
	Where results make sense

Topic: Birthday problem

Related Topics

List of mathematical coincidences

Kalman filter

Inertial guidance system

In the News (Thu 24 Dec 09)

Northern Trust

Marvin Harrison

Peter Chernin

Gary Locke

Match Play

Penelope Cruz

Mickey Rourke

AR Rahman

Danny Boyle

Hugh Jackman

	The Birthday Problem (Site not responding. Last check: 2007-11-01)
		The Birthday Problem: A short lesson in probability.
		There are a number of ways to approach this problem.
		Below is a simulation of the birthday problem.
	www.mste.uiuc.edu /java/java/birthday/birthday.html (188 words)

	Wikinfo \| Birthday paradox (Site not responding. Last check: 2007-11-01)
		The birthday paradox states that if there are 23 people in a room then there is roughly a 50/50 chance that two of them have the same birthday.
		To compute the approximate probability that in a room of n people, at least two have the same birthday, we disregard leap years and twins, and assume that the 365 possible birthdays are equally likely.
		The birthday problem used to be a splendid illustration of the advantages of pure thought over mechanical manipulation; the inequalities can be obtained in a minute or two, whereas the multiplications would take much longer, and be much more subject to error, whether the instrument is a pencil or an old-fashioned desk computer.
	www.wikinfo.org /wiki.php?title=Birthday_paradox (633 words)

	Illuminations: Birthday Paradox
		For example, in the birthday paradox, a group of twenty-three randomly selected persons must be selected to have a greater than 50 percent chance that any two of them share the same birthday (Lesser, 1999).
		After the students explore whether twenty-three birthdays are sufficient to get a match 50 percent or more of the time, they can explore the percent of times that matches occur with groups of different sizes by running additional trials with a new number.
		The probability that two strangers do not share a birthday is 364/365, assuming that neither of them was born in a leap year, with the probability of a match being the complement of this event, that is, 1 -(364/365), or.00274.
	illuminations.nctm.org /index_d.aspx?id=299 (1961 words)

	THE BIRTHDAY PROBLEM (Site not responding. Last check: 2007-11-01)
		In the numerator, there are 365 possible birthdays for the first person, and once that birthday is chosen there are 364 birthdays for the second person.
		In the denominator, there are are 365 possible birthdays for the first person and 365 for the second person.
		For arbitrary k, the probability that at least two of them have the SAME birthday is 1 - (365 X 364 X … X (365 – k + 1)) / (365 ^ k), or 1 – (365Pk) / (365 ^ k).
	users.aol.com /fcas/bday.html (263 words)

	The Birthday Problem Programmed in Visual Basic - R. Berman
		Example 0.5: Compute the probability that of k individuals, no two will have the same birthday; also find the probability that there will be at least 1 match.
		Problem 0.5: Do the analogous problem to Example 0.5 using the 12 astrological signs in place of the 365 days of the year.
		Do the analogous problem to Example 0.5 using the 12 astrological signs in place of the 365 days of the year.
	www.math.wayne.edu /~rberman/syllabi/ex_0_5.htm (605 words)

	The Birthday Problem
		There is a problem in mathematics relating to birthdays.
		And therefore, the probability that no two people would have the same birthday out of a group of three people is the number of ways that their birthdays could be different divided by the total number of ways in which their birthdays could occur or (366)(365)(364) / (366)(366)(366).
		Upon an examination of their birthdays, we find that the eleventh President of the United States, James K. Polk, was born on November 2, 1795, and that the twenty-ninth President, Warren G. Harding, was born on November 2, 1865.
	www.pen.k12.va.us /Div/Winchester/jhhs/math/lessons/stat/birthday.html (1108 words)

	FAQ-13 What is the answer to the Monty Hall, Envelope, or Birthday problem?
		In a variant of this problem, you are allowed to peek into the envelope you chose before finally settling on it.
		A good reference to the problem is Christensen, R. and Utts, J. "Bayesian Resolution of the 'Exchange Paradox,'" The American Statistician, 46(4), 274-276.
		A nice summary of this problem with extensions into non-uniform birthdates is Nunnikhoven, T.S. (1992) "A Birthday Problem Solution for Nonuniform Birth Frequencies," The American Statistician, 46(4), 270-274.
	www.childrens-mercy.org /stats/faq/faq13.asp (575 words)

	Coincidence
		Thus we have a person with a birthday which falls onto one of the 365 days and ask what is the probability that another person has a different birthday.
		In particular, some of the articles were concerned with the birthday problem and its generalizations.
		By the time 23rd person walks through the door, their chances would be 22 in 365, but added to all the previous chances, the total chance of coincidence in the group would be better than 50%.
	www.cut-the-knot.org /do_you_know/coincidence.shtml (1715 words)

	Birthday Paradox: Formula, Probability, Combinatorics
		The birthday paradox calculates that the probability to get the same pick-3 combination at least two times in 100 trials is 99.4%.
		The cold truth is that the famous and appealing Birthday Paradox merely shows the percentage of sets with duplicate elements in the total elements of an exponential set.
		Relation of Birthday Paradox to roulette: The connection between Birthday paradox and the game of roulette is extraordinarily appealing to some gamblers.
	www.saliu.com /birthday.html (4164 words)

	The Birthday Problem
		The birthday problem is one of the most famous problems in combinatorial probability.
		The problem is famous, in part, because the answer is a bit surprising.
		The counting that we need to do for the birthday problem is easy, using the multiplication principle of combinatorics.
	www.ship.edu /~deensl/mathdl/stats/Birthday.html (782 words)

	Research in Psychology - THE BIRTHDAY PROBLEM (ANSWER 20)
		After the first person has given his or her birthday, the second person will have one chance in 365 of having the same birthday as the first.
		By the same logic, the fourth person has 3 chances of having the same birthday as any of the first three so this needs to be added to the previous 3 to get 6 chances out of 365, and so on.
		Although N=20 is an incorrect answer to the original problem it does suggest an alternate approach for betting purposes.
	www.psy.pdx.edu /PsiCafe/Research/BDay-B.htm (458 words)

	Common Sense Problem Page (Site not responding. Last check: 2007-11-01)
		Ernest Davis, who contributed many of these problems, has suggested that very few of these problems are currently solvable without considerable simplification.
		Two problems that he believes are solvable are The Surprise Birthday Present and the first half of Wolves and Rabbits.
		The problem is to formalise some aspect of this problem: e.g.
	www-formal.stanford.edu /leora/commonsense (3858 words)

	[No title]
		Problem B: In a group of 6 people, what is the probability that at least two of them have the same birthday?
		Assume that each of the 365 days in a year is equally likely to be the birthday of a randomly selected person.
		Note that “at least two share a common birthday” could mean that all six have the same birthday, or that two have one birthday, three another, etc. That is, every way in which at least one birthday is shared is included.
	math.usask.ca /~oshaughn/103Birthday.doc (328 words)

	Birthday Type Problems (Site not responding. Last check: 2007-11-01)
		Marilyn vos Savant, who holds the distinction of having the highest IQ in the world, discussed the birthday problem in her column in Parade Magazine.
		While the problems involve birthdays and, on the surface, appear to be similar to the original birthday problem, not all are mathematically similar to the birthday problem, and some (e.g., Problem 2) require a completely different argument.
		It is essential that you see beyond the superficial similarity of these problems and spot the mathematical differences between the various problems.
	www.math.uiuc.edu /~hildebr/361/birthday.html (406 words)

	Three Easy Pieces
		The birthday problem is famous because the event of interest has much higher probability than one might first expect.
		Buffon's needle problem is surprising in its very concept, has great historical interest, and illustrates the beautiful interplay between probability and statistics.
		Yet the three problems share a couple of key properties: each is mathematically simple, but still leads to results that are interesting and a bit surprising.
	www.ship.edu /~deensl/mathdl/stats/index.html (1053 words)

	The Birthday Problem - Probabilities (Site not responding. Last check: 2007-11-01)
		The third person would then have a 363/365 probability of not sharing a common birthday with the first two, since there are 363 possible birthdates leftover from which to choose.
		Therefore, the complement of no common birthdays, which is at least one pair of shared birthdays, is found by subtracting the first probability from 1.
		Therefore, the probability that there is a pair of shared birthdays in a group of 22 is 1 - 0.524 = 0.476, or 47.6%, awful close to the halfway mark!
	fym.la.asu.edu /~fym/mat114/projects/birthday.html (410 words)

	The Birthday Problem (Site not responding. Last check: 2007-11-01)
		By same birthday we mean same day and month but not necessarily the same year.
		We assume further that the birthdays of the k people are unrelated and each equally likely to be in any of the 365 days of the year.
		So the problem reduces to the computation of the number of ways in which k people can have different birthdays.
	omega.albany.edu:8008 /mat465dir/birthday-m2h.html (196 words)

	The Birthday Problem Revisited (Site not responding. Last check: 2007-11-01)
		One lesson that has a safe place in the teaching calendar for me is the Birthday Problem, tackled fairly early on in the study of probability.
		We supposed that the number of people with a birthday on a particular day was the random variable X. Now X ~ B(365, 1/365), which is close to the Poisson distribution with mean 1.
		We were also saying (as of course we say for the basic Birthday Problem) that every day is equally likely as a birthday, which is not quite true.
	homepage.ntlworld.com /jonny.griffiths/birthday/birthday.html (1190 words)

	BirthdayProblemCalculator
		The relative likelihood of this sort of coincidence comes from the fairly large number of possible pairings: in order for there to be no shared birthday, each of the 506 possible pairs of people has to have a different birthday.
		where n is the number of possible choices (365 days for the birthday problem) and m is the number of selections we make (23 people, for example).
		In this case she might just be able to ignore the problem altogether.
	pluralsight.com /wiki/default.aspx/Craig/BirthdayProblemCalculator.html (1136 words)

	Ivars Peterson's MathTrek - Birthday Surprises
		For a second person to have a birthday that doesn't match that of the first, he or she must be born on any one of the other 364 days of the year.
		The probability that a third person has a birthday that differs from the other two distinct birthdays is 363/365.
		Variations of the birthday problem serve as useful models for analyzing coincidences, says statistician Persi Diaconis of Stanford University.
	www.maa.org /mathland/mathtrek_11_23_98.html (1007 words)

	The Birthday Problem- explanation (Site not responding. Last check: 2007-11-01)
		For two people, there are 364 different ways that the second could have a birthday without matching the first.
		If there is no match after two people, the third person has 363 different birthdays that do not match the other two.
		This leads to the following formula for calculating the probability of a match with N birthdays is 1 - (365)(364)(363)...(365 - N + 1)/(365)^N. Running this through a computer gives the chart below.
	www.csulb.edu /~cwallis/170/explanation.html (113 words)

	The Birthday Paradox
		A favorite problem in elementary probability and statistics courses is the Birthday Problem: What is the probability that at least two of N randomly selected people have the same birthday?
		The original problem can be solved with a slide rule, which is exactly what I did when I first heard it many, many years ago.
		The Birthday Paradox shows that the probability that two or more items will end up in the same bin is high even if the number of items is considerably less than the number of bins.
	efgh.com /math/birthday.htm (774 words)

	Birthday Problem
		The chance that you and the other person have the same birthday is approximately 1 in 365.
		The chance that the birthday of the third person will match yours or the other person's is 2 in 365.
		We got the exact reference from Linda Wagner, who encountered the text in the course EDUC W554 (Creative Problem Solving and Metacognition) at Indiana University-Purdue University at Ft. Wayne.
	isds.bus.lsu.edu /chun/teach/reading-a/birthday.htm (290 words)

	The birthday problem (Site not responding. Last check: 2007-11-01)
		I am not sure that there is a way to explain the birthday problem that is intuitive.
		The birthday problem is then, for each y, find the smallest n that gives p(y,n) > 0.5.
		If this number is called f(y) then the traditional birthday problem has answer f(365) = 23 and your "12 day year" problem gives f(12) = 5.
	mathcentral.uregina.ca /QQ/database/QQ.09.98/cooke1.html (346 words)

	The Birthday Problem and OFB (Site not responding. Last check: 2007-11-01)
		This is a famous problem in probability theory.
		There is a very closely related problem that Schneier and Ferguson refer as "meet in the middle".
		This is like having two classes of N and M students respectively with no two students in the same class having the same birthday.
	www2.ics.hawaii.edu /~wes/ICS623/Birthday.html (1015 words)

	Birthday!
		If we neglect February 29th as a birthday, assume that all birthdays are equally likely, and therefore, the probability for any birthday is 1/365.
		Anyway, our assumptions are that each of the 365 birthdays is equally likely (our data from 1978 does not really support this conclusion) and that the people gathered have unrelated birthdays.
		As always, if you have problems with any of these files, you can email me a short description of your troubles and I will try to help.
	birthday-surprise.blogspot.com (1426 words)

	Probability Theory - The Laymans guide to probability (Site not responding. Last check: 2007-11-01)
		When the first person enters the room and announces their birthday, the probability of the second person sharing the same birthday is 1/365.
		When two birthdays are known, the probability of the third being different is 363/365, as there are now two 'favourable' outcomes among 365.
		Each of these problems is an exercise in calculating combinations, although British football pools companies and pools journalists always refer for some reason to the second example, as a permutation.
	www.probabilitytheory.info - !http: //www.peterwebb.co.uk/probability.htm (6661 words)

	The Birthday Problem (Site not responding. Last check: 2007-11-01)
		"Let's see, there are about 20 of us here, out of 365 possible birthdays, 20/365........that's about 5%.........5% chance that one of the other people in line with me at the bank has the same birthday." Sounds reasonable, after all, most people probably don't know more than one other person with their birthday.
		He just calculated the probability of one person having one of 20 different birthdays.
		That is more than half of 23, so the number of days to choose from increases slower than the sample size.
	student.fortlewis.edu /JRJONES1/birthday.htm (477 words)

	Mobility Today - Birthday reminder - problem
		09-13-2004 10:05 PM If you have a lot of contacts with Birthdays entered, it can take quite some time for the Calendar to be populated with the Birthdate appoinment info as birthdates are treated like an appointment.
		If it appears, it may be a problem with the birthdate appointment not being processed as a recurring appointment.
		I have a problem with doubling of Birthday appointments.
	mobilitytoday.com /forum/printthread.php?t=3825 (484 words)

Try your search on: Qwika (all wikis)