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Most background checks are generated automatically from data repositories maintained by private companies and state and federal governments. When people say public records, they are usually referring to civil issues such as lawsuits, traffic tickets, vital records such as marriage, divorce, birth and death , bankruptcies and the like. When people say background check, they usually have in mind criminal records, warrant searches, arrest records and other criminal information.
Believe it or not, almost all of this information is public record by law, and is available to anyone who is willing to make the effort to search for it. Some jurisdictions make it incredibly simple, and have dedicated web portals that allow you to search. Others are stuck in the paper age, and often require an on site search in the county courthouse. Should I search myself or use an online background check provider?
Online background check services really shine in two ways. They spend a fortune sucking up all the data available, something that might take you years or you might not even be able to accomplish with a limited budget at all. We try to be unbiased about these things.
How To Find out Someone’s Birthday
Every provider has its strengths and weaknesses, depending on how they collect their data and what kind of algorithms they use. Free background check - does such a thing really exist? One could also use this chart to determine the minimum hash size required given upper bounds on the hashes and probability of error , or the probability of collision for fixed number of hashes and probability of error.
The argument below is adapted from an argument of Paul Halmos. This yields. Therefore, the expression above is not only an approximation, but also an upper bound of p n. The inequality. Solving for n gives. Now, ln 2 is approximately Therefore, 23 people suffice. Mathis cited above. This derivation only shows that at most 23 people are needed to ensure a birthday match with even chance; it leaves open the possibility that n is 22 or less could also work. In other words, n d is the minimal integer n such that.
The classical birthday problem thus corresponds to determining n The first 99 values of n d are given here:. A number of bounds and formulas for n d have been published. In general, it follows from these bounds that n d always equals either. The formula. Conversely, if n p ; d denotes the number of random integers drawn from [1, d ] to obtain a probability p that at least two numbers are the same, then.
This is exploited by birthday attacks on cryptographic hash functions and is the reason why a small number of collisions in a hash table are, for all practical purposes, inevitable. The theory behind the birthday problem was used by Zoe Schnabel  under the name of capture-recapture statistics to estimate the size of fish population in lakes.
The basic problem considers all trials to be of one "type". The birthday problem has been generalized to consider an arbitrary number of types. Shared birthdays between two men or two women do not count. The probability of no shared birthdays here is.
A related question is, as people enter a room one at a time, which one is most likely to be the first to have the same birthday as someone already in the room? The answer is 20—if there is a prize for first match, the best position in line is 20th. In the birthday problem, neither of the two people is chosen in advance. By contrast, the probability q n that someone in a room of n other people has the same birthday as a particular person for example, you is given by. Another generalization is to ask for the probability of finding at least one pair in a group of n people with birthdays within k calendar days of each other, if there are d equally likely birthdays.
Thus in a group of just seven random people, it is more likely than not that two of them will have a birthday within a week of each other.click
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The expected total number of times a selection will repeat a previous selection as n such integers are chosen equals . In an alternative formulation of the birthday problem, one asks the average number of people required to find a pair with the same birthday. If we consider the probability function Pr[ n people have at least one shared birthday], this average is determining the mean of the distribution, as opposed to the customary formulation, which asks for the median.
The problem is relevant to several hashing algorithms analyzed by Donald Knuth in his book The Art of Computer Programming. An analysis using indicator random variables can provide a simpler but approximate analysis of this problem. An informal demonstration of the problem can be made from the list of Prime Ministers of Australia , of which there have been 29 as of [update] , in which Paul Keating , the 24th prime minister, and Edmund Barton , the first prime minister, share the same birthday, 18 January.
An analysis of the official squad lists suggested that 16 squads had pairs of players sharing birthdays, and of these 5 squads had two pairs: Argentina, France, Iran, South Korea and Switzerland each had two pairs, and Australia, Bosnia and Herzegovina, Brazil, Cameroon, Colombia, Honduras, Netherlands, Nigeria, Russia, Spain and USA each with one pair. Voracek, Tran and Formann showed that the majority of people markedly overestimate the number of people that is necessary to achieve a given probability of people having the same birthday, and markedly underestimate the probability of people having the same birthday when a specific sample size is given.
The reverse problem is to find, for a fixed probability p , the greatest n for which the probability p n is smaller than the given p , or the smallest n for which the probability p n is greater than the given p. Some values falling outside the bounds have been colored to show that the approximation is not always exact. A related problem is the partition problem , a variant of the knapsack problem from operations research. Some weights are put on a balance scale ; each weight is an integer number of grams randomly chosen between one gram and one million grams one tonne.
The question is whether one can usually that is, with probability close to 1 transfer the weights between the left and right arms to balance the scale. In case the sum of all the weights is an odd number of grams, a discrepancy of one gram is allowed. If there are only two or three weights, the answer is very clearly no; although there are some combinations which work, the majority of randomly selected combinations of three weights do not.
If there are very many weights, the answer is clearly yes. The question is, how many are just sufficient?