Chapter 51 · Case Study
Sales Calls to First Win: the Geometric ☎️
How many calls until a rep lands the first sale? That waiting time is the geometric distribution. We fit the success rate, compare lead segments, and use the memoryless property to plan staffing.
Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026
⚙️ Setup & data¶
In [1]:
import numpy as np, pandas as pd, matplotlib.pyplot as plt
from scipy import stats
AMBER="#d97706"; TEAL="#0d9488"; INK="#1a2138"; GRID="#e6e9f2"; PINK="#db2777"
plt.rcParams.update({"figure.facecolor":"white","axes.facecolor":"white","figure.dpi":110,"font.size":11,
"axes.edgecolor":GRID,"axes.grid":True,"grid.color":GRID,"axes.axisbelow":True,"axes.spines.top":False,
"axes.spines.right":False,"axes.titlesize":12,"axes.titleweight":"bold","legend.frameon":False})
BASE="https://raw.githubusercontent.com/johnfisher-ai/Statistics-Data-Science-AI-Visual-Book/main/data/"
try: df = pd.read_csv("../../data/telemarketing_sales_outreach.csv")
except FileNotFoundError: df = pd.read_csv(BASE+"telemarketing_sales_outreach.csv")
print("loaded:", df.shape)
df.head()
loaded: (1000, 4)
Out[1]:
| agent_id | call_date | lead_segment | calls_to_first_success | |
|---|---|---|---|---|
| 0 | AGT_DOE | 2026-03-01 | Mid-Market | 1 |
| 1 | AGT_JONES | 2026-03-01 | Mid-Market | 5 |
| 2 | AGT_SMITH | 2026-03-01 | Mid-Market | 8 |
| 3 | AGT_SMITH | 2026-03-01 | SMB | 18 |
| 4 | AGT_SMITH | 2026-03-01 | SMB | 2 |
BEAT 1 · FIT THE SUCCESS RATE
Mean wait gives p
The geometric counts trials up to the first success, PMF (1-p)^(k-1) p. Its mean is 1/p, so the per-call success rate is just one over the average number of calls.
In [2]:
x = df["calls_to_first_success"]
p = 1/x.mean()
print(f"records : {len(x):,}")
print(f"mean calls to win : {x.mean():.3f}")
print(f"per-call rate p = 1/mean = {p:.4f} ({p:.1%})")
print(f"Geometric mean = 1/p = {1/p:.3f}")
print(f"Geometric variance = (1-p)/p^2 = {(1-p)/p**2:.3f}")
records : 1,000 mean calls to win : 6.449 per-call rate p = 1/mean = 0.1551 (15.5%) Geometric mean = 1/p = 6.449 Geometric variance = (1-p)/p^2 = 35.141
In [3]:
fig,ax=plt.subplots(figsize=(7,3.4))
kmax=30; k=np.arange(1,kmax+1)
obs=x.clip(upper=kmax).value_counts(normalize=True).sort_index()
ax.bar(obs.index,obs.values,color=AMBER,alpha=0.7,width=0.9,label="observed calls-to-win")
ax.plot(k, stats.geom.pmf(k,p),"o-",color=TEAL,ms=4,label=f"Geometric(p={p:.3f})")
ax.axvline(1/p,color=PINK,ls="--",lw=2,label=f"mean 1/p = {1/p:.1f}")
ax.set_xlabel("calls until first success"); ax.set_ylabel("probability"); ax.set_title("Calls to first win: observed vs geometric"); ax.legend()
plt.tight_layout(); plt.show()
In [4]:
# Formal goodness-of-fit: chi-square on observed vs expected counts.
# Rule of thumb: merge adjacent bins so every EXPECTED count is at least 5, then
# chi-square = sum((obs-exp)**2 / exp), with df = (bins - 1 - number of fitted parameters).
def chisq_gof(counts, pmf, n_params):
N = len(counts); kmax = int(counts.max())
obs = np.array([(counts == k).sum() for k in range(kmax + 1)], float)
exp = np.array([N * pmf(k) for k in range(kmax + 1)], float)
obs = np.append(obs, 0.0); exp = np.append(exp, N - exp.sum()) # tail bin for k > kmax
mo, me, co, ce = [], [], 0.0, 0.0
for o, e in zip(obs, exp):
co += o; ce += e
if ce >= 5:
mo.append(co); me.append(ce); co = ce = 0.0
if ce > 0 and mo:
mo[-1] += co; me[-1] += ce
mo, me = np.array(mo), np.array(me)
chi2 = ((mo - me) ** 2 / me).sum(); dof = len(mo) - 1 - n_params
return chi2, dof, stats.chi2.sf(chi2, dof)
chi2, dof, pval = chisq_gof(x.values, lambda k: stats.geom.pmf(k, p), 1)
print(f"Chi-square goodness-of-fit: chi2 = {chi2:.2f}, df = {dof}, p = {pval:.3f} -> "
+ ("cannot reject the fit (Geometric is a good model)" if pval > 0.05 else "reject: not Geometric"))
Chi-square goodness-of-fit: chi2 = 12.57, df = 23, p = 0.961 -> cannot reject the fit (Geometric is a good model)
To validate rather than assume, a chi-square goodness-of-fit test compares the observed call-counts to a Geometric(p) (rare tail bins merged so expected counts stay at least 5): chi-square = 12.6 on 23 df, p = 0.96, so we comfortably cannot reject the Geometric fit.
The data is the classic geometric shape: success on the first call is the single most likely outcome, but a long right tail of stubborn leads drags the average wait to about 6.5 calls. One rate p ≈ 0.155 describes the whole funnel.
BEAT 2 · WHICH SEGMENT IS EASIEST?
A rate per lead type
Each lead segment is its own geometric process. Comparing 1/p across segments shows where reps land sales fastest, a direct read on lead quality.
In [5]:
seg = df.groupby("lead_segment")["calls_to_first_success"].agg(["mean","count"])
seg["p_per_call"] = 1/seg["mean"]
print(seg.sort_values("mean").round(4))
mean count p_per_call lead_segment SMB 6.2733 494 0.1594 Mid-Market 6.3706 402 0.1570 Enterprise 7.5865 104 0.1318
Segments with a smaller average wait have a higher per-call success rate p, easier leads. The spread across segments is exactly the kind of efficiency signal a sales manager acts on: route reps toward the high-p segments, or invest training where p is low.
BEAT 3 · MEMORYLESS: THE SUNK-COST TRAP
Past calls do not change the future
The geometric is MEMORYLESS: after k failures, the expected remaining calls is still 1/p. Simulating confirms it, and it warns against the gambler's fallacy of 'we are due for a yes'.
In [6]:
rng = np.random.default_rng(51)
sim = rng.geometric(p, size=2_000_000)
print(f"overall expected remaining wait = {1/p:.2f} calls")
cond = sim[sim>5] - 5 # already failed 5 times: how many MORE calls?
print(f"expected remaining AFTER 5 failures = {cond.mean():.2f} calls (still ~1/p, memoryless)")
print(f"P(win within 5 calls) = {1-(1-p)**5:.3f}")
print(f"P(win within 10 calls) = {1-(1-p)**10:.3f}")
overall expected remaining wait = 6.45 calls expected remaining AFTER 5 failures = 6.46 calls (still ~1/p, memoryless) P(win within 5 calls) = 0.569 P(win within 10 calls) = 0.815
Whether a rep is on call 1 or call 6, the expected number of remaining calls to a sale is the same 6.5: the geometric has no memory. There is a 57% chance of a win within 5 calls and 82% within 10. The lesson for managers is hard but useful: a string of rejections does not make the next yes any more "due", so set call budgets from p, not from a hot or cold streak.
✅ Case closed
Calls to the first sale are geometric. We fit p ≈ 0.155 (mean wait 6.5), compared lead segments by their per-call rate, and used the memoryless property to set honest call budgets. The single rate p turns a noisy sales floor into a plan.
Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher