Measure CPU speed by counting assembly instructions

Edit: My original example had a silly mistake. After fixing it I still get weird results, though.

In my naive attempt to measure my CPU speed the "brute-force" way, I made the program below:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#pragma comment(linker, "/entry:mainCRTStartup")
#pragma comment(linker, "/Subsystem:Console")

int mainCRTStartup()
{
    char buf[20];
    clock_t start, elapsed;
    unsigned long count = 0;
    start = clock();
    __asm
    {
        mov EAX, 0;
    _loop:
        add EAX, 3; // accounts for itself and next 2 instructions
        cmp EAX, 0xFFFFFFFF - 0x400;
        jb _loop;
        mov count, EAX;
    }
    elapsed = clock() - start;
    _gcvt(count * (long long)CLOCKS_PER_SEC / (elapsed * 1000000000.0), 3, buf);
    puts(buf);
}

Which disassembles into something like:

mainCRTStartup:
  push   ebp
  mov    ebp,esp
  sub    esp,28h
  mov    dword ptr [count],0
  call   dword ptr [_clock]
  mov    dword ptr [start],eax
  mov    eax,0

_loop:
  add    eax,03h
  cmp    eax,0FFFFFBFFh
  jb     _loop

  mov    dword ptr [count],eax
  call   dword ptr [_clock]
  sub    eax,dword ptr [start]

  ...    // call _gcvt, _puts, etc.

  mov    esp,ebp
  pop    ebp
  ret

Notice that the loop is 3 instructions, so the final value of eax should be the total number of instructions.

Why do I get 4.2 when I run this?