It's not clear what you're asking. I see two ways to interpret your question.
Given a GUID g
, what is the probability of someone guessing it? Let's assume for simplicity that all 128 bits of a GUID are available. Then the probability of guessing g
is 2^-128
. That's small. Let's get some intuition around that. Let's assume that our attacker can generate one billion GUIDs per second. To have a 50% chance of guessing g
, our attacker would have to generate 2^127 GUIDs. At a rate of one billion per second, it would take 5391448762278159040348 years to generate 2^127 GUIDs.
We are generating a collection of guids. What is the likelihood of a collision? That is, what is the likelihood that we generate two guids with the same value? This is the birthday paradox. If you generate a sequence of n GUIDs randomly, then the probability of at least one collision is approximately p(n) = 1 - exp(-n^2 / 2 * 2^128)
(this is the birthday problem with the number of possible birthdays being 2^128).
n p(n)
2^30 1.69e-21
2^40 1.77e-15
2^50 1.86e-10
2^60 1.95e-03
So, even if you generate 2^60 GUIDs, the odds of a collision are extremely small. If you can generate one billion GUIDs per second, it would still take 36 years to have a 1.95e-03 chance of a collision.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…