Top Special Offer! Check discount

Get 13% off your first order - useTopStart13discount code now!

251 views
3 pages ~ 804 words

Get a Custom Essay Writer Just For You!

**Experts** in this subject field are ready to write an original essay following your instructions to the dot!

In probability theory, the birthday paradox, is based on the underlying probability principle that in a set of n randomly selected individuals, a pair from the group would have a 50 percent chance of having the same birthday (Suzuki, Tonien, Kurosawa, & Toyota, 2008). Through pigeonhole principle, the probability stretches to 100% only if the sum of people is 367. The only explanation is because there are only 366 possible birthdays, inclusive of February 29. However, with 70% people a 99.9% probability would be achieved, while for 23 people only 50% chance. The birthday paradox conclusions are founded on the theory that each day of the year excluding a leap year is equally probable for a birthday. In a group of 23 people, there are a total of 253 combinations with each combination having the same odd, 99.72% of not matching. Therefore, a random group of people for instance 30, there is a 50% chance of two of the individuals sharing a birthday.

Calculating the Birthday Paradox

Making the assumption that the year always has a 365 days, and that one has an equal chance of having his or her birthday on any day of the year, despite the fact that some birthdays may be to some extent more likely unlike others, the math would be simplified without affecting the result. Determining the likelihood of two people having the same birthday would require understanding that the first person could have any birthday. Based on this understanding, the first person has a total of 365 possible birthdays owing to 365 days. Therefore, the chance of the first person having the ‘right’ birthday would be 365/365 equivalent to 100%. The likelihood of the second person having the same birthday would be 1/365. Calculating the probability that the first and second person have this birthday it would be a multiplication of their separate probabilities that is (365/365) * (1/365) = 1/365.

For three people, the first and second person having the same birthday date would still be 1/365. There is a chance that the first and third individuals sharing a birthday. Hence, the probability of 1/365 for the first and third person. There is also the probability of all three having the same birthday, and the second and third person sharing a birthday.

Solving the birthday paradox would require using one of the fundamental guidelines of probability; the summation of the probability that an event may take place and the probability it will not take place which is one. That is the chance of anything happening or not happening is equal to 100%. There working out the probability that no two people from the randomly selected group share birthday, then this rule is applicable in finding when they share a birthday. Determining the probability of at least two people in the room having same birthday is 1 minus the likelihood of every person in the room having a different birthday.

In summary, birthday paradox involve computing an approximation probability where in a group having n people at least two share birthday. To simplify the probability, variation in the distribution for instance, twins and a leap year are assumed. Moreover, it is presumed all the possible 365 birthdays are correspondingly probable. It is however important to note that real-life birthday distributions do not have a even pattern because not all dates are similarly probable. The aim is computing P(A), which is the probability of two people in a room sharing a birthday. However, computing the probability that no two people in the room share a birthday is easy by simply computing P(A’). Given that there are only two possibilities, A, and A’ there exists a mutual exclusivity, P(A) = 1 – P(A’).

Numbering the 30 people in the room from 1 to 30, the event that all 30 people have different birthdays would be equivalent to the probability of second person not have same birthday as person one. It would also be similar as the event that person 3 not having same birthday as person 2 and 1 which progresses up to the 30th person. It would be the same for person 30 not having same birthday as person 1 through 29. These events would be called Event 2, Event 3 and so on. Solving the problem would involve adding Event 1 which represent the occurrence of person 1 having a birthday, occurring with probability 1. Conditional probability is therefore used to compute the concurrence of this events. It would be the probability of Event 2 as 364/365, taking into account the probability of person 2 having same birthday as any other but not person 1. Correspondingly, the possibility of Event 3 factoring that Event 2 happened is 363/365, for person 3 cannot have same birthday as person 1 and 2. The permutation progresses until the final probability of Event 30 taking into account all the prior events happening as 335/343. In conclusion, the principle of conditional probability infers P(A’) is equivalent to the product of the individual probabilities.

References

Suzuki, K., Tonien, D., Kurosawa, K., & Toyota, K. (2008). Birthday paradox for multi-collisions. IEICE Transactions on Fundamentals of Electronics, Communications and Computer Sciences, 91(1), 39-45.

September 25, 2023

Eliminate the stress of Research and Writing!

Hire a Pro
Hire one of our **experts** to create a completely original paper even in **3 hours**!