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Expected tosses until a six

Short answer

The number of tosses until a six follows a geometric distribution with p=1/6, so expected value = 1/p = 6 tosses.

A fair dice was rolled until one gets a Six; find the expected number of toss required?

Let X be the number of tosses needed to get the first six. Each toss is an independent trial with success probability p = 1/6, so X follows a geometric distribution:

P(X = x) = q^(x−1) p = (5/6)^(x−1) (1/6), x = 1, 2, 3, …

Expected value

E(X) = Σ x · q^(x−1) p = p (1 + 2q + 3q² + …) = p · 1/(1 − q)² = p / p² = 1/p

E(X) = 1 / (1/6) = 6

Variance (for reference) = q/p² = (5/6)/(1/36) = 30.

Answer: on average 6 tosses are required to get a six.

✓ Answer checked by Claude Sonnet 5.5

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