Why set the probability density for finding a falling stone as proportional to the time interval it spends at a given location? Today a friend of mine, who has started out with Griffiths, asked me a question about one of the examples (1.1) in Griffiths' QM book. The example basically says that a rock is dropped from a cliff, and as it falls the observer clicks a large number of photos. On each picture, he measures the distance fallen by the rock. We are required to find the average of all these distances.
Now, Griffiths calculates the probability of camera flashing in the interval $dt$, which is of course $dt/T$. Then he proceeds to write this as a function of $x$, and the rest is simple. The problem is, why do you need to start out writing the probability equation in terms of $t$. I know that you can write the probability density function in terms of $t$ and get the result you want, but why? I mean, why can't we start out with an $x$ expression? Something like $dx/H$ (which obviously doesn't work out correctly). What motivation does he have to start out with t?
In other words, could someone clearly explain what is going on in this problem? (now I seem to have gotten confused about it )
 A: Consider a random variable $x$ (Griffiths doesn't use this language. I only say 'random' variable to connect with language you hear in statistics courses). The average value $\langle x \rangle$ is given by $\langle x \rangle = \int x\rho(x)dx$ where $\rho$ is the notation Griffiths uses for probability density. A function of a random variable is also a random variable. It looks like you are asking whether $\langle f(x) \rangle = \int f(x)\rho(x)dx$ or $\langle f(x) \rangle = \int f\;\rho(f)df$. The answer, as you know, both are correct. The former is usually much easier than the latter. 
To rephrase the example problem: $x(t) = \frac{1}{2}gt^2$, what is $\langle x(t) \rangle$? This is what Griffiths is asking. Following the (first) formula, $\langle x(t) \rangle = \int_0^T x(t) \rho(t)dt = \int_0^T x(t) (1/T)dt$. And that's it, you're done. The probability density $\rho(t)$ is $1/T$ - it's a constant function. Or said in another way, the probability of choosing a $t$ (a picture) in the range $t$ to $t + dt$ is $(1/T)dt$ (this is just like saying that the amount of mass in a region $V$ to $V + dV$ is $dm = \rho_m dV$ where $\rho_m$ is mass density playing the role analogous to probability density). A constant probability density function (pdf) is the simplest one you can make. This is why you should use $t$. I don't know why Griffiths does the integration the other way because it takes more thinking power in my opinion. Griffiths uses the (second version) formula: $\langle x \rangle = \int_0^H x \rho(x) dx$, even though he knows that $x$ is a function of $t$. And notice how he got there. He got to this integral with $x$'s by going through $t$. The reason why you can't just start off with $x$'s is because it's not immediately clear what $\rho(x)$ is. In your question, you say that $\rho(x) = 1/H$. In other words, you say that every value of $x$ is equally likely as $\rho$ is a constant function. It seems like you figured out why this can't be the case (the rock spends more time at the top than at the bottom - so $\rho(x)$ can't be constant. In other words, the probability of getting a result in an interval $dx$ (which is $\rho(x)dx$) can't be the same across all intervals. Saying $(1/H)dx$ puts the probability of getting a result in an interval $dx$ the same value across all $dx$ intervals when we know that the probability has to be higher at intervals towards the top and lower at intervals towards the bottom). Just to be clear, $x$ in this problem means the distance from the top of the cliff. Each photo shows the distance from the top of the cliff. Short answer: you can use $x$'s, but $\rho(x)dx$ is not $(\text{const})dx = (1/H)dx$. I don't know what $\rho(x)$ is easily. It's much easier to use $t$'s because it has a simple straight forward probability density function $\rho(t) = 1/T$. 
Likewise, in early calculus courses, we learn that the average value of a function $f(x)$ is $\langle f(x) \rangle = \frac{1}{b-a}\int_a^b f(x)dx.$ This is because the probability density function of $x$ is simply the constant probability density function $1/(b-a)$. It's also true that $\langle f(x) \rangle = \int_{f(a)}^{f(b)} f \; \rho (f) df$. This is not very convenient. For instance, $\langle f = x^2 \rangle = \int_a^b x^2 \frac{1}{b-a}dx$. This is equivalent to, if you work it out, $\langle f = x^2 \rangle = \int_{a^2}^{b^2} f \frac{1}{2(b-a)\sqrt{f}}df$. All I did was rewrite $dx/(b-a)$ in terms of $df$ and $f$.
Side note: In statistics, we learn about many different probability density functions. Constant probability density function, gaussian pdf, gamma pdf, beta pdf. For a given real life situation/problem, we have to hope that our situation/problem matches one of these pdf's or hope that we can figure out the pdf from scratch (through our heads or through experimentation). Else we won't be able to find, say, averages and whatnot. I think the Schrodinger equation is pretty cool because no matter the situation (as long as you can formulate the problem), it will generate a pdf for you. You don't have to go matching problems to pdf's. For any situation, you write down it's Schrodinger equation and solve it for the pdf of that situation. 
A: I think that Griffiths is trying to get the reader to understand what is meant by a probability distribution function.  
The graph of distance (fallen) by the rock $x$ against time taken $t$ with the total height of fall $H$ which hakes time $T$ looks like this.  

From the graph you can see that the the rock spends more time falling less that $\frac H 2$ than falling more than $\frac H 2$.  
The photographs are taken at random with all times equally likely so you would expect the average distance to be less than $\frac H 2$. 

The easiest way to produce a solution is to define a probability distribution function $f(t)$ which depends on time and that will look like this.  

The most important thing to remember is that the probability distribution function is not the probability rather it is the area under the probability distribution function which is the probability.
So here the area under the graph from $t=0$ to $t=T$ is $\frac 1 T \times T=1$.
The probability of the rock being between time $t$ and $t+dt$ is $\frac 1 T \times dt=\frac {dt} {T}$. 
Now $x = \frac 12 gt^2$ so the average value of $x$ is $\displaystyle \int _0^T x\,f(t)\,dt = \int _0^T \frac 12gt^2\,\frac 1 T \,dt = \frac{\frac12gT^2}{3}= \frac H3$

However $f(t)$ is not the only probability distribution function which can be used in this example and Griffiths introduces $g(x)$ which depends on the distance the rock has fallen and he shows that $g(x) = \frac{1}{2\sqrt {Hx}}$

The area under the graph is the probability.
Evaluating the area from $x=0$ to $x=H$ gives a value of $1$.  
The area under the graph from $x$ to $x+dx$ is the probability of finding the rock between falling $x$ and falling $x+dx$.  
Note if the interval $dx$ is kept constant there is less probability of finding the rock between $y$ and $y+dx$ than between $x \,(<y)$ and $x+dx$.  
Put another way if you want the probability of finding the rock between $y$ and $y+dy$ to be the same as between $x \,(<y)$ and $x+dx$ you must have $dx<dy$.
Evaluation of $\displaystyle \int _0^H x\,g(x)\,dx$ gives $\dfrac H3$ as before.
A: To first answer your question in the comment (which other answers have clarified too) $x$ is the total distance from the starting point (viz. peak of the cliff) that the rock has fallen through, and not the distance traveled between successive snapshots.

Which probability distribution you assume depends on what is given in the statement of the problem. Griffiths speaks of the observer who stands on the ground and clicks many snapshots of the object while it is falling. It is perhaps much easier to think that the observer takes one snapshot randomly in time (Griffiths probably doesn't state the latter point this precisely, but this is what he means) during the fall of the rock. This means that if the rock takes time $T$ to fall from the cliff to the ground, then he clicks the snapshot at a random instant in the time interval $[0,T]$. While this is not the only assumption that can be made (see next paragraph), this is the most natural one. And given the problem statement, there is no reason for us to think (notice the epistemic viewpoint) that he prefers to click at one instant of time instead of another. Therefore the reasonable and unbiased thing for us to do would be to assign probability of clicking the snapshot uniformly over the entire time interval $[0,T]$.

However you can contrive a photographing machine/robot whose probability of clicking a snapshot is uniformly distributed over the total distance of fall instead of the total time of fall. It would of course have greater probability of clicking the snapshot when the object is closer to the top than to the bottom (but hey, that doesn't violate any laws of physics).
Or you can have more fun by bringing relativity into picture. Consider an observer who jumps off the cliff (say he is bungee jumping) exactly when the rock is also thrown off the cliff. Suppose this observer clicks a snapshot randomly in time as shown by his wristwatch. That is, if $\tau$ is his proper time recorded during his fall from cliff to ground, then probability is uniformly distributed over $[0,\tau]$. Since rate of passage of time for this observer is different from the ground-based observer and because this rate keeps changing (due to acceleration), this does not translate into uniform probability over $[0,T]$.

In summary, what probability distribution you assume depends on what information you have available, which information in your case comes from the problem statement.
