CareerCup

def rand5():
    return ((3 * (rand3() - 1) + rand(3)) % 5) + 1

The initial part -

(3 * (rand3() - 1) + rand(3)

- generates numbers uniformly from 0-9. Mod by 5 now means the numbers are uniform from 0-4. Add 1 and now they're uniform 1-5.

Maximum and average number of trials is 1.

this wont work:
rand5(){
return 2*rand3()%5;
}

number 3,5 will never show up

expected number of trials is 2.4

think bit manipulation

@win : Can you elaborate ??

rand5(){  return (rand3()+rand3())%(3+3); }

rand5(){
return 2*rand3()%5;
}

i think this should work...