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: sum 11 has 27 ways, sum 14 has 15 ways, 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, the number of ways to get sum 11 is 27 and sum 14 is 15 (out of 216 total outcomes, though we only care about the ratio). 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 $q = 1 - p$.
First, count the dice outcomes. For three dice, the number of ways to roll sum $k$ can be computed via the coefficient of $z^k$ in $(z + z^2 + \cdots + z^6)^3$: - Sum = 11: 27 ways - Sum = 14: 15 ways
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.