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Comparing p(n) = probability of a birthday match with q(n) = probability of matching your birthday. In the birthday problem, neither of the two people is chosen in advance. By contrast, the probability q(n) that at least one other person in a room of n other people has the same birthday as a particular person (for example, you) is given by
Another reason hash collisions are likely at some point in time stems from the idea of the birthday paradox in mathematics. This problem looks at the probability of a set of two randomly chosen people having the same birthday out of n number of people. [5] This idea has led to what has been called the birthday attack.
Probability of a random day of the year being your birthday (for all birthdays besides Feb. 29) 4×10 −3: Probability of being dealt a straight in poker 10 −2: Centi-(c) 1.8×10 −2: Probability of winning any prize in the UK National Lottery with one ticket in 2003 2.1×10 −2: Probability of being dealt a three of a kind in poker 2.3× ...
(1/365! is the probability that you take 365 people with distinct birthdays and, picking them one at a time, correctly pick them in birthday order). Let's work with smaller numbers: assume a 3-sided coin (it's more interesting than a two-sided, but the numbers are small).
With SUHA however, we can state that because of an assumed uniform hashing, each element has an equal probability of mapping to a slot. Since no particular slot should be favored over another, the 30 elements should hash into the 10 slots uniformly. This will produce a hash table with, on average, 10 chains each of length 3
A birthday attack is a bruteforce collision attack that exploits the mathematics behind the birthday problem in probability theory. This attack can be used to abuse communication between two or more parties. The attack depends on the higher likelihood of collisions found between random attack attempts and a fixed degree of permutations ...
The natural extension to 64-bit integers is by use of a table of 2 8 ×8 64-bit random numbers. This kind of function has some nice theoretical properties, one of which is called 3-tuple independence , meaning that every 3-tuple of keys is equally likely to be mapped to any 3-tuple of hash values.
This probability can be computed precisely based on analysis of the birthday problem. [26] For example, the number of random version-4 UUIDs which need to be generated in order to have a 50% probability of at least one collision is 2.71 quintillion, computed as follows: [27]