4, 6–6 give NDD: the race ends on roll three. Another non-double would then form DDN, but would arrive too late.

The unequal race comes from overlapping patterns and stopping at the first occurrence, with fair dice throughout. In our article about CUBE Backgammon’s dice, we explain how the app generates its rolls and how to check a completed match’s record.

The pattern-race idea is Walter Penney’s. We consulted Ben Lynn’s explanation of Penney’s game and the paper A Markov Chain Analysis of a Pattern Matching Coin Game. Our contribution here is the two-dice presentation, the selected DDN/NDD comparison, its explanation and the playable example; we do not claim the underlying mathematics is new.

How we checked the two-dice result

We counted all 36 ordered outcomes per roll: six doubles and 30 non-doubles. Exhaustively checking the 36³ fixed windows gives 1,080 for each pattern, or 5/216. A separate exact absorbing-state calculation gives NDD a first-arrival chance of 35/36, agreeing with the opening-DD argument. The interactive sample uses actual pairs of dice generated from random bytes with rejection sampling. A cap of 10,000 rolls per sampled race prevents unbounded work; any unfinished races are reported separately, never assigned a winner. The opening-rule distinction follows the USBGF beginner’s guide.