Implement rand3() using rand2()

Given a function rand2() that returns 0 or 1 with equal probability, implement rand3() using rand2() that returns 0, 1 or 2 with equal probability. Minimize the number of calls to rand2() method. Also, use of any other library function and floating point arithmetic are not allowed.
 

The idea is to use expression 2 * rand2() + rand2(). It returns 0, 1, 2, 3 with equal probability. To make it return 0, 1, 2 with equal probability, we eliminate the undesired event 3.
Below is the implementation of above idea – 
 

Output : 
 

2111011101112002111002020210112022022022211100100121202021102100010200121121210122011022111020

Another Solution - 
If x = rand2() and y = rand2(), x + y will return 0 and 2 with 25% probability and 1 with 50% probability. To make probability of 1 equal to that of 0 and 2 i.e. 25%, we eliminate one undesired event that's resulting in x + y = 1 i.e. either (x = 1, y = 0) or (x = 0, y = 1). 
 

int rand3()
{
 int x, y;

 do {
 x = rand2();
 y = rand2();
 } while (x == 0 && y == 1);

 return x + y;
}

Please note above solutions will produce different results every time we run them.