Two players, A and B, take turns rolling a fair six-sided die. Player A goes first. The game ends as soon as someone rolls a 6 -- that player wins. 1. What is the probability that Player A wins? 2. What is the expected total number of rolls before the game ends?

First-to-Roll-Six Dice Game

Probability · Medium · Free problem

Two players, A and B, take turns rolling a fair six-sided die. Player A goes first. The game ends as soon as someone rolls a 6 -- that player wins.

What is the probability that Player A wins?

What is the expected total number of rolls before the game ends?

Hints

Player A gets the first roll. If both players miss, the game resets to the same starting position -- this self-similarity is key.
Write a recursion: $P(A) = 1/6 + (5/6)^2 \cdot P(A)$. The factor $(5/6)^2$ is the probability both players miss in one round.
For the expected number of rolls, note that each individual roll has the same
/6$ chance of being a 6 regardless of whose turn it is. What does that tell you about $E[N]$?

Worked Solution

How to Think About It: This is a classic first-passage game with geometric waiting times. Player A has the advantage of going first, so your gut should say A wins more than half the time. How much more? Well, A gets first crack (probability

/6$), and if both miss (probability $(5/6)^2 \approx 0.694$), the game resets to the same position. So A's edge comes entirely from moving first in a symmetric game. You would guess something modestly above

/2$ -- maybe $0.55$.

Quick Estimate: Each round (one roll by A, one by B) where nobody wins has probability $(5/6)^2 = 25/36 \approx 0.694$. The probability *someone* wins in a given round is

1/36 \approx 0.306$. Within each round, A wins with probability

/6$ and B wins with probability $(5/6)(1/6) = 5/36$. So A's share of wins within a round is $(1/6) / (11/36) = 6/11 \approx 0.545$. That matches the gut feel -- A wins about $54.5\%$ of the time.

For the expected number of rolls: the game lasts a geometric number of rounds, each round containing at most 2 rolls. On average, roughly $36/11 \approx 3.27$ rounds are needed, and each round averages slightly less than 2 rolls (because sometimes A wins on roll 1 of the round). A cleaner approach uses the per-roll miss probability.

Approach: Sum the geometric series for A's winning probability directly, then compute the expected total rolls.

Formal Solution:

*Part 1: Probability A wins.*

A wins on roll 1 with probability

/6$. If A misses (probability $5/6$) and B misses (probability $5/6$), the game resets to the same state. So:

$P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 P(A)$

Solving:

$P(A) - \frac{25}{36} P(A) = \frac{1}{6}$

$P(A) \cdot \frac{11}{36} = \frac{1}{6}$

$P(A) = \frac{6}{11}$

Alternatively, write it as an infinite geometric series:

$P(A) = \sum_{k=0}^{\infty} \left(\frac{5}{6}\right)^{2k} \cdot \frac{1}{6} = \frac{1/6}{1 - 25/36} = \frac{6}{11}$

*Part 2: Expected total number of rolls.*

Let $N$ be the total number of rolls. Each individual roll is independent, and someone rolls a 6 with probability

/6$. But the rolls alternate between A and B, so we need to be a bit careful.

Let $E$ be the expected number of rolls. Condition on the first roll:

With probability
/6$, A rolls a 6 on roll 1. Total rolls: 1.
With probability $5/6$, A misses. Then B rolls. With probability
/6$, B gets a 6 on roll 2. Total rolls: 2. With probability $5/6$, B misses and we are back to the start having used 2 rolls.

$E = \frac{1}{6}(1) + \frac{5}{6}\left[\frac{1}{6}(2) + \frac{5}{6}(2 + E)\right]$

$E = \frac{1}{6} + \frac{5}{36}(2) + \frac{25}{36}(2 + E)$

$E = \frac{1}{6} + \frac{10}{36} + \frac{50}{36} + \frac{25}{36}E$

$E = \frac{6}{36} + \frac{10}{36} + \frac{50}{36} + \frac{25}{36}E$

$E = \frac{66}{36} + \frac{25}{36}E$

$E\left(1 - \frac{25}{36}\right) = \frac{66}{36}$

$E = \frac{66}{36} \cdot \frac{36}{11} = \frac{66}{11} = 6$

This makes intuitive sense: each roll independently has a

/6$ chance of being a 6, so the expected number of rolls until the first 6 is

/(1/6) = 6$, regardless of who is rolling.

Answer: Player A wins with probability $P(A) = 6/11 \approx 0.545$, and the expected total number of rolls is $E = 6$.

Intuition

The first-mover advantage in this game comes purely from acting before your opponent in a symmetric setup. Both players have identical

/6$ success probabilities on each turn, but A gets the first shot. The recursion $P(A) = 1/6 + (5/6)^2 P(A)$ captures the self-similar structure: if neither player wins in a round, the game resets perfectly. This "memoryless restart" pattern appears constantly in quant problems -- any time you have independent geometric trials with a repeating structure, look for a one-step recursion rather than summing an infinite series by hand.

The expected number of rolls being exactly 6 is a nice sanity check. It does not matter that two players alternate -- each roll is still an independent Bernoulli trial with success probability

/6$, so the waiting time for the first success is $\text{Geometric}(1/6)$ with mean 6. The alternation only affects *who* wins, not *when* someone wins. This separation between "who" and "when" is a useful trick: you can often compute the total waiting time ignoring player identity, then figure out the allocation of wins separately.

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