First let's consider a fixed $t$.
After we have picked $t$ numbers, we will almost surely have picked $t$ different real numbers, so we can assume that this is always the case without disturbing any probabilities or expectations.
Now, instead of picking $t$ numbers one by one, we could start by picking an unordered set of $t$ numbers, and then afterwards decide a random order to present them in.
Since the $t$ picked numbers are being presented in a random order (and it ought to be clear that no order is more probable than any other), it is easy to see that the $t$th number is a new maximum with probability $\frac1{t}$. On the other hand given whatever the $t$th number is, the first $t-1$ numbers are also presented in a uniformly random order, so the $(t-1)$th number was a new maximum with probability $\frac 1{t-1}$, independently of whether the $t$th one is.
What this argument shows is that the $k$th number is a new maximum with probability $\frac 1k$, and that the new-maximumness at different positions in the sequence are all independent events.
So the expectation of $n(t)$ is simply $\sum_{k=1}^t \frac 1k$. For large $t$ this sum approaches $\gamma+\log t$, where $\gamma$ is a constant known as the Euler-Mascheroni constant.
Finally un-fixing $t$, your limit ought to be (insert handwavy appeal to the law of large numbers here):
$$\lim_{t\to \infty} \sqrt[\gamma+\log t]{t} = \lim_{t\to\infty} t^{\frac1{\gamma+\log t}} =
\lim_{t\to\infty} e^{\frac{\log t}{\gamma+\log t}} = \lim_{t\to\infty} e^1 = e$$
There are $6$ equally-probable orders:
- $a_1<a_2<a_3$
- $a_1<a_3<a_2$
- $a_2<a_1<a_3$
- $a_2<a_3<a_1$
- $a_3<a_1<a_2$
- $a_3<a_2<a_1$
In $3$ of them, the first number is smaller than the second number:
- $a_1<a_2<a_3$
- $a_1<a_3<a_2$
- $a_3<a_1<a_2$
In $1$ of them, the third number lies between the first number and the the second number:
Hence the probability is $\dfrac13$.
Best Answer
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $\frac{1}{3}$ .. so therefore the probability that C is between A and B is also $\frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?