When the count is too hard, let a thousand random trials do the talking — and watch the frequency close in on the truth.
Some probabilities you can simply count: P(sum = 7) on two dice is 6/36, no contest. But many real questions resist a clean count — the geometry is awkward, the trial has a dozen stages, or "favorable" is genuinely hard to pin down. When the formula stalls, you can still act it out. Build a machine that imitates one trial using random numbers, press the button thousands of times, and record how often the thing you care about happens. That long-run share is exactly the frequency from Lesson 33.1 — and it steadies near the true P. This lesson turns that idea into a tool sharp enough to estimate even π.
Every simulation starts from one humble ingredient: a random number drawn uniformly from the interval [0, 1). "Uniform" means no stretch of the interval is favored — a value near 0.03 is exactly as likely as one near 0.91. A calculator's RAND key, a spinner, or a line of code all produce the same thing: an unpredictable decimal between 0 and 1.
That single uniform number is enough to imitate almost any simple trial. To flip a fair coin, draw u and call it heads when u < 0.5. To imitate a biased coin that lands heads with probability 0.7, just move the cutoff: heads when u < 0.7. To roll a fair die, slice [0, 1) into six equal strips and read off which strip u lands in.
A uniform random number on [0, 1) is a universal trial-maker. Compare it to a cutoff p: the chance it falls below p is exactly p, because the favorable strip [0, p) takes up the fraction p of the whole interval. That is geometric probability on a line.
Conclusion. One uniform draw on [0, 1), compared against a cutoff, reproduces any single yes/no trial with the probability you choose.
Here is the whole method in one breath. (1) Describe one trial in terms of random numbers. (2) Run that trial a large number n of times. (3) Count the m runs in which your event happened. The estimate is the frequency mn. By the law of large numbers — the very heart of Lesson 33.1 — that frequency wobbles less and less as n grows, closing in on the true P.
Watch it happen. Below, a fair coin (p = 0.5) is flipped over and over, and we plot the running frequency of heads after each flip. The first few flips are jumpy — a couple of heads in a row sends the curve high. But the dashed line at 0.5 is a magnet: the more flips, the tighter the curve hugs it.
Suppose we do not know a thumbtack lands point-up with what probability. We toss it 1000 times and it lands point-up 372 times. Our estimate is 3721000 = 0.372. Toss it 1000 more and we might get 0.361 — a slightly different estimate. Neither is "the" answer; both are honest readings of the same hidden P, and averaging more tosses narrows the spread.
Conclusion. Simulation converts a hard probability into a tally: imitate one trial, repeat it many times, and the success frequency mn estimates P, sharpening as n grows.
Now a question with a story. A family has three children; assume each child is independently a girl or boy with probability 0.5. What is the chance of at least one girl? We can answer this exactly with the complement shortcut from Lesson 33.3: the only way to fail is three boys, and
P(at least one girl) = 1 − P(all boys) = 1 − (0.5)3 = 1 − 18 = 7/8 = 0.875.
But pretend we did not see that trick. We can still simulate: each family is three coin flips, "at least one girl" means at least one of the three is a head, and we tally the share of families that qualify. The simulator below does exactly this. Watch the green estimate creep toward the dashed exact 7/8 as you run more families — and notice it lands a little differently each time.
A simulation estimates — it never proves an exact value. More trials make the estimate tighter, but a run of 7/8 ≈ 0.875 might read 0.862 today and 0.889 tomorrow. If you need the exact number and a count is available, count; use simulation when the count is out of reach, or to check a count you are unsure of.
Conclusion. A real, multi-stage question becomes a sequence of random draws; the simulated frequency approaches the exact 7/8, confirming a count we could also do by hand.
The crown jewel of simulation is estimating a number that has nothing obviously to do with chance: π. Picture the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 with the quarter-circle x² + y² ≤ 1 tucked in one corner. The square has area 1; the quarter-circle has area π4. So if we drop a point at random into the square, by the geometric-probability idea of Lesson 33.2,
P(point lands inside the quarter-circle) = area insidearea of square = π/4.
Turn that around. Scatter n random points; count the inside ones (where x² + y² ≤ 1). The fraction insiden estimates π/4, so
π ≈ 4 · insidetotal.
Each point is green inside or red outside, decided by the honest test x² + y² ≤ 1 — never by the eye. Crank up n and the estimate tightens around the true π = 3.1416…, but it never sits exactly on it.
One scatter of 500 darts lands 382 inside. Then the estimate is π ≈ 4 · 382500 = 4 · 0.764 = 3.056 — close to 3.1416, off by about 3%. With 5000 darts the wobble typically shrinks to about a percent. The error falls roughly like 1/√n: to halve the typical error you need about four times as many points.
Conclusion. Counting darts inside a quarter-circle estimates π/4, so π ≈ 4 · inside / total — a geometric probability harvested by pure randomness, sharper with every extra point yet never exact.
Every simulated answer comes with a built-in disclaimer: it is a sample, not a certainty. Two ideas keep you honest.
More trials, tighter answer. Doubling n does not double your accuracy — the typical error shrinks like 1/√n, so you need roughly four times the trials to halve the error. That is why simulations run thousands or millions of times, not dozens.
Different runs, different answers. Two honest simulations of the same question will disagree slightly. Neither is wrong; both are unbiased glimpses of the same hidden P. Reporting a single decimal as if it were exact is the trap — a simulation gives you a neighborhood of the truth, and a bigger n shrinks that neighborhood.
Conclusion. A simulation reports a value near P with a spread that shrinks like 1/√n; trust the size of n, never a single lucky run.
| Idea | How it works | What you read off |
|---|---|---|
| Random number | a uniform draw on [0, 1); success when it is below the cutoff p | imitates any single yes/no trial |
| Simulation | imitate one trial, run it n times, count the m successes | P ≈ mn |
| Geometric estimate | scatter points; share inside the target estimates the area ratio | π ≈ 4 · insidetotal |
| Accuracy | typical error shrinks like 1/√n | 4× the trials → half the error |
| Honesty | it is an estimate, never an exact value | count when you can; simulate when you can't |
Two of the six faces are favorable, so success should take up 26 = 1/3 of the interval. Let "success" be u < 1/3 (the strip [0, 1/3), of length 1/3 ≈ 0.333). Any strip of length 1/3 works equally well.
Estimate = 328800 = 0.41. This is below 0.5, but a single run of 800 can wander; it is evidence, not proof. More flips would tell us whether 0.41 holds up or drifts back toward 0.5.
π ≈ 4 · 15262000 = 4 · 0.763 = 3.052. The true π is 3.1416, so the estimate is low by about 0.090 — roughly a 2.8% error, typical for a couple thousand darts.
Simulate each family as four fair coin flips; "success" = at least three heads. Tally the share over many families. Exactly: P(exactly 3) + P(exactly 4) = 416 + 116 = 516 = 0.3125. The simulated frequency should settle near 0.3125.
A simulation always estimates; matching π to five digits would be an extraordinary fluke, and the very next run would disagree. The estimate is 4 · inside / total — a ratio of whole numbers — so it can only hit certain rational values, almost never π itself. Tight, yes; exact, never.
Error shrinks like 1/√n, so cutting the error to one-third needs about 3² = 9 times the trials: roughly 9000 trials. (To merely halve an error you need about 4× the trials.)
Six questions to lock it in. Tap the answer you think is right.
This lesson develops simulation as a way to estimate probabilities when an exact count is hard, and to check theoretical answers empirically — touching CCSS HSS-IC.A.2 (decide if a model is consistent with data, e.g. using a simulation) and CCSS HSS-MD.A.3 (compare an empirical frequency to a theoretical probability). The Monte-Carlo π widget and the "at least one girl" simulator both end at a known exact value (π/4 and 7/8), so the running frequency can be seen converging on the truth. The honest throughline: every quoted probability comes from a counted ratio of random trials, and the estimate tightens like 1/√n — it is never exact. A good follow-up discussion: when is it smarter to count than to simulate?