FriarFritz Posted October 15, 2020 Report Share Posted October 15, 2020 (edited) The trouble is that I needed to graph some functions. I think I got it, though. To have a 98% chance of someone having the same birthday as you, you have to go through at least 1,426 people. For a 99% chance, you need at least 1,679. Yes! I did it! Huzzah! Thanks for the problem, I really enjoyed it. Edited October 15, 2020 by FriarFritz Oop, mistyped a number 1 Quote Link to comment Share on other sites More sharing options...
Condensation Posted October 15, 2020 Report Share Posted October 15, 2020 2 minutes ago, FriarFritz said: The trouble is that I needed to graph some functions. I think I got it, though. To have a 98% chance of someone having the same birthday as you, you have to go through at least 1,246 people. For a 99% chance, you need at least 1,679. Yes! I did it! Huzzah! Thanks for the problem, I really enjoyed it. Nice! 0 Quote Link to comment Share on other sites More sharing options...
theTruthshaper Posted November 4, 2020 Report Share Posted November 4, 2020 (edited) I don't get why everyone was so troubled? Danex was right the first time around. Oh and yes; I have a different version of this problem for y'all: What is the minimum number of people required in a group so that the probability of at least one pair having the same birthday is greater than 99%? Edited November 4, 2020 by The_Truthwatcher 0 Quote Link to comment Share on other sites More sharing options...
FriarFritz Posted November 5, 2020 Report Share Posted November 5, 2020 On 11/4/2020 at 4:09 AM, The_Truthwatcher said: I don't get why everyone was so troubled? Danex was right the first time around. Well, his initial idea wasn't correct, because it led to a probability of 1 at 365 people. But the probability can't be 100%, ever, because none of the individual people have a 100% probability of the same birthday as you, and the probability of each one is independent from the rest. Simplifying and modeling it as a coin flip actually works pretty well as a starting point, though, because a coin flip is essentially rolling a two-sided die. On 11/4/2020 at 4:09 AM, The_Truthwatcher said: Oh and yes; I have a different version of this problem for y'all: What is the minimum number of people required in a group so that the probability of at least one pair having the same birthday is greater than 99%? I thought I had this figured out, but I somehow muddled myself up. I'll be trying this later, though. 0 Quote Link to comment Share on other sites More sharing options...
revelryintheart Posted November 6, 2020 Report Share Posted November 6, 2020 (edited) On 11/4/2020 at 5:09 AM, The_Truthwatcher said: What is the minimum number of people required in a group so that the probability of at least one pair having the same birthday is greater than 99%? Two if you make sure beforehand to get two people with the same birthday *slinks away* Edited November 6, 2020 by revelryintheart 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 6, 2020 Report Share Posted November 6, 2020 13 hours ago, revelryintheart said: Two if you make sure beforehand to get two people with the same birthday *slinks away* 'Tis true. Good idea! 0 Quote Link to comment Share on other sites More sharing options...
Inkspren_K Posted November 26, 2020 Report Share Posted November 26, 2020 The easiest way to do this type of problem is to approach it backwards. You want the probability of all of them having a different birthday than you too be less than 1 or 2 percent. The probability of them having a different birthday is 364/365, so the probability of two different people having different birthdays than you is (364/365)^2, for 100 people it's (364/365)^100. (364/365)^1425 is greater than 2%, but (364/365)^1426 is less than 2% so for a 98% probability you would need 1426 people. The same process can be used to find that you need 1679 people for a 99% probability. Now to make things more interesting let's include leap years. Now (assuming you weren't born on leap day) the probability of someone having a different birthday is 1457/1461 (there are 1461 days in 4 years, and 4 are your birthday). So we want to solve (1457/1461)^n <0.02. Now you need 1427 people for a 98% probability and 1680 people for a 99% probability. Last but not least, to address the problem of how many people are needed for a 98-99% probability of any two having the same birthday, for one person there is a 365/365 change of having a unique birthday, for 2 people there is a 364/365 chance that their birthday is unique, for the third person, multiply 364/365x363/365 (there are 363 remaining unique birthdays did the third person, for the fourth, multiply by 362/365, etc. With 53 people, there is a >98% chance two share a birthday, with 57 people there is a >99% probability two share a birthday. 4 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 27, 2020 Report Share Posted November 27, 2020 Wow, that's impressive. They even included leap years! @Inkspren_K, you deserve a round of applause for that. *applause* 0 Quote Link to comment Share on other sites More sharing options...
FriarFritz Posted November 28, 2020 Report Share Posted November 28, 2020 On 11/25/2020 at 4:56 PM, Inkspren_K said: The easiest way to do this type of problem is to approach it backwards. You want the probability of all of them having a different birthday than you too be less than 1 or 2 percent. The probability of them having a different birthday is 364/365, so the probability of two different people having different birthdays than you is (364/365)^2, for 100 people it's (364/365)^100. (364/365)^1425 is greater than 2%, but (364/365)^1426 is less than 2% so for a 98% probability you would need 1426 people. The same process can be used to find that you need 1679 people for a 99% probability. Now to make things more interesting let's include leap years. Now (assuming you weren't born on leap day) the probability of someone having a different birthday is 1457/1461 (there are 1461 days in 4 years, and 4 are your birthday). So we want to solve (1457/1461)^n <0.02. Now you need 1427 people for a 98% probability and 1680 people for a 99% probability. Last but not least, to address the problem of how many people are needed for a 98-99% probability of any two having the same birthday, for one person there is a 365/365 change of having a unique birthday, for 2 people there is a 364/365 chance that their birthday is unique, for the third person, multiply 364/365x363/365 (there are 363 remaining unique birthdays did the third person, for the fourth, multiply by 362/365, etc. With 53 people, there is a >98% chance two share a birthday, with 57 people there is a >99% probability two share a birthday. Whoa! My hat is off! That solution is much more elegant than mine. 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 28, 2020 Report Share Posted November 28, 2020 Ooh, here's a question. What's 0 divided by 0? And is 0.99999(repeating) equal to 1? 0 Quote Link to comment Share on other sites More sharing options...
theTruthshaper Posted November 28, 2020 Report Share Posted November 28, 2020 3 hours ago, Ookla the Grammatical said: Ooh, here's a question. What's 0 divided by 0? And is 0.99999(repeating) equal to 1? 0 divided by o isn't anything. It's undefined. Yes. 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 29, 2020 Report Share Posted November 29, 2020 17 hours ago, Ookla of Truthshapers said: 0 divided by o isn't anything. It's undefined. Yes. Thank you. @Vapor, here's your answer. 0 Quote Link to comment Share on other sites More sharing options...
Vapor Posted November 29, 2020 Report Share Posted November 29, 2020 It's weird. 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 29, 2020 Report Share Posted November 29, 2020 4 minutes ago, Vapor said: It's weird. No. It's true. 0 Quote Link to comment Share on other sites More sharing options...
FriarFritz Posted November 29, 2020 Report Share Posted November 29, 2020 Oh-ho! Back in middle school, I had an argument with every one of my math teachers about this. I was convinced that 0/0 was actually the set of all numbers. And technically it is, but it's treated as undefined because that's not useful in actual math. 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 30, 2020 Report Share Posted November 30, 2020 10 hours ago, Ookla the Monk said: Oh-ho! Back in middle school, I had an argument with every one of my math teachers about this. I was convinced that 0/0 was actually the set of all numbers. And technically it is, but it's treated as undefined because that's not useful in actual math. Smart. I wish it was useful, though. 0 Quote Link to comment Share on other sites More sharing options...
Experience Posted November 30, 2020 Report Share Posted November 30, 2020 1 minute ago, Ookla the Grammatical said: Smart. I wish it was useful, though. Smart! I wish it was useful, though. There, I corrected your punctuation correctly after nov 23. 0 Quote Link to comment Share on other sites More sharing options...
Condensation Posted November 30, 2020 Report Share Posted November 30, 2020 Just now, Ookla the Shadowed said: Smart! I wish it was useful, though. There, I corrected your punctuation correctly after nov 23. Very funny. 0 Quote Link to comment Share on other sites More sharing options...
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