The Archer's Trials — Geometric Distribution

PRO0

Prologue — The Question in the Woods

Every great unit begins with a story.

In the village of Elmswood, there lived an archer named Robin, famous for one strange habit: whenever Robin practiced, the whole village gathered — not to watch the skill, but to count. "How many arrows until the first bullseye?" they would whisper, placing bets. Robin's bow never changed. The wind never changed the odds. Each shot was its own small story, unconnected to the last. And yet — the villagers noticed something magical: no matter how unpredictable each single shot was, the pattern of waiting for that first success always followed the same secret law. That law is what mathematicians call the Geometric Distribution — and this book is Robin's diary of discovering it.

What you will learn in this tale

4.1

Geometric ConditionsWhat makes a situation "geometric" in the first place.

4.2

The Probability FormulaHow to calculate the chance the first success lands on a chosen trial.

4.3

Mean & Std. DeviationHow many shots to expect on average, and how much that varies.

4.4

Calculator MagicUsing geometpdf and geometcdf to do the heavy lifting.

4.5

Geometric vs. BinomialTwo cousins that are easy to confuse — until now.

4.1

Chapter 1 — The Four Rules of the Range

Geometric Conditions

Before Robin could count arrows scientifically, the village elder set down four rules. "Break any one of these," she said, "and your counting is no longer geometric." Robin nicknamed them the B.I.T.S. of shooting.

The B.I.T.S. Checklist

B

Binary outcomeEach trial is either a "success" (bullseye) or a "failure" (miss) — nothing in between.

I

Independent trialsOne shot never affects the next. Robin's bad shot doesn't curse the next arrow.

T

Trials share the same pThe probability of success, p, stays constant from shot to shot.

S

Stop at first successWe are counting trials until the first success — then we stop, no fixed number of trials.

If a situation satisfies all four, the number of trials needed for the first success, X, is a geometric random variable.

Example 1.1

Robin hits the bullseye with probability $p = 0.3$ on every shot, and shots are independent. Robin keeps shooting until the first bullseye. Is $X = $ "number of shots until the first bullseye" geometric?

Reasoning

✅ Binary: hit or miss. ✅ Independent: one shot doesn't change the next. ✅ Same p: 0.3 every time. ✅ We stop counting at the first success.

Yes — this is geometric ✓

Example 1.2 — A Trick from the Elder

Robin now draws arrows from a small quiver of exactly 10, where 3 are enchanted "lucky" arrows that always hit. Robin shoots until the first lucky arrow is drawn (without replacing arrows back into the quiver). Is this geometric?

No. Because arrows are not replaced, the probability of drawing a lucky arrow changes after every draw — the trials are no longer independent, and p is not constant. (This situation is closer to a hypergeometric setting, a tale for another day.)

4.2

Chapter 2 — The Weight of Waiting

Geometric Probability Formula

"If I miss, and miss, and miss again — and only then succeed," Robin wondered, "what is the chance of that exact story unfolding?" The elder smiled and wrote a single elegant formula in the dirt.

The Formula

If $X$ = the trial number of the first success, and each trial succeeds with probability $p$ (so fails with probability $1-p$), then:

$$P(X = k) = (1-p)^{\,k-1}\, p, \qquad k = 1, 2, 3, \dots$$

In words: (miss) repeated $k-1$ times, then one hit. Every earlier trial must be a failure, and the very last one a success.

Example 2.1

Robin's chance of a bullseye each shot is $p = 0.3$. What is the probability that the first bullseye happens on exactly the 3rd shot?

Solution

We need: miss, miss, then hit → $k = 3$.

$$P(X=3) = (0.7)^{2}(0.3) = (0.49)(0.3) = 0.147$$

P(X = 3) ≈ 0.147

Example 2.2 — Beyond the Bow

A locksmith's apprentice picks a lock correctly with probability $p = 0.2$ on each attempt, attempts are independent. Find the probability the lock first opens on the 2nd attempt.

$$P(X=2) = (0.8)^{1}(0.2) = 0.16$$

So there is a 16% chance the very first success happens on the second try.

4.3

Chapter 3 — The Long Run of Arrows

Mean & Standard Deviation

Robin practiced for a hundred days, tallying every session. "On average," Robin asked, "how many arrows do I need before my first bullseye — and how much does that number typically wander?"

The Formulas

$$\mu_X = E(X) = \dfrac{1}{p}$$

$$\sigma_X = \sqrt{\dfrac{1-p}{p^{2}}}$$

The rarer the success (small $p$), the larger the expected wait — and the more that wait swings from attempt to attempt.

Example 3.1

Robin's bullseye probability is $p = 0.3$. On average, how many shots until the first bullseye, and what is the standard deviation?

Solution

$$\mu_X = \frac{1}{0.3} \approx 3.33 \text{ shots}$$

$$\sigma_X = \sqrt{\frac{0.7}{0.3^{2}}} = \sqrt{7.78} \approx 2.79 \text{ shots}$$

μ ≈ 3.33, σ ≈ 2.79

Example 3.2 — The Rare Golden Arrow

Only 1 in 20 arrows in Robin's quiver is a rare "golden" arrow drawn at random with replacement ($p = 0.05$). On average, how many draws until a golden arrow appears?

$$\mu_X = \frac{1}{0.05} = 20 \text{ draws}$$

Rare events take a long average wait — this matches intuition: a 1-in-20 chance takes about 20 tries on average.

4.4

Chapter 4 — The Elder's Enchanted Scroll

Calculator Use: geometpdf & geometcdf

The elder handed Robin an enchanted scroll (which Robin's friends in town simply called a "calculator"). "Speak the right words," she said, "and it will count arrows faster than any villager."

Two Spells

Command	What it finds	Meaning in words
`geometpdf(p, k)`	$P(X = k)$	Probability the first success is exactly on trial k
`geometcdf(p, k)`	$P(X \le k)$	Probability the first success happens on or before trial k

On a TI-84: press 2nd → VARS (DISTR), scroll to geometpdf( or geometcdf(, then enter p and k.

Example 4.1

With $p = 0.3$, find the probability the first bullseye is exactly on shot 3, using the scroll.

Solution

$$\texttt{geometpdf}(0.3,\,3) = 0.147$$

This matches Example 2.1 exactly — the calculator just saves the by-hand work!

P(X = 3) = 0.147

Example 4.2 — "At Most" Questions

Still with $p = 0.3$, find the probability Robin's first bullseye happens within the first 3 shots (on shot 1, 2, or 3).

$$\texttt{geometcdf}(0.3,\,3) = 1-(0.7)^3 = 1 - 0.343 = 0.657$$

There is about a 65.7% chance the bullseye happens within the first three shots. Note the shortcut: $P(X \le k) = 1-(1-p)^k$.

Quick tip: "more than k" shots

Sometimes you'll be asked for $P(X > k)$ — the probability it takes more than k shots. Since all failures happened for the first k shots:

$$P(X > k) = (1-p)^{k}$$

4.5

Chapter 5 — Two Cousins of the Range

Comparison with Binomial Distribution

One day, Robin's twin cousin Casey visited — also an archer, but Casey always shot exactly 5 arrows and counted how many hit the bullseye, no matter what. The villagers were confused: "Aren't you two doing the same thing?" Robin and Casey laughed. "Not quite," they said, and explained the difference.

Robin (Geometric) vs. Casey (Binomial)

	Geometric (Robin)	Binomial (Casey)
Number of trials	Not fixed — keep going until success	Fixed number, n
What X counts	The trial number of the first success	The total number of successes in n trials
Possible values of X	$1, 2, 3, \dots$ (no upper limit)	$0, 1, 2, \dots, n$
Formula	$(1-p)^{k-1}p$	$\binom{n}{k}p^{k}(1-p)^{n-k}$
Typical question	"On which shot is the first success?"	"Out of n shots, how many succeed?"

Example 5.1 — Same Numbers, Different Question

Both Robin and Casey have $p = 0.3$. Casey always shoots $n = 5$ arrows.

(a) Casey (Binomial): probability of exactly 2 bullseyes out of 5 shots?

(b) Robin (Geometric): probability the first bullseye is on shot 3?

Solution

(a)

$$P(X=2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10(0.09)(0.343) = 0.309$$

(b)

$$P(X=3) = (0.7)^2(0.3) = 0.147$$

Same $p$, completely different questions — and completely different answers!

(a) 0.309 (b) 0.147

Example 5.2 — Spot the Type

A basketball player makes free throws with $p=0.6$. Decide: is each scenario geometric or binomial?

① "Probability the player makes exactly 7 of the next 10 shots."
② "Probability the player's first miss happens on the 4th shot."

① Binomial — fixed number of trials (10), counting total successes.

② Geometric — counting trials until the first occurrence of an outcome (here, "success" = miss, with p = 0.4).

EPI∞

Epilogue — Robin's Diary, Summarized

Everything the range taught us

Years later, Robin's diary became required reading for every apprentice archer in Elmswood. Here is the single page every student folds the corner of.

4.1 CONDITIONS — B.I.T.S.

Binary outcome · Independent trials · Trials share the same p · Stop at first success.

4.2 PROBABILITY FORMULA

$$P(X=k)=(1-p)^{k-1}p$$

4.3 MEAN & STD. DEVIATION

$$\mu=\frac{1}{p} \qquad \sigma=\sqrt{\frac{1-p}{p^2}}$$

4.4 CALCULATOR

geometpdf(p,k) → $P(X=k)$
geometcdf(p,k) → $P(X\le k)$
$P(X>k) = (1-p)^k$

4.5 GEOMETRIC vs. BINOMIAL — the one-line test

Ask: "Is the number of trials fixed?" — Yes → Binomial (counting successes). No, we stop at the first success → Geometric (counting the waiting trial).

Robin's Final Words

"Every arrow is independent. Every wait has a pattern. Learn the pattern, and even uncertainty becomes something you can calculate." — Robin of Elmswood