The probability of a random variable being larger than a sequence of random values Suppose we have a fixed, known, $n$, and each $x_1 \ldots x_n$ is a random number generated uniformly over $[0,1]$. What is the probability that $x_n$ is the largest value in the sequence?
 A: Uppercase $X$ denotes random variables, lowercase $x$ denotes realizations of random variables.  
If the realization of the sequence is known up to $n-1$, then we know what the value of the maximum is up to then, say $x_{(n-1)}$. Then we want the probability
$$\Pr\big(X_n > x_{(n-1)}) = 1-x_{(n-1)}$$
If we don't know  the actual realizations, then the probability we are after is the probability of the difference of two independent random variables:
$$\Pr\big(X_n > X_{(n-1)} \big) = \Pr\big(X_{(n-1)} -X_n \leq 0 \big)$$
(since we have continuous RVs the appearance of weak inequality to conform with standard expressions is neutral to the results).  
The distribution of $X_n$ is $U(0,1)$. $X_{(n-1)}$ is the maximum order statistic from a sample of $n-1$ independent $U(0,1)$ random variables so its distribution and probability density functions are
$$F_{X_{(n-1)}} (x_{(n-1)}) = x_{(n-1)}^{n-1},\;\;\; f_{X_{(n-1)}} (x_{(n-1)}) = (n-1)x_{(n-1)}^{n-2}$$
Deriving the distribution of their difference $Z = X_{(n-1)} -X_n$ requires some care. $Z$ ranges in $(-1,1)$ while its defining components are non-negative. We can instead opt for the specific probability we are after: in order for $Z\leq 0$ it must be the case that $X_{(n-1)} \leq  X_n$. The general formula for calculating such a probability is
$$P(Y < W) = \int_{S_w}\int_{\{y<w\}}f_{YW}(y,w) {\rm d}y{\rm d}w$$ 
which in our case becomes (decomposing the joint density into the product of the marginals due to independence)
$$\Pr(Z\leq 0) = \int_0^1\int_0^{x_n}(n-1)x_{(n-1)}^{n-2}\cdot 1 {\rm d}x_{(n-1)}{\rm d}x_n$$
$$= \int_0^1x_n^{n-1} {\rm d}x_n = \frac 1n$$
That this probability decreases with $n$ (and so with $n-1$) is intuitive: the more chances the sequence has of obtaining the theoretical maximum, the more probable it is that it will get closer and closer to it, and so the less probable is that the next value will be even bigger.  
Simulations verify the above result.
A: Not quite sure of the setup. Here are some answers.
If you observe $x_1, x_2, \ldots, x_{n-1}$, and wonder if the next observation will top the series, then the answer is $1-F(M)$, where $F$ is the cumulative distribution function and $M$ is the maximum of the first $n-1$ observations. 
On the other hand, if you only observe $x_n$ and wonder about the probability that this was the maximum of the series, then the answer is $$F(x_n)^{n-1}$$.
And on the third hand, if you observe nothing at all, but wonder about the probability that the maximum will occur on the last observation, then the probability is $1/n$.
A: disclaimer: I'm not a statistician or algebrist.
So I discovered that one has to work very hard to get a large number of samples that are sequentially larger from a uniform distribution.  Perhaps Chebychev speaks to this.
My result was this plot:

Here is the summary for a decent fit (the red line):
Call:
lm(formula = I(log10(y3)) ~ x)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.13084 -0.06066  0.01368  0.03426  0.11696 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.199568   0.056497  -3.532  0.00771 ** 
x            0.383633   0.009105  42.133 1.11e-10 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.0827 on 8 degrees of freedom
Multiple R-squared:  0.9955,    Adjusted R-squared:  0.995 
F-statistic:  1775 on 1 and 8 DF,  p-value: 1.11e-10

That means this is the relationship:
$ y = 10^{0.383\cdot x-0.2}$
And here is my code:
#my CV 132949

#housekeeping
rm(list = ls())

#I am going to draw a random number from a uniform distribution.
# I am then going to count how many other numbers I have to draw until one is bigger than it.
# I am going to do this 10 times in a row.
# I am going to repeat this 300 times

nbig <- 300
nsmall <- 10

nmax <- 300000

#pre-declare storage
data <- matrix(0,nrow=nsmall,ncol=nbig)

#big for
for (i in 1:nbig){

  print(i)

  #little for
  for (j in 1:nsmall){

    #if this is the first time
    #draw a random number 
    if (j==1){
      y <- 0.5
    }
    #if this is not the first time
    #update the random number
    if (j>1){
      y <- z
    }

    #initialize z and n
    z <- y
    n <- 0

    #draw random numbers until the next one is bigger
    # recording how long it takes
    while ((z <= y)&(n < nmax)) {
      z <- runif(1);
      n <- n+1;
    }


    #store in a variable
    if (n < nmax){
      data[j,i] <- n
    } else {
      #this is to retain the median value below
      data[j,i] <- 1e12
    }
  }
}

y3 <- (rowMeans(data))

for (i in 1:nsmall){
  y3[i] <- median(data[i,])
}

x <- 1:nsmall

plot(x,log10(y3),
     xlab=c("number of terms in sequence"),
     ylab=c("log_{10} median number new samples"))

est <- lm( I(log10(y3))~x )
summary(est)

lines(x,predict(lm( I(log10(y3))~x ) ),col="Red")

Comments: 
This was the median value, not an upper percentile.  If I was gambling then I shoot for the most likely outcome.  If I was building a business security approach, then I would like to envelope all bad things and would look at the 99th (or more) percentile and not the median.
