Best way to calculate if there is a 1/4 chance something will happen in C++?

Question

Hi, I was wondering if there is a smart way to find out

There is a 1/4 chance something happens.

I know we can do this with rand() % 4 and checking if it is equal to 0, but is there a way without using rand()? In c++, thanks.

Answer 1

You could write your own rand. ( dont do it ).
You could grab the tickcount. ( dont do it too often ).
You could just count, and every fourth call return true.

You should probably just call rand().

Answer 2

Umm... write your own rand()? You will need some kind of random function!

Answer 3

Why not use rand()? If you are concerned about "true" randomness vs. pseudo randomness, you can try using physical sources of random bits. Much more complicated, and usually unnecessary.

Answer 4

If you mean you want to avoid the inherent crappiness of many rand() implementations, you should probably look into the Boost Random library, which has several high-quality pRNGs (pseudo-random number generators), and many ways to control the output. This library is also present in slightly modified form in std::tr1.

Answer 5

rand() < RAND_MAX/n;

pick a better rand() than C's if you don't like C's standard rand().

Answer 6

Never ever use % for truncating a PRNG value into a range. Most PRNGs have relatively non-random lower order bits.

For your case, use a division (RAND_MAX / n) like BCS suggests.

Answer 7

You could use another type of RNG such as the Mersenne twister which has better overall entropy. I also hear good thing about Multuply with Carry RNGs.

Answer 8

Try:

static int r = 0;
: : :
if ((r = (r+1)%4) == 0) {
    // do something.
}

Then you'll find it gives you a perfect 25% probability of something happening (assuming you execute the if-statement a multiple of four times.

</humor>

Answer 9

I don't know much C++, so I might be wrong. But it seems rand() return a value between 0 and RAND_MAX. So maybe you could do something like this:

double odds = .25;

if(rand() <= RAND_MAX * odds) {
    // there should be .25 chance of entering this condition
}

PS: Maybe this requires some casting.

Answer 10

4 is a special case. You can assume that your PRNG has got 50% chances of outputting an even number, which is the case - I think - for the LCG of the libc (rand). The probability of outputting an even number twice is therefore 25%.

Therefore...

bool rand_afourth(void)
{
    return !!((rand() & 1) & (rand() & 1));
}

And now for the pedantic...

What you want to do is to have an uniform random generated, but restricted to a certain range, in this case an entropy of 4. If your PRNG has, say, an entropy of 32-bit, you cannot be certain that computing the output mod 4 will work as expected. This require a bit more work.

Fortunately, this work has already been implemented in the boost library.

boost::uniform_int<> aFourth(1,4)

And you would for example say "ok" everytime you get 1 (or 2, 3, 4, as you fancy).

But you may not want to use the boost library. Then, simply look at the code of uniform_int and reproduce the behaviour. Talents imitate, geniuses steal. ;)