Help with math problem

[FriarFritz]

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

[Condensation she/her]

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!

[theTruthshaper]

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

[FriarFritz]

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.

[revelryintheart she/her]

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

[Condensation she/her]

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!

[Inkspren_K she/her]

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. 

[Condensation she/her]

Wow, that's impressive. They even included leap years!

@Inkspren_K, you deserve a round of applause for that. *applause*

[FriarFritz]

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.

[Condensation she/her]

Ooh, here's a question. What's 0 divided by 0? And is 0.99999(repeating) equal to 1?

[theTruthshaper]

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.

[Condensation she/her]

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.

[Vapor she/her]

It's weird.

[Condensation she/her]

4 minutes ago, Vapor said:

It's weird.

No. It's true.

[FriarFritz]

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.

[Condensation she/her]

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.

[Experience he/him]

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.  

[Condensation she/her]

Just now, Ookla the Shadowed said:

Smart! I wish it was useful, though.

There, I corrected your punctuation correctly after nov 23.  

Very funny.