Probability of arriving at a number after a certain number of tries with the following criteria I'm not sure what field of math this is, I'm just interested in mathematics in daily life, here's a question that kept me thinking:
Lets say for a range between $1$ to $200$, I randomly pick a number, for example I pick $63$, then all numbers lesser than or equals to $63$ will be discarded, next iteration I will be picking from $64$ to $200$, what is the probability that I will arrive at $200$ in some number of picks?
E.g. What's the probability I will pick $200$ after $5$ picks
 A: This is a nice question. Let $X=\{\text{number of picks needed to get to }200\}$. First question is how many possible outcomes are there? 
Well, if we play the game until we get to $200$, then, we could either pick $200$ on the first try ($1$ outcome), or pick $1$ number less than $200$ and then pick $200$ ($199$ outcomes), et cetera. 
If you pick need to $k$ numbers $$1\leq n_1< n_2<\cdots < n_k<200$$ before you pick $200$, this is the same as choosing $k$ distinct numbers less than or equal to $199$, so there are ${199\choose k}$ possible games that end after $k+1$ picks. Since we have to win in at most $200$ turns, the total number of games is 
$$ 1+199+\cdots+{199\choose k}+\cdots +{199\choose 198}+1=2^{199},$$
where I used the binomial theorem. Then the probability that you arrive at $200$ in $k$ picks is the number of such outcomes divided by the total number of outcomes, so that
$$P(X=k) =\frac{{199\choose k-1}}{2^{199}}.$$
In particular, the probability that you arrive at $200$ in exactly $5$ picks is 
$$P(X=5) = \frac{{199\choose 4}}{2^{199}}\sim 7.88\times 10^{-53}.$$
