Gambler's Ruin with Biased Dice
Two players each start with 12 tokens. They repeatedly roll three dice. If the sum is 14, player $A$ gives a token to player $B$; if the sum is 11, player $B$ gives a token to player $A$. All other sums are ignored, and they keep rolling. The game ends when one player has all 24 tokens.
What is the probability that player $A$ wins?
Hints
- Ignore rolls that are neither 11 nor 14 -- the game reduces to a biased random walk. What is the probability $p$ that $A$ gains a token on each relevant roll?
- This is the Gambler's Ruin problem. For a biased walk with $P(\text{step up}) = p$, the probability of reaching $+a$ before $-a$ starting at 0 is $\frac{1}{1 + (q/p)^a}$.
- Count the dice outcomes by inclusion-exclusion: $N(k) = \sum_j (-1)^j \binom{3}{j}\binom{k-1-6j}{2}$ gives $N(11) = 45 - 18 = 27$ and $N(14) = 78 - 63 = 15$, so $p = 9/14$ and $q/p = 5/9$. Plug into the Gambler's Ruin formula with $a = 12$.
Worked Solution
How to Think About It: This is a classic Gambler's Ruin problem in disguise. Strip away the dice and tokens -- the core question is: in a biased random walk on $\{0, 1, \ldots, 24\}$ with absorbing barriers at 0 and 24, starting at position 12, what is the probability of reaching 24 before 0? The only thing you need from the dice is the probability $p$ that $A$ gains a token on each relevant roll.
Quick Estimate: Rolling three dice gives $6^3 = 216$ equally likely outcomes; sum $11$ occurs $27$ ways and sum $14$ occurs $15$ ways (both derived below). On each relevant roll, $A$ wins the token with probability $p = 27/(27+15) = 27/42 = 9/14 \approx 0.643$. Since $A$ has a substantial edge and needs to win a net 12 tokens from a symmetric start, $A$ should win with very high probability. Think of it as a biased coin that lands heads 64% of the time -- reaching +12 before $-12$ is nearly certain.
Approach: Apply the standard Gambler's Ruin formula for a biased random walk.
Formal Solution:
Let $x$ denote $A$'s token count minus 12, so $x \in \{-12, \ldots, 12\}$ with $x = 0$ at the start. $A$ wins when $x = 12$, loses when $x = -12$. At each step, $x$ increases by 1 with probability $p$ or decreases by 1 with probability $q = 1 - p$.
*Counting the dice outcomes.* The number of ways three dice total $k$ is the coefficient of $z^k$ in $(z + z^2 + \cdots + z^6)^3$. Equivalently, substituting $y_i = x_i - 1$ turns "solve $x_1 + x_2 + x_3 = k$ with $1 \le x_i \le 6$" into "solve $y_1 + y_2 + y_3 = k - 3$ with $0 \le y_i \le 5$", which inclusion-exclusion on the upper bounds counts as $$N(k) = \sum_{j \ge 0} (-1)^j \binom{3}{j}\binom{k - 3 - 6j + 2}{2}.$$ - $k = 11$: $\binom{10}{2} - 3\binom{4}{2} = 45 - 18 = 27$. - $k = 14$: $\binom{13}{2} - 3\binom{7}{2} = 78 - 63 = 15$.
Cross-check by symmetry: replacing every die $x_i$ by $7 - x_i$ is a bijection sending total $k$ to total $21 - k$, so the distribution is symmetric about $10.5$ and $N(14) = N(7) = \binom{6}{2} = 15$ (no correction term, since $7 - 3 = 4 < 6$). That matches, and the $27$ for sum $11$ is likewise $N(10)$, the joint peak of the distribution.
So on each relevant roll, $p = 27/42 = 9/14$ and $q = 5/14$.
The Gambler's Ruin probability starting at position $x$ (measuring $A$'s surplus tokens, so starting at 0) with absorbing barriers at $-a$ and $+a$ is:
$$R_x = \frac{1 - (q/p)^{x+a}}{1 - (q/p)^{2a}}$$
With $a = 12$, $q/p = 5/9$, and $x = 0$:
$$R_0 = \frac{1 - (5/9)^{12}}{1 - (5/9)^{24}}$$
Since $(5/9)^{12}$ is very small, we can simplify. Note that $1 - (5/9)^{24} = (1 - (5/9)^{12})(1 + (5/9)^{12})$, so:
$$R_0 = \frac{1}{1 + (5/9)^{12}}$$
Now compute $(5/9)^{12}$. We have $\ln(5/9) = \ln 5 - \ln 9 \approx 1.609 - 2.197 = -0.588$, so $12 \times (-0.588) = -7.06$, giving $(5/9)^{12} \approx e^{-7.06} \approx 0.00086$.
Answer:
$$P(A \text{ wins}) = \frac{1}{1 + (5/9)^{12}} \approx 0.99914$$
Player $A$ wins with probability approximately $99.91\%$.
Intuition
This problem is a clean application of Gambler's Ruin, one of the most fundamental results in probability. The key step is recognizing that all the dice mechanics boil down to a single number: the probability $p$ that $A$ wins each contested round. Once you have that, the token exchange is just a biased random walk with absorbing barriers, and the closed-form formula applies immediately.
The practical takeaway is how dramatically a small edge compounds. Here $p \approx 0.643$ -- not overwhelmingly biased -- but over a game that requires a net swing of 12 tokens, the probability that the advantaged player loses is less than 0.1%. This is the same reason that a casino with a 2% edge on each bet is virtually guaranteed to take your money if you play long enough. In trading, this is the intuition behind why even a modest positive-EV strategy, run consistently, dominates over time.