How many people do you need in a group to ensure at least a 50 percent probability that 2 people in the group share a birthday?<\/p>\n\n\n\n
Let’s take a show of hands. How many people think 30 people is enough? 60? 90? 180? 360?<\/p>\n\n\n\n
Surprisingly, the answer is only 23 people to have at least a 50 percent chance of a match. This goes up to 70 percent for 30 people, 90 percent for 41 people, 95 percent for 47 people. With 57 people there is better than a 99 percent chance of a birthday match!<\/p>\n\n\n\n
Presentation Suggestions:<\/strong> The Math Behind the Fact:<\/strong> To calculate the probability of a match, calculate the probability of no match and subtract from 1. But the probability of no match among n people is just (365\/365)(364\/365)(363\/365)(362\/365)…((366-n)\/365), where the k-th term in the product arises from considering the probability that the k-th person in the group doesn’t have a birthday match with the (k-1) people before her.<\/p>\n\n\n\n If you want to do this calculation quickly, you can use an approximation: note that for i<\/em> much smaller than 365<\/em>, the term (1-i\/365) can be approximated by EXP(-i\/365). Hence, for n much smaller than 365, the probability of no match is close to For still more fun, if you know some probability: to find the probability that in a given set of n people there are exactly M matches, you can use a Poisson approximation. The Poisson distribution is usually used to model a random variable that counts a number of “rare events”, each independent and identically distributed and with average frequency lambda.<\/p>\n\n\n\n Here, the probability of a match in a given pair is 1\/365. The matches can be considered to be approximately independent. The frequency lambda is the product of the number of pairs times the probability of a match in a pair: (n choose 2)\/365. Then the approximate probability that there are exactly M matches is: (lambda)M<\/sup> * EXP(-lambda) \/ M! which gives the same formula as above when M=0 and n=-365.<\/p>\n\n\n\n How to Cite this Page:<\/strong>\u00a0 Fun Fact suggested by:<\/strong> How many people do you need in a group to ensure at least a 50 percent probability that 2 people...<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[40,3,45],"class_list":["post-206","page","type-page","status-publish","hentry","tag-demonstration","tag-easy","tag-probability"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=206"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/206\/revisions"}],"predecessor-version":[{"id":1363,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/206\/revisions\/1363"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=206"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
If you have a large class, it is fun to try to take a poll of birthdays: have people call out their birthdays. But of course, whether or not you have a match proves nothing…<\/p>\n\n\n\n
Most people find this result surprising because they are tempted to calculate the probability of a birthday match with one particular person. But the calculation should be done over all pairs of people. Here is a trick that makes the calculation easier.<\/p>\n\n\n\n
EXP( – SUMi=1 to (n-1)<\/sub> i\/365) = EXP( – n(n-1)\/(2*365)).
When n=23, this evaluates to 0.499998 for the probability of no match. The probability of at least one match is thus 1 minus this quantity.<\/p>\n\n\n\n
Su, Francis E., et al. “Birthday Problem.”\u00a0Math Fun Facts<\/em>. <http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n
Francis Su<\/p>\n","protected":false},"excerpt":{"rendered":"