probability - Using random number generator to generate a number with a given expected value -

I came to the following question:

Using the rand () function, generate the expected value number Do k The options are:

1)

  int GetRandom (int k) {v = 0; While (rand () <1.0f / (float) a) v ++; Return vi; }   
 2)  
  int GetRandom (int k) {v = 0; While (rand ()   
 3)  
  int GetRandom (int k) {v = 0; While (rand ()> (1-1.0 f / (float) (k + 1)) v ++; Return vi; }   
 1) The correct answer seemed like. Investigating the result for specific values ​​of  k  indicates that this is not the case. (I set  k = 3 . In the image below, the frequency distribution of values ​​for  100000  tests can be seen  
 you want (2 ) A  k  with these samples.  
 A geometrical distribution indicates such an experiment:  
 
 A certain event will result in a result Along with that which happens either 0 or 1  
 The result of an incident is 1 probability with  p  and 0 What is the index of the first event with the result of  1-p   
 1 with the possibility?   
 So if  X ~ G (p) , where  X  is a random variable and the possibility of  p  is above, then represents  X  "What is the index of the first event with the result of 1?" Hopefully  e [x] = 1 / p .  
 Looking at this information, it should be clear now That of the following with the random variable  X  with  p = 1 / k  (and (2a) Equivalent)).  
  int sample (int k) {int v = 1; While (true) {// the result is true with probability p = 1 / k bool result = rand () & lt; 1 / (double) K; If (results) return v; Other V ++; }}   
 Keep in mind that seeing peak (mode) and distributing is not the same thing as the peak of geometrical distribution will always be at 1!