Probability, randomness, and the discrete distributions that show up everywhere in ecology

Sample space: the four equally likely birth orders are {BB, BG, GB, GG}, each with probability 1/4.

(a) "At least one girl" rules out BB. We are left with {BG, GB, GG}, three equally likely outcomes. Only GG is "both girls". So $P(\text{both} \mid \text{at least one girl}) = \tfrac{1}{3}$.

(b) "Older child is a girl" rules out BB and GB. We are left with {BG, GG}, two equally likely outcomes. So $P(\text{both} \mid \text{older is girl}) = \tfrac{1}{2}$.

(c) Knowing which child is a girl (a more specific piece of information) shrinks the sample space more than just knowing one of them is.

Check in R:

# Monte Carlo check
set.seed(1)
fams <- matrix(sample(c("B","G"), 2*1e6, replace=TRUE), ncol=2)
mean(rowSums(fams=="G")==2 & rowSums(fams=="G")>=1) /
 mean(rowSums(fams=="G")>=1) # ~ 1/3
mean(fams[,1]=="G" & fams[,2]=="G") /
 mean(fams[,1]=="G") # ~ 1/2

Let $B$ = "taxi is blue", $W_B$ = "witness says blue".

• $P(B) = 0.01$, $P(\bar B) = 0.99$
• $P(W_B \mid B) = 0.99$
• $P(W_B \mid \bar B) = 0.05$ (the 5% misclassification of green)

Total probability of "says blue":

Bayes:

So only ~17% — even though the witness is "99% reliable" on blue taxis.

Where the small answer really comes from — the odds form. Don't stop at the plug-in; the same problem read in odds shows exactly why a "99% reliable" witness still loses. Divide Bayes' rule for $B$ by Bayes' rule for $\bar B$ — the shared denominator $P(W_B)$ cancels — and you are left with a rule worth memorizing:

Posterior odds = prior odds × likelihood ratio: the evidence only ever multiplies the odds you started with. Here the prior odds are $1:99$ and the likelihood ratio is $0.99/0.05 = 19.8$, so the posterior odds are $19.8/99 = 0.2$, i.e. $1:5$, which is $P = \tfrac{1}{1+5} = \tfrac{1}{6} \approx 0.167$ — the same answer, now with the mechanism exposed. To reach even money (odds $1:1$) the test would need a likelihood ratio near 99 — it would have to be about five times sharper than it is — just to overcome a base rate that starts you 99-to-1 in the hole. That is the general lesson: against a rare event, the base rate sets the odds you must climb out of, and even a very accurate test barely dents it.

p_B <- 0.01; p_W_B <- 0.99; p_W_notB <- 0.05
(p_W_B * p_B) / (p_W_B * p_B + p_W_notB * (1 - p_B)) # 0.1667

Let $X_i$ be 1 if person $i$ is shorter than both neighbors, else 0. The total count is $X = \sum X_i$. By linearity of expectation, $E[X] = \sum E[X_i] = n \cdot P(\text{person 1 shorter than both neighbors})$.

Consider person $i$ together with their two neighbors. Because everyone is seated in random order and all heights are distinct, each of the $3! = 6$ height orderings of these three people is equally likely. Person $i$ is shorter than both neighbors exactly when $i$ is the smallest of the three. Fixing $i$ as the smallest leaves the two neighbors free to be arranged in $2! = 2$ ways, so 2 of the 6 orderings qualify. Hence $P(\text{person } i \text{ shorter than both neighbors}) = 2/6 = 1/3$.

Therefore $E[X] = n/3$.

n <- 50; reps <- 1e5
sapply(1:reps, function(i){
 h <- sample(n) # distinct heights at table positions
 sum(h < c(h[n], h[-n]) & h < c(h[-1], h[1]))
}) |> mean() # ~ 16.67 = 50/3

Punnett: CC : Cc : Cc : cc = 1:1:1:1, each with probability 1/4. Unaffected = CC, Cc, or Cc.

(a) $P(\text{unaffected}) = 1/4 + 1/4 + 1/4 = 3/4$. (Or just $1 - P(cc) = 1 - 1/4 = 3/4$.)

(b) Children are independent: $(3/4)^4 = 81/256 \approx 0.316$.

(3/4)^4 # 0.3164
dbinom(0, size=4, prob=1/4) # same thing

Approximate values (one seed) — yours will be close:

Those numbers are the theoretical values the simulation fluctuates around, and none of them is arbitrary — each column is one formula evaluated, so you can predict the table before you run a single draw.

The variance column. Every row has mean 10 by construction; the story is entirely in the variance. Start from what the NB literally is — a Poisson whose rate is itself random (gamma-distributed) across transects, the mathematical version of "animals clump." That mixing gives the mean-variance law

Read it as two pieces: the $\mu$ is the irreducible Poisson noise, and the $\mu^2/k$ is the extra variance clumping adds. That second term is the whole point — it is what a plain Poisson cannot produce. The general move is to see $k$ as a clumping dial: as $k \to \infty$ the extra term vanishes and you recover $\mathrm{Var}=\mu$ (the Poisson is just NB with the clumping turned off), and as $k$ shrinks the extra term explodes. Only now plug in, with $\mu=10$:

Halving $k$ from 1 to 0.5 doubles the excess term (100 → 200) — clumping bites hardest at small $k$.

The $P(Y=0)$ column. A zero transect means "not one whale," and the NB pmf evaluated at zero collapses to a clean closed form,

so the same clumping dial controls how common empty transects are. Plug in $\mu=10$: $k=5$ gives $(5/15)^5=(1/3)^5=1/243\approx 0.004$; $k=1$ gives $(1/11)^1\approx 0.091$; $k=0.5$ gives $(0.5/10.5)^{0.5}=\sqrt{1/21}\approx 0.218$. The Poisson is the limit again — $(k/(k+\mu))^k \to e^{-\mu}$ as $k\to\infty$, and $e^{-10}\approx 0.0000454$, which rounds to the $0.000$ in the table. So a Poisson essentially never gives an empty transect, while NB($k=0.5$) gives one more than a fifth of the time.

(d) Small $k$ inflates both tails at once — the $P(Y=0)$ and $P(Y>20)$ columns rise together — which is exactly the "many zeros, a few huge counts" signature. The dataset described matches NB with small $k$; most ecological count data are like this.

Conceptual takeaway: the Poisson is a one-parameter distribution and is forced to have variance equal to its mean. Real ecological counts almost never satisfy that. Use the negative binomial unless you have a specific reason not to.

k <- 0:5
probs <- dbinom(k, size=5, prob=0.7)
round(probs, 3)
# 0.002 0.028 0.132 0.309 0.360 0.168
barplot(probs, names.arg=k, ylab="P(survivors)")

This is the Binomial distribution, $X \sim \text{Bin}(n=5, p=0.7)$. dbinom is not a black box — it evaluates the binomial pmf, and it is worth seeing where one entry comes from so you could reproduce any of them by hand:

Two factors, pulling in opposite directions. The $p^{k}(1-p)^{n-k}$ is the probability of one specific survivor/casualty pattern; the $\binom{n}{k}$ counts how many distinct patterns give exactly $k$ survivors. Take the peak value, $k=4$:

Now the mode makes sense. Each extra survivor multiplies the probability by $\tfrac{p}{1-p}=\tfrac{0.7}{0.3}=2.33$ but by a shrinking combinatorial factor; the two forces balance and the pmf turns over at $k=\lfloor(n+1)p\rfloor=\lfloor 6\times0.7\rfloor=\lfloor 4.2\rfloor=\mathbf{4}$ — so even though each marten is more likely to live than die, "all five survive" ($p=0.168$) is less likely than "four of five" ($p=0.360$).

(a) $Y \sim \text{Bin}(n=24, p=0.12)$.

dbinom(3, 24, 0.12) # P(Y=3) = 0.2387
pbinom(4, 24, 0.12) # P(Y<=4) = 0.8471
dbinom(0, 24, 0.12) # P(Y=0) = 0.0465

(b) Same pmf as Exercise 6, just bigger numbers — $P(Y=3)=\binom{24}{3}(0.12)^3(0.88)^{21}=2024\times0.001728\times0.0683=0.239$.

(d) The zero-case probability is the one you can do in your head, because at $k=0$ the combinatorial factor is $\binom{24}{0}=1$ and the whole pmf collapses to "every single deer must dodge infection":

That is the load-bearing quantity for surveillance, so make it general. The chance of catching at least one case in $n$ samples is its complement,

and flipping it around tells a program how many deer to sample: to be $95\%$ sure of detecting disease at prevalence $p$, solve $1-(1-p)^n \ge 0.95$ for $n$. Here, with $p=0.12$ and $n=24$, $P(Y\ge1)=1-0.0465=0.95$ — so $n=24$ is almost exactly the sample size that buys you 95% detection power at 12% prevalence. Even at that prevalence, a smaller sample would leave a real chance of a clean-but-wrong "no cases found." This is precisely the arithmetic behind "freedom from disease" sample-size calculations.

p <- c(0.20, 0.45, 0.25, 0.10)
dmultinom(c(18, 38, 18, 6), prob=p) # ~0.0011

set.seed(1)
flights <- rmultinom(5000, size=80, prob=p)
hist(flights[3,], main="Calves out of 80", xlab="count")

(a) Where the 0.0011 comes from. The multinomial pmf is the binomial's "many boxes" generalization: one multinomial coefficient counting the orderings, times each category's probability raised to its observed count,

With $\mathbf{x}=(18,38,18,6)$, $n=80$, that is $\dfrac{80!}{18!\,38!\,18!\,6!}\,(0.20)^{18}(0.45)^{38}(0.25)^{18}(0.10)^{6}\approx 0.0011$. It reads small, but every specific outcome is small — there are thousands of plausible flights, and even the single most likely one carries only about 1% of the probability.

(c) Why each margin is Binomial — the collapse trick. Don't take it on faith; derive it, because it is the move that lets you reason about one category at a time. Fix your attention on calves and lump the other three classes into a single "not-calf" box. Each of the 80 elk is now just calf (prob $0.25$) or not-calf (prob $0.20+0.45+0.10=0.75$), independently — and "count of successes in $n$ independent yes/no trials with fixed $p$" is the definition of the Binomial. So

The general rule: any single category of a Multinomial$(n,\mathbf{p})$ is Binomial$(n,p_i)$ — collapse the rest into "other" and the multi-box problem becomes the two-box problem you already know. (Note $\mathrm{Var}=15$ is below the mean 20: unlike the whales, a multinomial margin is under-dispersed, because a fixed total $n$ caps how large any one category can run.)

p <- c(.30, .20, .15, .20, .15)
rmultinom(1, 80, p)
sims <- rmultinom(500, 80, p)
summary(sims[5,])

Same collapse trick as the elk (Exercise 8): lump the other four responses into "not strongly-support" and the marginal is $\text{Bin}(80, 0.15)$, with $E=np=12$ and $\mathrm{Var}=np(1-p)=80\cdot0.15\cdot0.85=10.2$, so $\mathrm{SD}=\sqrt{10.2}\approx 3.2$. That is the punchline made concrete: a typical single survey lands anywhere from about 6 to 18 "strongly support" votes (12 $\pm$ two SDs) even though the underlying rate never moved — the sample-to-sample range is large compared to the point estimate, which is exactly why one survey is weak evidence.

(a) $K \sim \text{Bin}(n, 0.25)$.

1 - dbinom(0, 5, 0.25) # 0.7627
# smallest n with P(>=1) >= 0.5
which(1 - dbinom(0, 1:10, 0.25) >= 0.5)[1] # 3

(b) The "at least one" complement. Never sum $P(1)+P(2)+\dots+P(5)$ — go through the one outcome that's a single term. "At least one affected" is the complement of "none affected," and "none affected" needs every child to independently escape, so the messy sum collapses to

At $n=5$ that is $1-(0.75)^5 = 1-0.2373 = 0.763$.

(c) Where 3 comes from — solve it, don't scan it. The which() above brute-forces a list, but the threshold has a closed form. You want the smallest $n$ with $1-(0.75)^n \ge 0.5$, i.e. $(0.75)^n \le 0.5$. Take logs (the log is negative, so the inequality flips):

You can see the crossing by hand: $(0.75)^2 = 0.5625$ (still more than half a chance of no affected child) but $(0.75)^3 = 0.422$ (now under half), so the coin flips between the second and third child — 3 children. The same $\ln 0.5 / \ln(1-p)$ formula gives the "how many trials until an even chance of at least one hit" threshold for any $p$: rare detections need many trials, common ones only a few.

(a) $E[X] = E[\sum Y_i] = \sum E[Y_i] = np$. Linearity of expectation holds even without independence.

(b) $\mathrm{Var}(X) = \mathrm{Var}(\sum Y_i)$. Under independence, variances add: $= \sum \mathrm{Var}(Y_i) = np(1-p)$.

(c) If the $Y_i$ were correlated, you would pick up covariance terms: $\mathrm{Var}(\sum Y_i) = \sum \mathrm{Var}(Y_i) + 2\sum_{i<j}\mathrm{Cov}(Y_i,Y_j)$. Positive correlation (the rule in ecology — birds in flocks, deer in herds) inflates the variance — this is exactly why over-dispersed count distributions exist.

Math. Start from the binomial pmf and substitute $p=\lambda/n$ (this is what "$np\to\lambda$" means — hold the rate fixed while $n$ grows):

The trick is to regroup into a $\lambda^k/k!$ that will survive, times three pieces that each have a clean limit. Expand $\binom{n}{k}=\dfrac{n!}{k!\,(n-k)!}=\dfrac{n(n-1)\cdots(n-k+1)}{k!}$ and split the last factor's exponent:

Now send $n\to\infty$ with $k$ and $\lambda$ fixed, and take the three pieces in turn:

• (1) is a product of $k$ fixed factors $\tfrac{n}{n}\cdot\tfrac{n-1}{n}\cdots\tfrac{n-k+1}{n} = 1\cdot(1-\tfrac1n)\cdots(1-\tfrac{k-1}{n})$, each $\to 1$, so the whole thing $\to 1$.
• (2) is the definition of the exponential: $\left(1-\tfrac{\lambda}{n}\right)^{n}\to e^{-\lambda}$.
• (3) has a fixed exponent $-k$ and a base $\to 1$, so it $\to 1^{-k}=1$.

The three pieces collapse to $1\cdot e^{-\lambda}\cdot 1$, leaving

the Poisson pmf. Each piece carried one idea: (1) says "with $n$ huge, choosing $k$ of them barely dents the pool," (2) manufactures the $e^{-\lambda}$, and (3) is the correction that vanishes. Notice the mean-variance signature falls out for free: the binomial variance $np(1-p)=\lambda(1-\lambda/n)\to\lambda$, matching its mean — the equidispersion the Poisson is famous for is just the $p\to0$ limit of $np(1-p)$.

Conceptual takeaway. Poisson is the natural distribution for many independent rare events occurring in a fixed window of time/space, whose rate aggregates to $\lambda$. Whale sightings per transect, salmon redds per km, parasites per host (if independent) — all natural Poisson stories.

for(params in list(c(20,0.2), c(100,0.04), c(1000,0.004))) {
 x <- rbinom(1e4, params[1], params[2])
 cat("n=", params[1], " mean=", mean(x), " var=", var(x), "\n")
}
y <- rpois(1e4, 4); cat("Pois mean=", mean(y), " var=", var(y))

Switch. The intuition is that your initial guess (A) was right with probability 1/3, and opening C collapses the remaining 2/3 onto B — but "collapses" is doing a lot of work, so derive the 2/3 with Bayes instead of asserting it. Let the marten's true camera be $A$, $B$, or $C$ (prior 1/3 each), and condition on the event "technician opens C." The whole subtlety lives in the likelihoods — how the technician behaves given each truth:

• $P(\text{opens C}\mid \text{marten at A}) = \tfrac12$ — your guess A is safe and the marten's A is safe, so the technician is free to open B or C, and picks C half the time.
• $P(\text{opens C}\mid \text{marten at B}) = 1$ — he cannot open A (your guess) or B (the marten), so he is forced to open C.
• $P(\text{opens C}\mid \text{marten at C}) = 0$ — he never opens the marten's own camera.

Total probability of the reveal:

Now Bayes for the two cameras still in play:

There is the 2/3, and now you can see exactly where it comes from: B's likelihood is twice A's ($1$ vs $\tfrac12$), because the technician's forced move when the marten is at B pumps evidence toward B that your own guess A never receives. Switching doubles your odds.

To verify that by simulation you have to model the technician and the switch explicitly — draw the truth and your guess, have the technician open a camera that is neither, then move to whatever remains:

set.seed(1)
trials <- 1e5
truth <- sample(1:3, trials, replace=TRUE) # camera holding the marten
guess <- sample(1:3, trials, replace=TRUE) # your initial pick

# Technician opens a camera that is neither your pick nor the marten's
opened <- vapply(seq_len(trials), function(i) {
 choices <- setdiff(1:3, c(guess[i], truth[i]))
 if (length(choices) == 1) choices else sample(choices, 1)
}, numeric(1))

# You switch to the one remaining unopened camera
switched <- vapply(seq_len(trials), function(i) {
 setdiff(1:3, c(guess[i], opened[i]))
}, numeric(1))

mean(guess == truth) # staying wins:   0.33257
mean(switched == truth) # switching wins: 0.66743

Over 100,000 trials, staying won 0.33257 and switching won 0.66743 — matching the analytic 1/3 and 2/3.