tag:blogger.com,1999:blog-2893315201359920156.post4042376632367984020..comments2017-06-24T13:26:35.266+05:30Comments on Best Brain Teasers: Probability Of Having Same Birthdaylaveshhttp://www.blogger.com/profile/03731980294342074301noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-2893315201359920156.post-86529368981021026372014-05-23T06:53:23.637+05:302014-05-23T06:53:23.637+05:30LolLolAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-12833715250863668592014-04-13T09:28:49.543+05:302014-04-13T09:28:49.543+05:30Four because they said birthday not birth date. So...Four because they said birthday not birth date. So you only have to count Sunday through SaturdayAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-12933984293192605332014-03-05T08:40:19.823+05:302014-03-05T08:40:19.823+05:30Why not three for triplets? Etc.Why not three for triplets? Etc.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-60353603320222527792014-01-22T09:59:23.955+05:302014-01-22T09:59:23.955+05:30OMG, its two, if you get identical twins, then the...OMG, its two, if you get identical twins, then there really isn't any way they can't share a birthday. 100% they share a birthday.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-8668443733256466022013-11-18T18:39:53.581+05:302013-11-18T18:39:53.581+05:30prob of 2nd person having the same date = 1/365, p...prob of 2nd person having the same date = 1/365, prob of 3rd person havin same date as either that of 1st or 2nd person is 2/365, similarly prob of 4th person having the same bday as other 3 is 3/365.........now all these added together should be >= 0.5 .....so 1/365+2/365+3/365+........n/365 >= 0.5 ........to find minimum value of n ........1+2+3+4+....n >= 365/2 ........or n* (n+1)/2 >= 365/n........or n*(n+1) >= 365 or n should be minimum 19 considering the 1st person the final answer is 20 ........so with 20 people in a room the probability is greater than 0.5 actually its 0.52.....By KaustubhAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-20294811924440141772013-11-18T18:36:43.522+05:302013-11-18T18:36:43.522+05:30prob of 2nd person having the same date = 1/365, p...prob of 2nd person having the same date = 1/365, prob of 3rd person havin same date as either that of 1st or 2nd person is 2/365, similarly prob of 4th person having the same bday as other 3 is 3/365.........now all these added together should be >= 0.5 .....so 1/365+2/365+3/365+........n/365 >= 0.5 ........to find minimum value of n ........1+2+3+4+....n >= 365/2 ........or n* (n+1)/2 >= 365/n........or n*(n+1) >= 365 or n should be minimum 19 considering the 1st person the final answer is 20 ........so with 20 people in a room the probability is greater than 0.5 actually its 0.52 .....By KaustubhAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-58839219047140724672013-09-28T09:37:26.515+05:302013-09-28T09:37:26.515+05:30This is the correct answer, but I couldn't fol...This is the correct answer, but I couldn't follow your explanation. Wikipedia has an article on this. Probability is a very strange subject, with non-obvious answers.<br /><br />http://en.wikipedia.org/wiki/Birthday_problemAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-41512362697374569372013-09-12T01:32:43.887+05:302013-09-12T01:32:43.887+05:302 they either do or don't have the same birt...2 they either do or don't have the same birthday 50/50Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-30425577780098052102013-08-21T13:53:47.124+05:302013-08-21T13:53:47.124+05:30Why are you hereWhy are you hereAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-83709199976322809962013-04-08T08:58:51.230+05:302013-04-08T08:58:51.230+05:30Triplets. So 3. I still think this makes more sens...Triplets. So 3. I still think this makes more sense than the actual answerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-55940465918761842152013-04-02T06:01:30.511+05:302013-04-02T06:01:30.511+05:30at the babies' room at the hospital.at the babies' room at the hospital.99c797a6-9b2c-11e2-9f00-000f20980440https://openid.aol.com/opaque/99c797a6-9b2c-11e2-9f00-000f20980440noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-85799628737124150642013-02-10T05:29:36.431+05:302013-02-10T05:29:36.431+05:30just put twins in the room hahajust put twins in the room hahaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-56384484641814570202013-01-20T10:59:12.518+05:302013-01-20T10:59:12.518+05:30oops, i'm wrongoops, i'm wrongAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-16031036873173969142013-01-20T10:58:27.173+05:302013-01-20T10:58:27.173+05:303. If there are 2 people in the room, they either...3. If there are 2 people in the room, they either have the same birthday or they don't. 50%. So, if there are 3 people in the room, the chances are greater than 50/50Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-39517737060123050052013-01-20T10:49:03.545+05:302013-01-20T10:49:03.545+05:302. If you put 2 people in a room, they either sha...2. If you put 2 people in a room, they either share a birthday or they don't. 50%Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-52037708517039315332013-01-04T15:08:57.910+05:302013-01-04T15:08:57.910+05:30if there are 365 ppl inside
then 365 => 1/365 ...if there are 365 ppl inside<br /><br />then 365 => 1/365 (for having different birthday)<br />if 365 + 365 => 2 / 365<br />if 365 * 3 => 3 / 365<br />if 365 * 193 => 193 / 365 which is greater than 1/2<br /><br />total number of ppl should be 365 * 193 = 70445Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-38100742676108598482012-11-06T22:59:40.535+05:302012-11-06T22:59:40.535+05:30The answer is 3. The question is talking about pr...The answer is 3. The question is talking about probability not reality if that makes sense. If there are 2 people in the room then there is a 50/50 chance that that they are the same birthday but if there is 3 people there is a greater than 50/50 chance that 2 have the same birthday.Peter Pan - forever youngnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-28287738876225231482012-10-25T21:47:23.018+05:302012-10-25T21:47:23.018+05:30I would like this one better as minus the date. Yo...I would like this one better as minus the date. You exclude the math and add the confusion:-) Cskhttp://www.blogger.com/profile/11563682112863848676noreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-90524919462789609802012-10-19T00:57:57.147+05:302012-10-19T00:57:57.147+05:30366, one for each day of the year, + one more to t...366, one for each day of the year, + one more to turn it into a 100% chance of there being two people who were born on the same day, (wich is more then 50%!)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-11114005779096736002012-08-16T09:34:21.250+05:302012-08-16T09:34:21.250+05:30Yes, twins have a slight chance of being born on c...Yes, twins have a slight chance of being born on consecutive days. However, the chance of the twins being born on the same day is still /more/ than 50%. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-73367010052866019592012-08-12T04:28:38.706+05:302012-08-12T04:28:38.706+05:30Twins can have diff birth days - 1 11:59 pm, the o...Twins can have diff birth days - 1 11:59 pm, the other oo:01 amAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-65682937100897083522012-06-24T00:29:13.348+05:302012-06-24T00:29:13.348+05:30Well.. why did we not consider leap year???Well.. why did we not consider leap year???Manishhttp://magnificia.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-28909597756553474192012-04-11T23:34:58.290+05:302012-04-11T23:34:58.290+05:302 cuz just get twins to come in the room...2 cuz just get twins to come in the room...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-23678168500229746222011-07-06T17:42:21.480+05:302011-07-06T17:42:21.480+05:30184
Suppose, person X born on 1st Jan.
Probability...184<br />Suppose, person X born on 1st Jan.<br />Probability of another person to born on the same day would be 1/365. so, probability of n persons to born on same day will be n/365 (As all these events are mutually exclusive events).<br />so, n/365>0.5 ... n = 183.<br />Total persons = 184 (including first person X)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2893315201359920156.post-41989337657730368062011-07-06T17:11:40.916+05:302011-07-06T17:11:40.916+05:30Say one person already in room. Second enters. now...Say one person already in room. Second enters. now probability of him not sharing b'day is (364/365)=x. Now third one enters, probability now of not sharing is (x*363/365=x).<br />keep on going like this.<br />take the product for n persons. let it be y.<br />now (1-y >1/2) solve for y(in terms of n) to get 23.Anonymousnoreply@blogger.com