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This variation of the birthday problem is interesting because there is not a unique solution for the total number of people m + n. For example, the usual 50% probability value is realized for both a 32-member group of 16 men and 16 women and a 49-member group of 43 women and 6 men.
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.
The birthday attack can be modeled as a variation of the balls and bins problem. In this problem: Balls represent inputs to the hash function. Bins represent possible outputs of the hash function (hash values). A collision occurs when two or more balls land in the same bin (i.e., two inputs produce the same hash output).
A naive application of the even-odd rule gives (,) = = () ()where P(m,n) is the probability of m people having all of n possible birthdays. At least for P(4,7) this formula gives the same answer as above, 525/1024 = 8400/16384, so I'm fairly confident it's right.
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The birthday problem asks, for a set of n randomly chosen people, what is the probability that some pair of them will have the same birthday? The problem itself is mainly concerned with counterintuitive probabilities, but we can also tell by the pigeonhole principle that among 367 people, there is at least one pair of people who share the same ...