TITLE: How likely is it for a randomly picked number to be larger than all previously chosen numbers?
QUESTION [13 upvotes]: Suppose we pick a uniformly distributed number on the range [a,b]. Then we continue to pick more numbers on the same range. Let n(t) be the number of times we have found a number bigger than any previously found, after sampling t total numbers. The initial number picked is not counted.
Solve:
$$\lim_{t\to\infty}\sqrt[n(t)]{t}$$
For reference, the problem I'm trying to solve is the geometric mean of the amount of time it takes to find the next bigger number, relative to the amount of time it took for the previous bigger number. So say it finds a new best value at the 4th try, 11th try, and 29th try, I get:
$$\sqrt[3]{4*\dfrac{11}{4}*\dfrac{29}{11}}$$
which simplifies into the top equation.
Experiments seem to indicate the amount of time it takes to find the next number multiples by about 3 for each number found, but I'm curious if there might be tools to solve this analytically. Interestingly, the value seems to be the same even if I take the randomly chosen number and plug it into a function (i.e. I'm checking to see if f(x) is greater than any previously seen f(x) for random values of x).
Related questions:

Is there any way to guess at the probability of finding a new biggest number?
Can the geometric mean be shown to be a constant across all (or some) functions? Intuitively I feel like it should be.

I don't have too extensive a background in math, so I'm hoping this isn't an unsolved problem.

REPLY [7 votes]: Here is an outline of a proof of the almost sure convergence of $t^{\frac{1}{n(t)}}$ to $e$ (i.e. why the "handwavy appeal to the law of large numbers here" can be invoked). This is very far from trivial; to get an almost sure convergence you need to control how the process evolves, and how likely are exceptional values; a control on its average is not enough. The proof is not complete. I think that fixings the gaps in this would be very technical, without any gained insight. You may try to complete it if you are interested (but it is rather involved - I don't know if you know the methods used here, as you say that "[you] don't have too extensive a background in math").
To simplify the subject, I'll assume that we work with the uniform distribution on $[0,1]$, and that we look at the successive minima, and not the maxima. It changes nothing; as you had guessed, what works for a uniform distribution works for any non-atomic measure over $\mathbb{R}$ (just send it to the uniform measure via the distribution function).
Let $(X_n)$ be a sequence of random variables uniformly distributed in $[0,1]$. Let us define recursively two sequences of random variables $(Y_n)$ and $(\tau_n)$, with $X_0 = 1$, $\tau_0 = 0$, and:

$\tau_{n+1} = \inf \{ m > \tau_n : X_m < Y_n \}$,
$Y_{n+1} = X_{\tau_{n+1}}$.

In plain English, $(Y_n)$ is the sequence of minima and $\tau_n$ is the sequence of times these minima occur. If we have a new minimum $Y_n$, then the following minimum $Y_{n+1}$ will occurs when a point $X_m$ fall into the interval $[0, Y_n]$, which occurs with probability $Y_n$ at each step. Hence, 

$\tau_{n+1} - \tau_n$ has an exponential distribution of parameter $Y_n$,
$Y_{n+1}$ is uniformly distributed in $[0, Y_n]$.

Now, let me define $Z_n = \ln (Y_n)$. A short computation shows that:

$\mathbb{E} (Z_{n+1} | Z_n, \cdots, Z_0) = -1$;
$\text{Var} (Z_{n+1} | Z_n, \cdots, Z_0) = 1$.

This implies that the sequence $(Z_n)$ behaves likes a random walk with constant drift. In particular, for all $\varepsilon > 0$, almost surely, for large enough $n$,
$$-(1+\varepsilon) n \leq Z_n \leq -(1-\varepsilon) n,$$
or, in other words,
$$e^{-(1+\varepsilon) n} \leq Y_n \leq e^{-(1-\varepsilon) n}.$$
This is not a trivial result, but there is an abundant literature on the subject. Heuristically, $Y_{n+1}/Y_n$ is roughly $e^{-1}$, so $\tau_{n+2} - \tau_{n+1} \sim e (\tau_{n+1} - \tau_n)$. When you say that "Experiments seem to indicate the amount of time it takes to find the next number multiples by about $3$ for each number found", you are quite close to the truth. It multiplies by about $e$.
Anyway, for large enough $n$, we get $\mathbb{P} (\tau_{n+1} - \tau_n \leq e^{(1-2\varepsilon)n}) \leq 1-e^{-e^{-(1-\varepsilon)n} e^{(1-2\varepsilon)n}} \leq e^{- \varepsilon n}$ and $\mathbb{P} (\tau_{n+1} - \tau_n \geq e^{(1+2\varepsilon)n}) \leq e^{-e^{-(1+\varepsilon)n} e^{(1+2\varepsilon)n}} = e^{-e^{\varepsilon n}}$. The sequences $(e^{- \varepsilon n})$ and $(e^{-e^{\varepsilon n}})$ are both summable, so by Borel-Cantelli lemma, almost surely, for all large enough $n$,
$$e^{(1-2\varepsilon)n} \leq \tau_{n+1} - \tau_n \leq e^{(1+2\varepsilon)n}.$$
Now, we sum this inequality for $0 \leq n < N$. I'll skip a few technical details (I take slightly larger margins in the exponents so as to get rid of any annoying constant); you get that, almost surely, for all large enough $N$,
$$e^{(1-3\varepsilon)N} \leq \tau_N \leq e^{(1+3\varepsilon)N}.$$
Moreover, $\tau_N \leq C$ is equivalent to $n(C) \geq N$ and $\tau_N \geq C$ is equivalent to $n(C) \leq N$. I'll skip another round of technicalities, but, using the fact that the function $n$ is non-decreasing, you get that, almost surely, for large enough $t$,
$$\frac{\ln (t)}{1+4\varepsilon} \leq n(t) \leq \frac{\ln (t)}{1-4\varepsilon}.$$
This is equivalent to the fact that, almost surely, 
$$\lim_{t \to + \infty} \frac{n(t)}{\ln (t)} = 1,$$
thus $\lim_{t \to + \infty} t^{\frac{1}{n(t)}} = e$.