For a randomly chosen real number x on the interval (0,10) find the exact probability of each:

(1) That x and 2^x have the same first digit

(2) That x and x² have the same first digit

(3) That x² and 2^x have the same first digit.

(4) That x, x² and 2^x all have the same first digit.

First digit refers to the first non-zero digit of the number written in decimal form.

Corrections made below, in light of Jer's correction (comment 11) are made in bold:

x^2 and 2^x:

Above 1, this isn't too hard:

2 matches 2 from sqrt(2) to log(3)/log(2), the latter aka lb(3)
3 matches 3 from sqrt(3) to 2
4 matches 4 from 2 to sqrt(5)
5 matches 5 from lb(5) to sqrt(6)
6 matches 6 from lb(6) to sqrt(7)
7 matches 7 from lb(7) to sqrt(8)
8 has probability zero of matching 8 (3^2 = 8.9999... & 2^3 = 8 -- only 1 point)
1 matches 1 from lb(10) to lb(20)
2 matches 2 again from sqrt(20) to lb(30)
9 matches 9 from lb(900) to lb(1000)

Below 1, as before, 2^x begins with a digit 1, so the question becomes What fraction of the squares begin with a digit 1?

Out of the range .1 to .99..., those x's from .1 to sqrt(2)/10 have squares beginning with a 1. Thus the probability within this range is (sqrt(2)/10 - .1) / .9.

Also, from sqrt(.1) to sqrt(.2) the squares begin with a 1. This adds to the total probability within the 0 to 1 range, making it (sqrt(2)/10 - .1 + sqrt(.2) - sqrt(.1)) / .9.

 But this same probability also applies in each of the infinitely many orders of magnitude for x, and so (sqrt(2)/10 - .1 + sqrt(.2) - sqrt(.1)) / .9 is the correct probability for all of the 0 to .999999... range.

We thus get

((sqrt(2)/10 - .1 + sqrt(.2)) / .9 +lb(3) + 2 + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + lb(20) +lb(30) + lb(1000) - sqrt(2) - sqrt(.1) - sqrt(3) - 2 - lb(5) - lb(6) - lb(7) - lb(10) - sqrt(20) - lb(900)) / 10

= ((sqrt(2)/10 - .1 + sqrt(.2)) / .9 +lb(3) + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + lb(20) +lb(30) + lb(1000) - sqrt(2) - sqrt(.1) - sqrt(3) - lb(5) - lb(6) - lb(7) - lb(10) - sqrt(20) - lb(900)) / 10

=

((sqrt(2)/10 - .1) / .9 + sqrt(.2) +log(3)/log(2) + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + log(20)/log(2) +log(30)/log(2) + log(1000)/log(2) - sqrt(2)  - sqrt(.1) - sqrt(3) - log(5)/log(2) - log(6)/log(2) - log(7)/log(2) - log(10)/log(2) - sqrt(20) - log(900)/log(2)) / 10

~= .264795606189634 rather than .251697023241324

This now agrees with the simulation, to a reasonable statistical difference.

I suspect this is low for some reason, as a simulation of a million trials finds 267079 matches:

DEFDBL A-Z
RANDOMIZE TIMER

PRINT (1 / 10 + LOG(60) / LOG(2) - LOG(50) / LOG(2) + LOG(70) / LOG(2) - 6 + LOG(1000) / LOG(2) - LOG(900) / LOG(2)) / 10
FOR tr = 1 TO 1000000
   x = 10 * RND(1)
   y = 2 ^ x
   s1$ = STR$(x * x): s2$ = STR$(y)
   FOR i = 1 TO LEN(s1$)
     IF INSTR("123456789", MID$(s1$, i, 1)) THEN m1$ = MID$(s1$, i, 1): EXIT FOR
   NEXT
   FOR i = 1 TO LEN(s2$)
     IF INSTR("123456789", MID$(s2$, i, 1)) THEN m2$ = MID$(s2$, i, 1): EXIT FOR
   NEXT
   IF m1$ = m2$ THEN ct = ct + 1
   PRINT ct; tr,
   PRINT USING "#.#######"; ct / tr
NEXT tr

Edited on March 30, 2011, 2:45 pm

Posted by Charlie on 2011-03-29 22:27:42