Race Between Two Patterns on a Fair Die

Probability · Hard · Free problem

You roll a fair six-sided die repeatedly. The process stops the first time you see either the pattern $6, 6$ (two consecutive sixes) or the pattern

, 2, 3$ (three consecutive rolls of 1 then 2 then 3), whichever comes first.

What is the probability that the process stops because

, 2, 3$ appeared before $6, 6$?

Build a minimal finite-state Markov chain where each state tracks the longest suffix of the roll history that is a prefix of either target pattern, and solve exactly for the absorption probability.

Hints

  1. Track progress toward both patterns simultaneously. The state you need is the longest suffix of the roll history that matches a prefix of either $6, 6$ or
    , 2, 3$.
  2. There are exactly 4 transient states: no progress ($\emptyset$), last roll was
    $, last two rolls were
    , 2$, and last roll was $6$. Write the transition probabilities carefully -- note that rolling a
    $ from any state always resets to the $\langle 1 \rangle$ state.
  3. Let $p_i$ be the probability that
    , 2, 3$ wins from state $i$. You get a $4 \times 4$ linear system. Start by using the $S_0$ and $S_6$ equations to express $p_1$ and $p_6$ in terms of $p_0$, then substitute into the remaining equations.

Worked Solution

How to Think About It: This is a pattern-race problem -- two target strings are competing to appear first in a stream of i.i.d. die rolls. The standard approach is to build an Aho-Corasick-style automaton: each state represents the longest suffix of the roll history that is a prefix of either target pattern. Transitions are deterministic given the current state and the new roll. Then you solve a system of linear equations for the absorption probabilities.

Before doing any algebra, think about which pattern has the advantage. Pattern $6, 6$ needs only 2 specific rolls with probability

/6$ each, while
, 2, 3$ needs 3 specific rolls. But the key subtlety is overlap: after a failed attempt at $6, 6$ (say you rolled 6 then something else), you lose all progress. After a failed
, 2, 3$ attempt (say
, 2, 5$), you also lose progress. The shorter pattern has a big structural edge here.

Quick Estimate: Pattern $6, 6$ has expected first-arrival time around $6 \times 6 + 6 = 42$ rolls (geometric waiting for each 6 in sequence). Pattern

, 2, 3$ has expected first-arrival around $6^3 + 6^2 + 6 = 258$ rolls (roughly -- each step has a
/6$ chance of advancing). So $6, 6$ should win most of the time, and $P(1,2,3 \text{ first})$ should be well below $0.5$ -- probably in the range $0.1$-$0.2$.

Approach: Define states by the longest suffix of the history that is a prefix of either target, solve a $4 \times 4$ linear system.

Formal Solution:

The two target patterns are $6, 6$ and

, 2, 3$. The relevant prefixes are:

  • $\emptyset$: no useful suffix
  • $\langle 1 \rangle$: last roll was
    $ (prefix of
    ,2,3$)
  • $\langle 1,2 \rangle$: last two rolls were
    , 2$ (prefix of
    ,2,3$)
  • $\langle 6 \rangle$: last roll was $6$ (prefix of $6,6$)

Note that no roll can simultaneously advance progress toward both patterns (since

\neq 6$), so there is no state tracking joint progress.

The absorbing states are $\langle 1,2,3 \rangle$ (pattern

,2,3$ wins) and $\langle 6,6 \rangle$ (pattern $6,6$ wins).

Transition table (each row sums to

$; all transitions have probability
/6$ per die face):

| From | Roll 1 | Roll 2 | Roll 3 | Roll 4 | Roll 5 | Roll 6 | |------|--------|--------|--------|--------|--------|--------| | $S_0$ | $S_1$ | $S_0$ | $S_0$ | $S_0$ | $S_0$ | $S_6$ | | $S_1$ | $S_1$ | $S_{12}$ | $S_0$ | $S_0$ | $S_0$ | $S_6$ | | $S_{12}$ | $S_1$ | $S_0$ | ABS | $S_0$ | $S_0$ | $S_6$ | | $S_6$ | $S_1$ | $S_0$ | $S_0$ | $S_0$ | $S_0$ | ABS |

Let $p_i = P(\text{pattern } 1,2,3 \text{ wins} \mid \text{start in state } i)$. The system of equations:

$p_0 = \frac{1}{6}p_1 + \frac{1}{6}p_6 + \frac{4}{6}p_0$

$p_1 = \frac{1}{6}p_1 + \frac{1}{6}p_{12} + \frac{1}{6}p_6 + \frac{3}{6}p_0$

$p_{12} = \frac{1}{6}(1) + \frac{1}{6}p_1 + \frac{1}{6}p_6 + \frac{3}{6}p_0$

$p_6 = \frac{1}{6}(0) + \frac{1}{6}p_1 + \frac{4}{6}p_0$

From the first equation:

p_0 = p_1 + p_6$.

From the fourth: $p_6 = \frac{1}{6}p_1 + \frac{2}{3}p_0$.

Substituting into p_0 = p_1 + p_6$:

$2p_0 = p_1 + \frac{1}{6}p_1 + \frac{2}{3}p_0 = \frac{7}{6}p_1 + \frac{2}{3}p_0$

$\frac{4}{3}p_0 = \frac{7}{6}p_1 \implies p_1 = \frac{8}{7}p_0$

Then $p_6 = \frac{1}{6} \cdot \frac{8}{7}p_0 + \frac{2}{3}p_0 = \frac{4}{21}p_0 + \frac{14}{21}p_0 = \frac{6}{7}p_0$.

From the third equation: $p_{12} = \frac{1}{6} + \frac{1}{6} \cdot \frac{8}{7}p_0 + \frac{1}{6} \cdot \frac{6}{7}p_0 + \frac{1}{2}p_0 = \frac{1}{6} + \frac{5}{6}p_0$.

Substitute everything into the second equation:

$\frac{8}{7}p_0 = \frac{1}{6} \cdot \frac{8}{7}p_0 + \frac{1}{6}\left(\frac{1}{6} + \frac{5}{6}p_0\right) + \frac{1}{6} \cdot \frac{6}{7}p_0 + \frac{1}{2}p_0$

$\frac{8}{7}p_0 = \frac{8}{42}p_0 + \frac{1}{36} + \frac{5}{36}p_0 + \frac{6}{42}p_0 + \frac{1}{2}p_0$

Collecting the $p_0$ terms on the right with common denominator 52$:

$\frac{5}{36} + \frac{1}{7} + \frac{1}{2} = \frac{35 + 36 + 126}{252} = \frac{197}{252}$

And $\frac{8}{7} = \frac{240}{252} + \frac{8}{42} = $ ... Let me collect all $p_0$ terms:

$\left(\frac{240}{252} - \frac{197}{252}\right)p_0 = \frac{1}{36}$

$\frac{43}{252}p_0 = \frac{1}{36} \implies p_0 = \frac{252}{36 \times 43} = \frac{7}{43}$

The complete solution: $p_0 = \frac{7}{43}$, $p_1 = \frac{8}{43}$, $p_{12} = \frac{13}{43}$, $p_6 = \frac{6}{43}$.

Answer: The probability that the process stops because of

, 2, 3$ is

$P(1,2,3 \text{ first}) = \frac{7}{43} \approx 0.1628$

Pattern $6, 6$ wins with probability $\frac{36}{43} \approx 0.8372$. The shorter pattern has a decisive advantage.

Intuition

This is a classic pattern-race (or string-competing) problem, and the key insight is that shorter patterns have a structural advantage. Pattern $6, 6$ needs only two consecutive hits, while

, 2, 3$ needs three -- and since each step has the same
/6$ probability of advancing, the two-step pattern reaches its absorbing state much faster on average. The result $7/43 \approx 16\%$ reflects this: the three-step pattern is roughly 5 times less likely to win.

In quant interviews, pattern-race problems test whether you can build the right state space. The Aho-Corasick / suffix-prefix construction is the general technique: after each new symbol, ask "what is the longest suffix of what I have seen so far that is a prefix of any target pattern?" This gives you the minimal Markov chain. The same idea appears in sequential detection problems in trading -- for example, detecting a specific sequence of order flow events (uptick-uptick-downtick) versus another pattern, and computing which is more likely to trigger first.

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