An interactive exploration ofThe Extremes of Chance

Roll two dice. Keep the lower one — or the higher one. How does that simple choice reshape the entire distribution? This page builds the answer from scratch: one concept at a time, each demonstrated before it's named.

Start with one roll

A single uniform random number on [0, 1) is the simplest possible distribution. Every value is equally likely. Click to sample — watch how the histogram fills in flat.

Rolls: 0

Mean: —

Expected: 0.500

The expected value is E[X] = 1/2. Nothing surprising. But what happens when you roll two numbers and pick one?

Two rolls, one choice

Click "Roll" to generate two uniform values. The interface picks the minimum for you. Before looking at the histogram — just watch the pairs. Notice anything about where the chosen value tends to land?

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→

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Pairs: 0

Mean: —

Expected: 0.333

"Wait — why isn't the expected minimum just 0.25? Each die averages 0.5, and the minimum should be about half of that?"

Good instinct, but the minimum of two values isn't "half of one value." It's the probability-weighted outcome of always picking the lower. Low values are likely to be chosen because either die landing low is enough. The exact answer comes from the CDF — keep going.

The shape of min vs. max

Now let's see both distributions simultaneously, with the theoretical curves overlaid. The min distribution follows f(t) = n(1−t)^(n−1) and the max follows f(t) = nt^(n−1). Watch how they mirror each other.

min(X₁ … Xₙ) — Disadvantage

E: —

Med: —

max(X₁ … Xₙ) — Advantage

E: —

Med: —

n = 2

The identities: For n independent uniform [0,1) variables — E[min] = 1/(n+1) and E[max] = n/(n+1). They always sum to 1. Drag n above and watch the means converge toward the extremes.

Mean vs. Median — which matters?

The mean and median of these distributions differ because the shapes are skewed. For min(X,Y): the mean is 1/3 ≈ 0.333 but the median is 1 − 1/√2 ≈ 0.293. Which one should you care about?

For threshold checks (D&D attack rolls, skill checks), you're asking "do I beat the DC?" — the full CDF matters, and the median tells you your 50/50 breakpoint. For accumulated outcomes (damage over many rounds), the mean governs your long-run total. Click through to see the CDF with both marked.

n = 2

Mean: —

Median: —

Δ: —

"So for a d% roll with advantage, my median outcome is about 71 — but my average outcome is about 67. Why are those different?"

Because the max distribution is left-skewed: the long tail of low outcomes drags the mean below the median. You'll beat 67 more than half the time, but those occasional low rolls pull your average down. For pass/fail checks, the median is the more useful number. For damage accumulation, trust the mean.

The full picture

One last experiment. Vary n from 2 to 10 and watch how the distributions, means, and medians shift. With more dice, the extremes get more extreme — min clusters near 0, max clusters near 1 — but the rate at which they converge follows the elegant 1/(n+1) and n/(n+1) formulas.

n = 2

Built to demonstrate: bottom-up scaffolding, clickable simulations as primary explanation, side-by-side comparison, skeptical interlocutor, metric reveal after intuition, and controlled parameter variation.