Give the average a name: the mean $\mu$. Markov's inequality says that for a quantity that is never negative, the chance it lands at or above a level $t$ is at most $\mu/t$. A worked number: if the average wait at a clinic is $\mu=10$ minutes, then the chance any one patient waits $\ge 50$ minutes is at most $10/50 = 0.20$ — no more than 1 in 5, no matter how the waiting times are distributed. We used only the mean, so the bound is deliberately loose.

If we also know the spread — the variance $\sigma^2$, or the standard deviation $\sigma$ — we can do much better. Chebyshev's inequality says values rarely stray many standard deviations from the mean: $P(|X-\mu|\ge k\sigma)\le 1/k^2$. Landing 3 standard deviations away happens at most $1/9\approx 11\%$ of the time, again for any distribution with that variance.

The sharpest classical bound uses the entire moment-generating function $M_X(\lambda)=E[e^{\lambda X}]$. The Chernoff bound, $P(X\ge t)\le \min_{\lambda>0} e^{-\lambda t}M_X(\lambda)$, decays exponentially in $t$ rather than polynomially — which is why it, and its bounded-variable cousin Hoeffding, power learning-theory guarantees. Averaging helps too: for the sample mean of $n$ values, $\operatorname{Var}(\bar X_n)=\sigma^2/n$, so Chebyshev gives $P(|\bar X_n-\mu|\ge\varepsilon)\le \sigma^2/(n\varepsilon^2)\to 0$ — a clean proof of the Weak Law of Large Numbers. In the sim, the t slider sets the deviation you ask about and n sets the sample size for the LLN view.

(1) On "Bound Comparison," slide $t$ large and watch the true-probability curve plunge far below the Markov and Chebyshev lines — the bounds are guarantees, not estimates. (2) Switch to "LLN via Chebyshev" and raise $n$; see the bound fall like $1/n$. (3) Change the distribution between Exponential, Normal, and Bernoulli — the same bounds still hold, because they assume almost nothing about the distribution's shape.

By Chebyshev: $P(|\bar X_n-\mu|\ge\varepsilon)\le\frac{\sigma^2}{n\varepsilon^2}\to0$. This proves the Weak LLN with explicit convergence rate $O(1/n)$. The bound is simple but loose — true convergence is often exponentially fast.

For Exponential(1): Chebyshev gives $P(X>3)\le \sigma^2/(3-1)^2=0.25$. The optimal Chernoff bound is $te^{1-t}=3e^{-2}\approx0.406$, while the true value is $e^{-3}\approx0.050$. At this moderate $t$ Chebyshev is actually the tighter of the two; what makes Chernoff valuable is its exponential decay rate — for large $t$ it eventually overtakes every polynomial (variance-based) bound, which is exactly why Chernoff and Hoeffding bounds underpin machine-learning generalization guarantees.