Why is Min-entropy significant in cryptography? Let $H_{min}$ = Min-entropy and $H$ = Shannon entropy. 

Unless you're gaming the system or deliberately trying to be obtuse, very rarely is $ \frac{H}  {H_{min}} &gt; 10$ from any conceivable physical source. To achieve a ratio of >10 in hardware/ real life would be quite a feat of engineering and tantamount to a joke/NSA conspiracy.  For  worldly context, lift engineering commonly uses a safety factor of 7.  If the lift gear fails, real people die. Since arbitrarily, $ \frac{H}{100} &lt;&lt; H_{min} $ in the overwhelming majority of cases, why is $H_{min}$ significant in cryptography?



I know the definitions.
In most cases, cryptography requires values to be uniformly random (in which case discussions of min-entropy are moot) or unpredictable (in which case min-entropy isn't sufficient --- though "conditional" min-entropy might be --- since in these contexts an adversary typically has multiple guesses).

The context where I see min-entropy play the greatest role is in randomness extraction. Since physical sources of randomness rarely produce uniformly random bits, we need a way to transform the output of some physical system (presumably containing some amount of entropy, however measured) into a uniformly random bit string. At that point, we can start to do cryptography.

The Leftover Hash Lemma tells us how to take an input $X$ and transform it into a value $f_S(X) \in \{0, 1\}^n$ that is "close" to uniform. In particular, it tells us how to construct $f$ such that $n \leq H_{\infty}(X) - 2 \log(1/\epsilon)$, where $\epsilon$ is the statistical distance between the uniform distribution and $f_S(X)$ (technically, the distance between $(S, f_S(X))$ and $(S, U)$, where $U$ and $S$ are uniform and $S$ is implicitly a fixed, public value).

We're not making any assumptions on $X$ except for it's min-entropy. This quite nice because in practice we can't precisely characterize the distributions used to provide inputs to RNGs (which vary from device to device in the case of HW RNGs, and are not known a priori in the case of, e.g., /dev/urandom). Of course, estimating $H_{\infty}(X)$ is its own bag of worms...