Expected Rolls to See an Odd Number
You roll a fair six-sided die repeatedly until you see an odd number. What is the expected number of rolls?
Hints
- What fraction of the die's faces are odd? Each roll is an independent trial with that success probability.
- This is a geometric distribution problem -- the expected number of independent trials until the first success is $1/p$.
- Set up the first-step equation: $E[X] = \frac{1}{2}(1) + \frac{1}{2}(1 + E[X])$ and solve for $E[X]$.
Worked Solution
How to Think About It: This is a classic "how many tries until your first success" problem. Each roll is independent, and exactly half the faces (1, 3, 5) are odd. So every roll is like a coin flip -- you either get an odd number (success) or you don't. The expected number of trials until the first success in a sequence of independent Bernoulli trials is $1/p$. With $p = 1/2$, you should expect about 2 rolls.
Quick Estimate: Half the faces are odd, so on average you succeed every other roll. Gut answer: 2.
Approach: Model the number of rolls as a geometric random variable and compute its expectation directly.
Formal Solution:
Let $X$ be the number of rolls until the first odd number appears. Each roll shows an odd number with probability $p = 3/6 = 1/2$, and rolls are independent. So $X \sim \text{Geom}(1/2)$, meaning:
$$P(X = k) = (1 - p)^{k-1} \, p = \left(\frac{1}{2}\right)^k, \quad k = 1, 2, 3, \ldots$$
The expected value of a geometric random variable is:
$$E[X] = \frac{1}{p} = \frac{1}{1/2} = 2$$
Alternatively, by first-step analysis: on the first roll, with probability $1/2$ you see an odd number (done in 1 roll), and with probability $1/2$ you see an even number and start over. So:
$$E[X] = \frac{1}{2}(1) + \frac{1}{2}(1 + E[X])$$
Solving: $E[X] = 1 + \frac{1}{2} E[X]$, which gives $\frac{1}{2} E[X] = 1$, so $E[X] = 2$.
Answer: The expected number of rolls is $E[X] = 2$.
Intuition
This problem is the simplest possible case of the geometric distribution: you repeat independent trials until you get a success, and the expected wait time is $1/p$. The key insight is recognizing that each roll is memoryless -- no matter how many even numbers you have seen, the next roll still has a 50% chance of being odd. There is no "building up" toward a success.
This pattern shows up constantly in quant interviews and in practice. Whenever you are waiting for the first occurrence of an event that happens independently with probability $p$ on each trial -- first trade execution, first default in a portfolio, first tick crossing a threshold -- the expected wait is $1/p$. The first-step analysis method (condition on what happens on the first trial) is also a fundamental technique worth internalizing, since it generalizes to problems where the geometric formula does not directly apply.