Some probabilities are hard to count, like "how likely is it to collect all 3 prizes in 5 cereal boxes?" A simulation uses a simple random tool to act out the situation many times. It is an experiment that stands in for the real thing, built on compound events.
Idea clave
To simulate: (1) choose a random tool whose outcomes have the same probabilities as the real situation, (2) decide what counts as one trial and what counts as success, (3) run many trials, (4) estimate the probability as successes ÷ trials.
Por qué funciona
A good simulation behaves like the real situation, because the chances match. And with many trials, an experimental probability gets close to the true one. So the fraction of successful trials estimates the real probability, even when no one can count it directly.
| Probability | A tool that matches |
|---|---|
| $\frac{1}{2}$ | flip a coin: heads |
| $\frac{1}{6}$ | roll a number cube: 6 |
| $\frac{1}{3}$ | roll a number cube: 1 or 2 |
| $30\%$ | pick a random digit 0 to 9: 0, 1 or 2 |
| $\frac{1}{4}$ | spin a spinner with 4 equal parts |
Worked examples
Ejemplo resuelto: Designing a simulation
A player makes 70% of her free throws. How could you simulate one shot?
- Use random digits 0 to 9; each is equally likely.
- Let 0 through 6 (7 of the 10 digits) mean "makes it" and 7, 8, 9 mean "misses."
Respuesta: Random digits: 0–6 make, 7–9 miss
Ejemplo resuelto: Reading the results
In 50 trials of a simulation of two free throws, the player made both shots in 24 trials. Estimate the probability of making both.
- $\frac{24}{50} = 0.48$.
- (The exact answer is $0.7 \times 0.7 = 0.49$, so the simulation came close.)
Respuesta: $0.48$
Ejemplo resuelto: Collecting prizes
A cereal has 3 equally likely prizes. Each trial rolls a number cube (1–2, 3–4 and 5–6 stand for the prizes) until all 3 have appeared. In 20 trials, 12 needed 5 or fewer rolls. Estimate the probability of collecting all 3 prizes in 5 boxes or fewer.
- Each roll stands for one box.
- $\frac{12}{20} = 0.6$.
Respuesta: $0.6$