But do you have to be sure you have a duplicate, or do you only care if there can be a duplicate. To be sure that you have two people with the same birthday, you need 366 people (not counting leap year). For there to be a greater than 50% chance of having two people with the same birthday you only need 23 people. That's the birthday problem.
If you have 32 bits, you only need 77,163 values to have a greater than 50% chance of a duplicate. Try it out:
Random baseRandom = new Random(0);
int DuplicateIntegerTest(int interations)
{
Random r = new Random(baseRandom.Next());
int[] ints = new int[interations];
for (int i = 0; i < ints.Length; i++)
{
ints[i] = r.Next();
}
Array.Sort(ints);
for (int i = 1; i < ints.Length; i++)
{
if (ints[i] == ints[i - 1])
return 1;
}
return 0;
}
void DoTest()
{
baseRandom = new Random(0);
int count = 0;
int duplicates = 0;
for (int i = 0; i < 1000; i++)
{
count++;
duplicates += DuplicateIntegerTest(77163);
}
Console.WriteLine("{0} iterations had {1} with duplicates", count, duplicates);
}
1000 iterations had 737 with duplicates
Now 128 bits is a lot, so you are still talking a large number of items still giving you a low chance of collision. You would need the following number of records for the given odds using an approximation:
- 0.8 billion billion for a 1/1000 chance of a collision occurring
- 21.7 billion billion for 50% chance of a collision occurring
- 39.6 billion billion for 90% chance of a collision occurring
There are about 1E14 emails sent per year so it would be about 400,000 years at this level before you would have a 90% chance of having two with the same GUID, but that is a lot different than saying you need to run a computer 83 billion times the age of the universe or that the sun would go cold before finding a duplicate.
