Suppose that there are $n$ randomly ordered numbers. I would like to pick $m$ of them such that I have picked the largest number with a probability of $p$
As a numerical example; the number from 1 to 100 are ordered randomly in a set $S = \{s_1, s_2, \dots, s_n\}$. Suppose that there exists a set $Q \subseteq S$ in which the number $100$ exists by $90\%$ probability. What is the cardinality of $Q$?
If $Q = S$, then the probability is $100\%$. How can I approach further?
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You can think about this way: suppose there is a sequence of numbers $(s_1; s_2; \dots ; s_n)$, and say you pick first $p$ numbers. What is the probability, that the greatest number is in a set of numbers you picked? In other words what is the probability, that one of $(s_1; \dots; s_p)$ is the greatest number from the whole set?
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Without loss of generality, suppose the maximum is $s_1$. By taking uniformly at random a subset $Q$ of $m$ elements, the probability that $s_1\in Q$ will be $$\frac{\binom{n-1}{m-1}}{\binom{n}{m}} = \frac{m}{n}.$$ Indeed, the denominator is the number of all subsets of $m$ elements out of $n$; while the numerator is the number of subsets containing $s_1$, and $m-1$ other elements out of the $n-1$ (other than $s_1$) elements remaining.
To get a probability (at least) $0.9$, you then need to choose $m$ such that $\frac{m}{n} \geq 0.9$.
Make sure that $\frac mn\ge p$, i.e., let $m=\lceil np\rceil$. Indeed, whatever subset of size $m$ you pick, the probability that a specific (e.g., the largest) element is in that subset is $\frac mn$.