Chop a second into slots of width Δ and give each slot an arrival with probability λΔ. The chance of still waiting after t is (1 − λΔ)^(t/Δ), exactly. Shrink Δ and that expression walks toward e^(−λt) — the exponential distribution, arriving as a limit of a distribution you already have.
Core ideas
The geometric distribution from Discrete Mathematics is doing all the work here. The exponential is what it becomes when the slot stops mattering.
A rate of 1/2 arrivals a second is not a probability of 1/2. Chopped into slots of Δ = 1/10 second it becomes a probability of λΔ = 1/20 per slot, and that is the number a coin gets flipped against. Halve the slot and the per-slot probability halves with it; the rate does not move.
(1 − λΔ)^(t/Δ) is (1 − x/n)ⁿ with x = λt and n = t/Δ, and Algebra's the-number-e says where that goes. Nothing on this path writes a density or an integral: the exponential distribution is defined here by the limit of a geometric tail, and that is enough to compute with.
The conditional tail P(T > s + t | T > s) equals P(T > t) at every slot size, for every s, with no approximation anywhere. It is not something the exponential acquires in the limit; the geometric already has it, and the limit inherits it.
The material
The exact tail, the limit it approaches, the property both share, and why one of the two columns is printed as rounded.
Divide time into slots of width Δ. In each slot an arrival occurs with probability λΔ, independently. The waiting time T until the first arrival is geometric, and surviving t units of time means surviving t/Δ consecutive slots without one: P(T > t) = (1 − λΔ)^(t/Δ).
That expression is exact and rational. With λ = 1/2 per second, Δ = 1/10 second and t = 4 seconds it is (19/20)⁴⁰ — a ratio of two forty-digit integers, computed as a fraction and only then printed as a decimal.
Each tenfold refinement of the slot closes about nine tenths of the remaining gap, which is what “the error is of order Δ” looks like when it is printed rather than asserted. The exact column approaches the rounded one from below, because a slotted clock can only let an arrival happen at a slot boundary and so gives it slightly fewer chances.
For the geometric waiting time above, and for every whole number of slots s and t: P(T > s + t | T > s) = P(T > t). Having waited already tells you nothing about how much longer you will wait.
P(T > s + t | T > s) = P(T > s + t)/P(T > s) because the event T > s + t is contained in T > s. Substituting the tail formula gives (1 − λΔ)^((s+t)/Δ) / (1 − λΔ)^(s/Δ).
The powers subtract: the quotient is (1 − λΔ)^(t/Δ), which is P(T > t). Nothing about s survives the cancellation, which is the whole content of the property. It holds at every slot size, so it holds in the limit too.
□
Concretely: if the bus has not come for three minutes, the chance it takes another four is the same as it was when you arrived. Quiet time is not credit. This is the single most counter-intuitive fact on the course, and it is the assumption the M/M/1 model rests on — which is why it is worth being able to say precisely what it claims and what it does not.
What it does not claim is that arrivals are evenly spread. Memorylessness is exactly what makes bunching possible: nothing in the process discourages two arrivals landing in consecutive slots, because the process has no record of the first one.
The geometric column is a fraction. The exponential column is not: e^(−λt) is computed by summing a series and is printed as rounded, with the method named on the page. The lab bounds λt at 12, because the series it uses loses significant digits to cancellation beyond that and a confident wrong number is worse than a refusal. The exact column has no such limit, which is the argument for exactness stated as a number.
The mean of the waiting time is 1/λ — two seconds at λ = 1/2 — and it is worth noticing that the tail at the mean is not a half: e^(−1) ≈ 0.368. An exponential wait is longer than its mean about 37% of the time, not half of it, and that asymmetry is the shape behind every long tail in the previous course.
Interactive
The geometric tail exactly, and the exponential it becomes
Method
Decide which number you want exact before deciding which formula to use.
λΔ, with λ and Δ in the same unit of time. A per-slot probability above one is not a probability, which is the slot being too coarse for the rate.
t/Δ, a whole number. The exact tail is (1 − λΔ) raised to that power, and because both are rational nothing is approximated.
e^(−λt) is the limit, and it is quicker and independent of Δ. It is also rounded. Use it to reason about shape and use the exact column when the number goes into a decision.
P(T > s + t | T > s) should equal P(T > t) for any s you try. If it does not, the model in your head has memory in it that the arithmetic does not.
The same probability computed at two slot sizes and in the limit, with the memoryless check underneath.
The two exact lines are fractions with forty and four hundred digit denominators, and they are computed as fractions. Only the last digit printed is rounding, and the e^(−2) line says on its face that it is the rounded one.
The final block is the property, not a coincidence of these numbers. Change the three seconds to thirty and the conditional tail is unchanged, which is either obvious or deeply strange depending on how long you look at it.
For a faded rehearsal, take λ = 1 per second and t = 3. The supplied first move is λt = 3, so the limit is e^(−3). Predict whether the geometric tail at Δ = 1/10 is above or below it before computing either.
You should be able to turn a rate into a per-slot probability, compute P(T > t) exactly in slots, say where e^(−λt) comes from and which of the two is rounded, and show that the conditional tail does not depend on how long the wait has already lasted.
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