Finding an element in a sorted array with at most three queries to larger elements
If you have only one life only safe way is to check every element starting from minimal. It's $O(n)$

If you have two lives and limited with $k + 1$ comparisons the minimal element of array you can check first has index $k$, so we still have $k$ comparisons to check all previous elements one by one, if no lives was lost than next element to check has index $k + (k - 1)$ so total numbers of elements we can check is that way $n = k + (k-1) + (k-2) + ... + 1 = \frac{k(k+1)}{2}$ So if you have two lives you can find target element for ($k = \frac{1 + \sqrt{1 + 8n}}{2}$) $+1$ comparisons. 
It's $O(\sqrt n)$

So if we have three lives and $k + 2$ comparisons first we should check element of index $\frac{(k + 1)(k)}{2}$ than if we will lost life we will be able to check all earlier elements for $k + 1$ comparisons. So in full analogy $n = \frac{(k+1)(k)}{2} +   \frac{(k)(k-1)}{2} +  \frac{(k-1)(k-2)}{2} + \frac{(k-2)(k-3)}{2} +... + \frac{3\cdot2}{2} + \frac{2\cdot1}{2} = \frac{1}{6}(k+2)(k+1)(k)$

Therefore $n&gt; \frac{1}{6}(k)^3$ and $k &lt; \sqrt[3]{6n}$

So we can solve this for $O(\sqrt[3]{n})$

This is variation of quite known skyscraper and glass or crystal balls puzzle, though I may be didn't find correct name for it in English.