Birthday Present
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Math Forum: Ask Dr. Math FAQ: The Birthday Problem -- Ask Dr. Math: FAQ The Birthday Problem Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time. What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet? This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised? How large must a class be to make the probability of finding two people with the same birthday at least 50%? Let's forget about leap year when we solve this problem (no February 29 birthdays!) This way, we can assume that a year is always 365 days long. We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%. Now, what about three people ? The chance of the first and second person sharing a birthday is still 1/365. The first and third person might share a birthday instead. The probability of that is 1/365 as well. But what if the second and third person shared a birthday? And what if all three of them had the same birthday? Things are getting complicated fast. Four or five people would be even messier. Is there a simpler way? To solve the birthday problem, we need to use one of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won't happen is always 1. (In other words, the chance that anything might or might nothappen is always 100%.) If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn't happen) = 1 P(two people share birthday) + P(no two people share birthday) = 1 P(two people share birthday) = 1 - P(no two people share birthday). So, what is the probability that no two people will share a birthday? Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person. To find the probability that both the second person and the third person will have different birthdays, we have to multiply: (365/365) * (364/365) * (363/365) = 132 132/133 225, which is about 99.18%. If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125, or about 98.36%. We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365). If you know permutation notation, you can write this formula as (365_P_n)/(365^n). That's the same as 365! / ((365-n)! * 365^n). We've made some progress, but we still haven't answered the original question: how large must a class be to make the probability of finding two people with the same birthday at least 50%? We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday, and we know how to find the probability that everybody has a different birthday for any number of people. The easiest way to find the right class size is to use a calculator to try different numbers in the formula. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is... a class of 23 people . (The probability is about 50.73%.) From the Dr. Math archives: Probability Theory: Coincidental Birthday Probability of the Same Birthday within a Group Birthday Probabilities Three Share a Birthday The Birthday Problem; Queuing at a Bank Birthday Probability, Class of 25 One Person of Seven Born on Monday Odds of Left-Handedness in a Group From the Web: The Birthday Problem: A short lesson in probability , George Reese A Java applet that you can use to test different class sizes (it works better with small classes) and graphs of the probability for different numbers of people. The Law of Small Errors , Keith Devlin The birthday problem, and related questions - what's the probability that someone will have your birthday? Birthday Surprises, Ivars Peterson Birthday Problem, Eric Weisstein's World of Mathematics Coincidence, Alexander Bogomolny How to Read Mathematics, Shai Simonson and Fernando Gouveau This article uses an explanation of the birthday problem as an example. An Introduction to Mathematica and the "Birthday Problem," Louie Beuschlein For a general review of probability: Probability, Dr. Math FAQ Probability in the Real World, Dr. Math FAQ - Ursula Whitcher, for the Math Forum Submit your ownquestion to Dr. Math [ Privacy Policy ] [ Terms of Use ] Math Forum Home || Math Library || Quick Reference || Math Forum Search Ask Dr. Math ® © 1994-2005 The Math Forum http://mathforum.org/dr.math/
College Gift
Kenyon College - Gift funds new faculty award Calendar | Contact Kenyon | Search | Parents | Alumni | Current Students | Faculty & Staff | Community Fortnightly, September 12, 2005 Fortnightly, August 29, 2005 Fortnightly, May 16, 2005 Fortnightly, May 3, 2005 Fortnightly, April 18, 2005 Fortnightly, April 4, 2005 Fortnightly, March 21, 2005 Fortnightly, February 28, 2005 Fortnightly, February 16, 2005 Fortnightly, January 31, 2005 Fortnightly, January 17, 2005 Fortnightly, November 29, 2004 Fortnightly, November 8, 2004 Fortnightly, October 25, 2004 Fortnightly, October 11, 2004 Fortnightly, September 27, 2004 Fortnightly, September 13, 2004 Fortnightly, August 30, 2004 Fortnightly, May 17, 2004 Kenyon receives $1.5 million from Howard Hughes Medical Institute Treasury secretary to speak at Commencement Five faculty members named to endowed chairs Gift funds new faculty award Reunion Weekend coming Kenyon Beat People Searches Profile Sports: Carr wins all-district honors Submissions Fortnightly, May 3, 2004 Fortnightly, April 19, 2004 Fortnightly, April 5, 2004 Fortnightly, March 22, 2004 Fortnightly, March 1, 2004 Fortnightly, February 16, 2004 Fortnightly, February 2, 2004 Fortnightly, January 19, 2004 Fortnightly, December 1, 2003 Fortnightly, November 10, 2003 Fortnightly, October 13, 2003 Fortnightly, September 29, 2003 Fortnightly, September 15, 2003 Fortnightly, September 2, 2003 About Kenyon Academics Admissions Athletics Student Life News & Events Giving to Kenyon Home » Fortnightly, May 17, 2004 » Gift funds new faculty award Gift funds new faculty award Jon Chun, the cofounder and former chief executive officer of a Silicon Valley security software startup, has surprised and delighted Kenyon with a gift of more than $100,000 to endow a competitive award supporting faculty research. The Dr. Newton Chun Award will be granted every other year, funding scholarly investigation and artistic projects of exceptional merit and promise. "The hallmark of Kenyon is the quality of its teaching," said President S. Georgia Nugent. "For our faculty to continue to be effective teachers, they must be encouraged and supported in their own intellectual development. The Dr. Newton Chun Award is a most welcome addition in this quest. I look forward to the diversity of projects this fund will inspire." The College plans to make the first award in the spring of 2005. Preference will be given to projects that could not be accomplished without this additional financial support. The award, totaling approximately $10,000, will underwrite expenses such as travel, materials and equipment, and research-assistant costs. Chun named the award in honor of his father, Newton Chun, a physician who yearned to be an academic. A native of Dubuque, Iowa, Jon Chun attended the University of California at Berkeley as an engineering student and went on to study medicine. As a successful technology entrepreneur, he cofounded SafeWeb, which provides secure remote access into corporate networks using only a Web browser. Chun is now a director at Symantec, the world leader in Internet security, which acquired SafeWeb in October 2003. He is married to Katherine Elkins, Andrew W. Mellon assistant professor of IPHS at Kenyon. Kenyon College. Gambier, Ohio 43022-9623 Phone: 740-427-5000
Birthday Present
Math Forum: Ask Dr. Math FAQ: The Birthday Problem -- Ask Dr. Math: FAQ The Birthday Problem Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time. What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet? This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised? How large must a class be to make the probability of finding two people with the same birthday at least 50%? Let's forget about leap year when we solve this problem (no February 29 birthdays!) This way, we can assume that a year is always 365 days long. We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%. Now, what about three people ? The chance of the first and second person sharing a birthday is still 1/365. The first and third person might share a birthday instead. The probability of that is 1/365 as well. But what if the second and third person shared a birthday? And what if all three of them had the same birthday? Things are getting complicated fast. Four or five people would be even messier. Is there a simpler way? To solve the birthday problem, we need to use one of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won't happen is always 1. (In other words, the chance that anything might or might nothappen is always 100%.) If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn't happen) = 1 P(two people share birthday) + P(no two people share birthday) = 1 P(two people share birthday) = 1 - P(no two people share birthday). So, what is the probability that no two people will share a birthday? Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person. To find the probability that both the second person and the third person will have different birthdays, we have to multiply: (365/365) * (364/365) * (363/365) = 132 132/133 225, which is about 99.18%. If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125, or about 98.36%. We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365). If you know permutation notation, you can write this formula as (365_P_n)/(365^n). That's the same as 365! / ((365-n)! * 365^n). We've made some progress, but we still haven't answered the original question: how large must a class be to make the probability of finding two people with the same birthday at least 50%? We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday, and we know how to find the probability that everybody has a different birthday for any number of people. The easiest way to find the right class size is to use a calculator to try different numbers in the formula. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is... a class of 23 people . (The probability is about 50.73%.) From the Dr. Math archives: Probability Theory: Coincidental Birthday Probability of the Same Birthday within a Group Birthday Probabilities Three Share a Birthday The Birthday Problem; Queuing at a Bank Birthday Probability, Class of 25 One Person of Seven Born on Monday Odds of Left-Handedness in a Group From the Web: The Birthday Problem: A short lesson in probability , George Reese A Java applet that you can use to test different class sizes (it works better with small classes) and graphs of the probability for different numbers of people. The Law of Small Errors , Keith Devlin The birthday problem, and related questions - what's the probability that someone will have your birthday? Birthday Surprises, Ivars Peterson Birthday Problem, Eric Weisstein's World of Mathematics Coincidence, Alexander Bogomolny How to Read Mathematics, Shai Simonson and Fernando Gouveau This article uses an explanation of the birthday problem as an example. An Introduction to Mathematica and the "Birthday Problem," Louie Beuschlein For a general review of probability: Probability, Dr. Math FAQ Probability in the Real World, Dr. Math FAQ - Ursula Whitcher, for the Math Forum Submit your ownquestion to Dr. Math [ Privacy Policy ] [ Terms of Use ] Math Forum Home || Math Library || Quick Reference || Math Forum Search Ask Dr. Math ® © 1994-2005 The Math Forum http://mathforum.org/dr.math/
College Gift
Kenyon College - Gift funds new faculty award Calendar | Contact Kenyon | Search | Parents | Alumni | Current Students | Faculty & Staff | Community Fortnightly, September 12, 2005 Fortnightly, August 29, 2005 Fortnightly, May 16, 2005 Fortnightly, May 3, 2005 Fortnightly, April 18, 2005 Fortnightly, April 4, 2005 Fortnightly, March 21, 2005 Fortnightly, February 28, 2005 Fortnightly, February 16, 2005 Fortnightly, January 31, 2005 Fortnightly, January 17, 2005 Fortnightly, November 29, 2004 Fortnightly, November 8, 2004 Fortnightly, October 25, 2004 Fortnightly, October 11, 2004 Fortnightly, September 27, 2004 Fortnightly, September 13, 2004 Fortnightly, August 30, 2004 Fortnightly, May 17, 2004 Kenyon receives $1.5 million from Howard Hughes Medical Institute Treasury secretary to speak at Commencement Five faculty members named to endowed chairs Gift funds new faculty award Reunion Weekend coming Kenyon Beat People Searches Profile Sports: Carr wins all-district honors Submissions Fortnightly, May 3, 2004 Fortnightly, April 19, 2004 Fortnightly, April 5, 2004 Fortnightly, March 22, 2004 Fortnightly, March 1, 2004 Fortnightly, February 16, 2004 Fortnightly, February 2, 2004 Fortnightly, January 19, 2004 Fortnightly, December 1, 2003 Fortnightly, November 10, 2003 Fortnightly, October 13, 2003 Fortnightly, September 29, 2003 Fortnightly, September 15, 2003 Fortnightly, September 2, 2003 About Kenyon Academics Admissions Athletics Student Life News & Events Giving to Kenyon Home » Fortnightly, May 17, 2004 » Gift funds new faculty award Gift funds new faculty award Jon Chun, the cofounder and former chief executive officer of a Silicon Valley security software startup, has surprised and delighted Kenyon with a gift of more than $100,000 to endow a competitive award supporting faculty research. The Dr. Newton Chun Award will be granted every other year, funding scholarly investigation and artistic projects of exceptional merit and promise. "The hallmark of Kenyon is the quality of its teaching," said President S. Georgia Nugent. "For our faculty to continue to be effective teachers, they must be encouraged and supported in their own intellectual development. The Dr. Newton Chun Award is a most welcome addition in this quest. I look forward to the diversity of projects this fund will inspire." The College plans to make the first award in the spring of 2005. Preference will be given to projects that could not be accomplished without this additional financial support. The award, totaling approximately $10,000, will underwrite expenses such as travel, materials and equipment, and research-assistant costs. Chun named the award in honor of his father, Newton Chun, a physician who yearned to be an academic. A native of Dubuque, Iowa, Jon Chun attended the University of California at Berkeley as an engineering student and went on to study medicine. As a successful technology entrepreneur, he cofounded SafeWeb, which provides secure remote access into corporate networks using only a Web browser. Chun is now a director at Symantec, the world leader in Internet security, which acquired SafeWeb in October 2003. He is married to Katherine Elkins, Andrew W. Mellon assistant professor of IPHS at Kenyon. Kenyon College. Gambier, Ohio 43022-9623 Phone: 740-427-5000
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