You suffer much of the criticism of people and need some time to feel quite yourself after such resentment. You have the qualities of a clairvoyant due to your sensitiveness and great intuition. Buy the way, your intuition is so strong that if you develop it and learn how to use it right you would become a good seer. As well, combination of your intuition and powerful inner energy lets you to become a healer.

Other people follow your ideas with readiness as they feel your inner power and they understand that very often your words become proved by some events. People love to communicate with you as you understand their thoughts and feel their emotions. It helps you in your daily life as you understand what may motivate other people for doing something and adopt the ideas and plans which you offer. You inspire them and they are ready to come to you with their happiness, sadness, success and failures.

You make them calm and ensured in principles which you consider right. Of cause, all these make you a good leader. You and your life becomes a good example for others. The opinion of others is important for you even if it doesn't influence the result of your activity. You are looking for acclaiming.

You are a good friend and a good colleague. You find mutual language with people easily and inspire them by devoting yourself to your aim. Therefore, its probability p n is. The following table shows the probability for some other values of n this table ignores the existence of leap years, as described above, as well as assuming that each birthday is equally likely :. Leap years.

## What Does Your Birthday Say About You? 11th - 17th February

The first expression derived for p n can be approximated as. According to the approximation, the same approach can be applied to any number of "people" and "days". The probability of no two people sharing the same birthday can be approximated by assuming that these events are independent and hence by multiplying their probability together. Since this is the probability of no one having the same birthday, then the probability of someone sharing a birthday is.

Applying the Poisson approximation for the binomial on the group of 23 people,. A good rule of thumb which can be used for mental calculation is the relation. In these equations, m is the number of days in a year. The lighter fields in this table show the number of hashes needed to achieve the given probability of collision column given a hash space of a certain size in bits row.

Using the birthday analogy: the "hash space size" resembles the "available days", the "probability of collision" resembles the "probability of shared birthday", and the "required number of hashed elements" resembles the "required number of people in a group". 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 [12] 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.

## Birthday Number 11

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. The expected total number of times a selection will repeat a previous selection as n such integers are chosen equals [15]. 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.

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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? That is, what is the number of weights such that it is equally likely for it to be possible to balance them as it is to be impossible?

Often, people's intuition is that the answer is above Most people's intuition is that it is in the thousands or tens of thousands, while others feel it should at least be in the hundreds. The correct answer is The reason is that the correct comparison is to the number of partitions of the weights into left and right. Arthur C.