Map a random number to pi

LabVIEW, 53 LabVIEW Primitives

I match Strings and put the number into an "empty" x.xxx string and remove the number from pi so it doesn´t show up again.

the random number and the single chars here are kinda visible, is that all right or do i have to redo the recording?

Mathematica, 105 or 147 characters

If random number "between 0 and 1" means 0 <= random <= 1, i.e. includes 0 & 1.

StringReplace[ToString@InputForm@N@Pi,
Thread[ToString/@Complement[Range@9,RandomInteger[{0,9},15]]->"x"]]

(105 characters)

Otherwise, taking random number "between 0 and 1" to mean 0 < random < 1.

Loop to get 15 random integers, not all zero. Select complement from 0 to 9 range, i.e. those numbers from 0 to 9 not in the random list. Convert those integers to strings and replace the matching characters in a pi string.

(147 characters)

While[True,r=RandomInteger[{0,9},15];
If[Union@r!={0},Break[]]];
StringReplace[ToString@InputForm@N@Pi,
Thread[ToString/@Complement[Range@9,r]->"x"]]

3.1x15x265358x7x3

Random digits:-

FromDigits[r]

820307536180783

Pyth, 25 bytes

 u&p?}HGH\x.-GH`.n0<`O017

Try it online: Demonstration or Test showing the random number

Explanation:

 u&p?}HGH\x.-GH`.n0<`O017  
                .n0         the constant pi
               `            convert it into a string
                     O0     random number in the range [0.0, 1.0)
                    `       convert to string
                   <   17   only use the first 17 chars (zero, point and 15 digits)
 u                          for each char H in the pi-string:
    ?}HGH\x                    if H in G (the random number string) then H else "x"
   p                           print this char without newline
  &                            and
           .-GH                remove the digit H once from G
<space>                     suppress the output (u returns the unused digits in G)