Why is the half-life 3.80 rolls and not 3 for p = 1/6?

After k rolls a fraction (5/6)ᵏ of the dice is expected to survive. Setting (5/6)ᵏ = 1/2 gives k = ln 2 / ln(6/5) = 0.693/0.182 ≈ 3.80. Three rolls leave about 58 %, four rolls about 48 %, so the half-way point lies between them.

My measured half-life came out quite different from 3.80. Is something wrong?

No — that is the lesson. Each roll is random, so a run with 20 dice can reach N₀/2 after 2 or after 6 rolls. The scatter shrinks roughly as 1/√N₀: switch to 1000 dice or turn on "Overlay 5 runs" and the curves bunch tightly around the theory line.

Why does the log scale give a straight line?

Taking the logarithm of N(k) = N₀(1 − p)ᵏ gives ln N = ln N₀ − λk, a straight line in k with slope −λ. On a log axis every half-life marker is the same vertical distance apart, which makes it easy to read T½ off the graph.

How does this relate to a real radioactive sample?

Each nucleus has a fixed probability per unit time of decaying, independent of its age, exactly like a die that does not remember earlier rolls. A real sample contains ~10²⁰ nuclei, so fluctuations are invisible and time is effectively continuous; the dice simply make the randomness visible at small N.

Half-Life with Dice: Random Decay

Roll N dice, remove the ones that "decay"; N vs roll with theory N₀(1−p)ᵏ, T½ = ln 2/λ, log scale.

À propos du modèle

This is the classroom dice experiment for radioactive decay, done with up to 1000 dice at once. Every die stands for a nucleus that has not decayed yet. On each roll every surviving die is thrown again and is removed if it shows the chosen decay face: a 6 (p = 1/6), a 5 or 6 (p = 1/3), or heads on a coin (p = 1/2). Nothing about a die's past changes its chance on the next roll — that memorylessness is the whole point, and it is the same assumption that makes real decay exponential. The expected number left after k rolls is N(k) = N₀(1 − p)ᵏ = N₀e^{−λk} with λ = −ln(1 − p) per roll, so the half-life in rolls is T½ = ln 2/λ: 3.80 rolls for one face, 1.71 for two faces and exactly 1 roll for coins. The simulator plots your actual count against this smooth curve, marks N₀/2, N₀/4, N₀/8 with the corresponding T½, 2T½, 3T½ positions, and reports the measured half-life as the first roll at which N ≤ N₀/2 (linearly interpolated). A log-scale toggle turns the exponential into a straight line of slope −λ, and "Overlay 5 runs" shows four extra independent runs so students can see run-to-run scatter shrink as N₀ grows (relative noise ≈ 1/√N). Time is discrete here — one roll is one time step — and the model omits daughter products, decay chains and detector effects. For the continuous version with N(t) = N₀e^{−λt}, activity and parent → daughter chains, see the companion simulator Radioactive Decay in this category (chemistry/radioactive-decay).

Public : Grades 7–11 chemistry and physics students meeting half-life for the first time, and teachers who want a quick virtual version of the dice or coin decay lab.

Notions clés

Half-life (T½)
Decay constant (λ)
Exponential decay
Decay probability p
Random (stochastic) process
Memorylessness
Log scale
Statistical fluctuation

Comment ça marche

Each die is a nucleus that has not decayed yet. On every roll it decays with the same probability p, whatever happened before — exactly the assumption behind radioactive decay. The expected number left after k rolls is N₀(1−p)ᵏ, an exponential; the half-life is the number of rolls after which half the dice are expected to be gone, T½ = ln 2 / (−ln(1−p)). Real dice fluctuate around that curve; with more dice the relative scatter shrinks.

Formules

N(k) = N₀ (1 − p)ᵏ = N₀ e^(−λk)

λ = −ln(1 − p) (per roll)

T½ = ln 2 / λ → 3.80 rolls for p = 1/6, 1.71 for 1/3, 1 for 1/2

Questions fréquentes

Why is the half-life 3.80 rolls and not 3 for p = 1/6?: After k rolls a fraction (5/6)ᵏ of the dice is expected to survive. Setting (5/6)ᵏ = 1/2 gives k = ln 2 / ln(6/5) = 0.693/0.182 ≈ 3.80. Three rolls leave about 58 %, four rolls about 48 %, so the half-way point lies between them.
My measured half-life came out quite different from 3.80. Is something wrong?: No — that is the lesson. Each roll is random, so a run with 20 dice can reach N₀/2 after 2 or after 6 rolls. The scatter shrinks roughly as 1/√N₀: switch to 1000 dice or turn on "Overlay 5 runs" and the curves bunch tightly around the theory line.
Why does the log scale give a straight line?: Taking the logarithm of N(k) = N₀(1 − p)ᵏ gives ln N = ln N₀ − λk, a straight line in k with slope −λ. On a log axis every half-life marker is the same vertical distance apart, which makes it easy to read T½ off the graph.
How does this relate to a real radioactive sample?: Each nucleus has a fixed probability per unit time of decaying, independent of its age, exactly like a die that does not remember earlier rolls. A real sample contains ~10²⁰ nuclei, so fluctuations are invisible and time is effectively continuous; the dice simply make the randomness visible at small N.

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Half-Life with Dice: Random Decay

À propos du modèle

Comment ça marche

Formules

Questions fréquentes

Simulations voisines

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Principe de Le Chatelier

Pile galvanique

Localisation d’Anderson dans une chaîne unidimensionnelle

Fonction de Wigner

Molécules en 3D

Half-Life with Dice: Random Decay

À propos du modèle

Comment ça marche

Formules

Questions fréquentes