Difference between Excel's RAND(), RAND()*RAND(), etc  I plotted below the standarized results of: 


*

*RAND()

*RAND() * RAND() 

*...

*RAND() * RAND()  * RAND() * RAND()  * RAND() * RAND()


It seems that the results are getting to zero, is that because you're multiplying a bunch of numbers close to zero together, or is there another explanation?

EDIT
There is no real motivation behind this. Working on a spreadsheet earlier today, I was in a situation where i was multiplying a constant * Rand() * Rand() .. each rand() corresponded to a different variable (column), so I wanted to find out what goes on if you multiply a bunch of random numbers together.
 A: Standardization is good, but it's not the right standardization for this situation.  It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent exponentiation).  Because the different calls to RAND() are supposed to be independent, those logarithms are still independently distributed.  As a simple calculation shows, their common distribution actually has a mean and variance.  (In fact, its negative is an exponential distribution.)  The Central Limit Theorem applies.  It says that the logs, suitably standardized, converge to a normal distribution.  We conclude that these products--standardized to have a constant geometric mean and constant geometric variance--are converging to the exponential of a normally distributed variable: that is, a lognormal distribution.
A: "In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http://www.techonthenet.com/excel/formulas/rand.php
Because RAND() is always less than one and greater than zero, multiplying it by itself will make it smaller. As you do that over and over, you will get closer to zero. If you want something that gives you a random number between 0 and a, you can do a*RAND() instead.
A: I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ for $u \in (0,1)$ and 1 otherwise.
When you multiply different iid realizations of the random draws you are essentially computing the following:
$Y = U^n$ where $n$ is the number of times you are multiplying the random draws.
Thus, the corresponding cdf is:
$F(y) = P(Y \le y)$
i.e., 
$F(y) = P(U^n \le y)$
i.e.,
$F(y) = P(U \le y^{1/n})$
i.e., 
$F(y) = y^{1/n}$ for $y \in (0,1)$ and 1 otherwise.
The above cdf of $Y$ converges to a dirac-delta function on $Y=0$ as $n \rightarrow \infty$. Thus, $E(y) \rightarrow 0$ as $n \rightarrow \infty$. 
The above convergence is also related to first-order stochastic dominance in the following sense: 
Suppose that $n_1 > n_2$. Then, it is the case that:
$F(y|n_1) \ge F(y|n_2)$
Intuitively, the above result states that: In visual terms as $n$ increases the cdf of $Y$ shifts to the right. This happens because the pdf associated with $Y$ starts concentrating at the lower end of the interval $[0,1]$ and asymptotically all the pdf concentrates at 0 which explains the observed behavior.
General Case
@whuber's comment to this answer gives the solution when $Y$ is the product of $n$ independent, different random variables drawn from [0,1].
A: There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be something close to (0.5^6), that is, 0.015625.
Be careful using Excel's RAND() function, though. It's not the best random number generator in the world.
