Zelly

Help with math problem

27 posts in this topic

So driving home today I was pondering.

With one other random person, there's a 1 in 365 chance of them sharing my birthday, correct? (If I'm wrong already we're in trouble).

But it's not...cumulative probability or whatever.  Like each person is a separate event, and 365 people together does not mean guarantee there will be someone with my shared birthday (right?)

So my main question is: How many people would I need to gather to have a 98-99% probability that at least 1 other person shared my birthday?

 

I'm not a mathy person, so if you're a math teacher (or want to be), this is your time to shine. Bonus points if you can explain with more words and less equations. :lol:

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I have no idea, but I remember another problem like that in a book I read.

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See if you can work out the answer to your question using this explanation (you really have everything you need, in there, except for the answer!): http://mathforum.org/dr.math/faq/faq.birthdayprob.html

The author does put aside the objections I made earlier to make the problem more manageable: 

Quote

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.

Also, let's assume that a person has an equal chance of being born on any day of the year, even though some birthdays may be slightly more likely than others. That will simpify the math, without changing the result signficantly.

 

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Does anyone know the explicit equation for the Fibonacci Sequence?

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7 hours ago, Condensation said:

Does anyone know the explicit equation for the Fibonacci Sequence?

Well... it's not as simple as the recursive: Fn = Fn-1 + Fn-2

The raw explicative equation is this: Fib(n) = (1.6180339..n – (–0.6180339..)n) / 2.236067977..


The other equation we can use is Fib(n) = (Phin–(–phi)n)/√5

where Phi =(1.618..) and -phi=-0.618..

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2 minutes ago, MusicalReader said:

Well... it's not as simple as the recursive: Fn = Fn-1 + Fn-2

The raw explicative equation is this: Fib(n) = (1.6180339..n – (–0.6180339..)n) / 2.236067977..


The other equation we can use is Fib(n) = (Phin–(–phi)n)/√5

where Phi =(1.618..) and -phi=-0.618..

I know the recursive, I just wanted to know the explicit. Thanks, that's great! We figured it out today in math.

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Just now, Condensation said:

I know the recursive, I just wanted to know the explicit. Thanks, that's great! We figured it out today in math.

Anytime! ^_^

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:) Phi's great too.

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21 hours ago, Condensation said:

:) Phi's great too.

Yeah, really is! ^_^

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Here's my question: Why did we make it Pi and Phi? *sighs*

In math, I kept mixing them up. It's weird, because I don't usually, but I just couldn't math that day.

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Pi is bad

Tau forever. 

On 10/7/2020 at 1:24 PM, Zelly said:

So my main question is: How many people would I need to gather to have a 98-99% probability that at least 1 other person shared my birthday?

As for the actual question, I think it is as simple as it sounds. 
If you choose one random date out of 365 possible dates, and then choose one truly random person, there is a 1/365 chance that they’d have that birthday. So if you gathered 365 people, you’d have a 1/365 * 365 chance. Which would be equal to 1, or 100%.
 

I think some people were referencing the famous birthday paradox scenario, which isn’t actually a paradox, just counterintuitive. It asks a similar question: How many people would you need to gather to be certain that any two of them shared a birthday. It’s a surprisingly low number. I don’t remeber the exact math, but it has to do with you not counting the number of people, you need to count the number of pairs. And the required number of pairs can be reached with a low number of people. Next part is me trying to figure it out from memory:

I think it’s 27. If you have 27 people, there is something like a 99% chance that 2 of them share a birthday. If you have 27 people, there are 351 possible pairings (I think) 351/365 equals 96%

 

 

hmmmm wait I just thought about the original question some more and I’m less sure. 

What if we simplify the question to a coin flip. Same principle really, just different numbers. How many times would you have to flip a coin to be reasonably certain it has landed on heads at least once? Two times? That doesn’t sound right. 

so flip #1, there’s a 50% chance it lands on heads. 
flip #2 is the same, but I don’t think we can add the probabilities together, you have to multiply them, and then add that to the original probability. 
So I’m flip #2 you have a 75% chance of having landed a heads at least once. Flip number three would add 12.5% so 87.5 total chance of having at least one heads. Yeah this is sounding better. What would that be as an equation...

P=0.5^n + 0.5^(n-1) + 0.5^(n-2)...

where n is the number of flips?

hmmm that’s not quite the notation, what happens when n is less than 0? I guess you could just add a clause “where n is not less than 0”. But I think there should be a better way to do that and I think there has to be a way to notate that without just doing ‘...’

maybe something to do with limits. 
ooooh yeah, limits are definitely the answer, but it’s 3:22 AM and I’m definitely overthinking this in the wrong direction so ima sleep and come back to this. 

Edited by Danex
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5 hours ago, Danex said:

Pi is bad

Tau forever. 

As for the actual question, I think it is as simple as it sounds. 
If you choose one random date out of 365 possible dates, and then choose one truly random person, there is a 1/365 chance that they’d have that birthday. So if you gathered 365 people, you’d have a 1/365 * 365 chance. Which would be equal to 1, or 100%.
 

I think some people were referencing the famous birthday paradox scenario, which isn’t actually a paradox, just counterintuitive. It asks a similar question: How many people would you need to gather to be certain that any two of them shared a birthday. It’s a surprisingly low number. I don’t remeber the exact math, but it has to do with you not counting the number of people, you need to count the number of pairs. And the required number of pairs can be reached with a low number of people. Next part is me trying to figure it out from memory:

I think it’s 27. If you have 27 people, there is something like a 99% chance that 2 of them share a birthday. If you have 27 people, there are 351 possible pairings (I think) 351/365 equals 96%

 

 

hmmmm wait I just thought about the original question some more and I’m less sure. 

What if we simplify the question to a coin flip. Same principle really, just different numbers. How many times would you have to flip a coin to be reasonably certain it has landed on heads at least once? Two times? That doesn’t sound right. 

so flip #1, there’s a 50% chance it lands on heads. 
flip #2 is the same, but I don’t think we can add the probabilities together, you have to multiply them, and then add that to the original probability. 
So I’m flip #2 you have a 75% chance of having landed a heads at least once. Flip number three would add 12.5% so 87.5 total chance of having at least one heads. Yeah this is sounding better. What would that be as an equation...

P=0.5^n + 0.5^(n-1) + 0.5^(n-2)...

where n is the number of flips?

hmmm that’s not quite the notation, what happens when n is less than 0? I guess you could just add a clause “where n is not less than 0”. But I think there should be a better way to do that and I think there has to be a way to notate that without just doing ‘...’

maybe something to do with limits. 
ooooh yeah, limits are definitely the answer, but it’s 3:22 AM and I’m definitely overthinking this in the wrong direction so ima sleep and come back to this. 

Why Tau?

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I know the algorithm! y=y=y=y=x=x=y=r=a=a=d=a=answer

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22 minutes ago, Condensation said:

Why Tau?

Tau = 2pi

and it makes everything so much better 

Why have circumference be 2*pi*r
when you cold just have Tau*R

but more importantly it makes radians and trig SOOO much easier 

because one whole circle equals one Tau
so the numbers actually go where you think they should go

 

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2 minutes ago, Danex said:

Tau = 2pi

and it makes everything so much better 

Why have circumference be 2*pi*r
when you cold just have Tau*R

but more importantly it makes radians and trig SOOO much easier 

because one whole circle equals one Tau
so the numbers actually go where you think they should go

 

Oh, that is rather nice.

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I like pie and I'm a Taurus. So.... there's that. :ph34r:

I'm trying to remember rolling probability and I can't remember how to do it with a 6 sided die,  let alone 365 outcomes. How many times do I need to plan to roll a die to be almost certain I will roll a 6 at some point? 

There's no 100%, because there's veeeeeery small unlikely chance I could technically never roll a six.

Edited by Zelly
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3 hours ago, Zelly said:

I like pie and I'm a Taurus. So.... there's that. :ph34r:

I'm trying to remember rolling probability and I can't remember how to do it with a 6 sided die,  let alone 365 outcomes. How many times do I need to plan to roll a die to be almost certain I will roll a 6? 

No idea.

Ha!

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So I think

14 hours ago, Danex said:

maybe something to do with limits. 
ooooh yeah, limits are definitely the answer

This is what I keeping running into.

Like if I don't define a set, there's just an infinite number of possible outcomes.  I could gather increasing numbers of people and never have someone with my birthday or on the other side of the spectrum I could have everyone share my birthday. And then all the possibilities inbetween the two.

So I'll revise my question with a limit.  Let's say I have 30 people (including myself).  At that number, we can be 99% sure that two random people share a birthday.  But what is the probability that one of the 29 others share my specific birthday?

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AGHHHHHHH

I think I finally got it.

I haven’t been able to think about literally ANYTHING else for the past 24 hours because this seems like it should be a really easy question to answer, but it was stumping me

heres the equation for the coin flip example 

P=1-0.5^x 

plug in the number of flips for x and you get the probability of having at least one heads by that many flips. 
and theoretically you should be able to replace the 0.5 with whatever the starting probability is if you did it once. So instead of 1/2 for the coin flip you’d do 1/365 for the birthday thing. 

EXCEPT THAT DOESNT WORK

AND I HAVE NO IDEA WHY

AT ALL

 

 

waitwaitwaitwaitwait

this is starting to sound kinda like the uhh whatsitcalled...Compund Interest formula

or no, what’s the other one 

Continuous interest?

 

 

update:

I AM TEACHING MYSELF ABOUT RECURSIVE SEQUENCES VIA AN OLD MATH TEXTBOOK

IM SO CLOSE

update:

AAAAAGGGHHH

I GOT IT

Its a recursive sequence.

P{n}=0.5^n+0.5{n-1}
with brackets as subscript
and you can successfully put in any original chance of something happening where the 0.5 is.

calculator notation is
u(n)=0.5^(n)+u(n-1)

So. To finally answer OP’s question.....

yeah idk. I guess it’s just guess and check by plugging in numbers to the calculator over and over until you get as close to 1 as possible, 
but my TI-84 Plus CE won’t even handle numbers that big. 
but I think I can answer the revised question, sorta. 
If you have 30 people, you can be 0.274% sure than one shares your birthday. Which isn’t the same as just, the odds of someone having your birthday, that would be 1/365 times 30, or about 8.2%

.......I really hope that’s all correct, otherwise I’m actually going to go insane. 
 

 

*deep breath*
its not. *internal shrieking*
i did further testing with an event that has a 0.25 chance of occurring 

instead of the horizontal asymptote being at 1, it’s at 0.333 repeating. So according to my calculator, with an event that has a 25% chance of happening, even if you run the event infinite times, you still can only be 33% sure that it actually happened. And that’s obviously not right.

and for the 1/365 it’s at like 0.0027 something. 

but it works for the lower numbers! I think? If you roll a 4 sided die twice, you have a 0.3125 chance of being sure that a certain thing was rolled. Right? Maybe all my base assumptions about this are wrong. How do you calculate the probability of an certain event occurring by a certain number of trials?? As opposed to just the probability of the event happening in general. That’s the whole question here. And I thought I was right but that means the numbers should all be approaching but never reaching 1, not 0.33 or 0.0027. Wagwhahfwhahwhffahwhwhfgagaaaaaa

can someone who’s actually smart come save me please

Edited by Danex
end my suffering I want to be able to think about other things again please
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Spoiler
9 hours ago, Danex said:

AGHHHHHHH

I think I finally got it.

I haven’t been able to think about literally ANYTHING else for the past 24 hours because this seems like it should be a really easy question to answer, but it was stumping me

heres the equation for the coin flip example 

P=1-0.5^x 

plug in the number of flips for x and you get the probability of having at least one heads by that many flips. 
and theoretically you should be able to replace the 0.5 with whatever the starting probability is if you did it once. So instead of 1/2 for the coin flip you’d do 1/365 for the birthday thing. 

EXCEPT THAT DOESNT WORK

AND I HAVE NO IDEA WHY

AT ALL

 

 

waitwaitwaitwaitwait

this is starting to sound kinda like the uhh whatsitcalled...Compund Interest formula

or no, what’s the other one 

Continuous interest?

 

 

update:

I AM TEACHING MYSELF ABOUT RECURSIVE SEQUENCES VIA AN OLD MATH TEXTBOOK

IM SO CLOSE

update:

AAAAAGGGHHH

I GOT IT

Its a recursive sequence.

P{n}=0.5^n+0.5{n-1}
with brackets as subscript
and you can successfully put in any original chance of something happening where the 0.5 is.

calculator notation is
u(n)=0.5^(n)+u(n-1)

So. To finally answer OP’s question.....

yeah idk. I guess it’s just guess and check by plugging in numbers to the calculator over and over until you get as close to 1 as possible, 
but my TI-84 Plus CE won’t even handle numbers that big. 
but I think I can answer the revised question, sorta. 
If you have 30 people, you can be 0.274% sure than one shares your birthday. Which isn’t the same as just, the odds of someone having your birthday, that would be 1/365 times 30, or about 8.2%

.......I really hope that’s all correct, otherwise I’m actually going to go insane. 
 

 

*deep breath*
its not. *internal shrieking*
i did further testing with an event that has a 0.25 chance of occurring 

instead of the horizontal asymptote being at 1, it’s at 0.333 repeating. So according to my calculator, with an event that has a 25% chance of happening, even if you run the event infinite times, you still can only be 33% sure that it actually happened. And that’s obviously not right.

and for the 1/365 it’s at like 0.0027 something. 

but it works for the lower numbers! I think? If you roll a 4 sided die twice, you have a 0.3125 chance of being sure that a certain thing was rolled. Right? Maybe all my base assumptions about this are wrong. How do you calculate the probability of an certain event occurring by a certain number of trials?? As opposed to just the probability of the event happening in general. That’s the whole question here. And I thought I was right but that means the numbers should all be approaching but never reaching 1, not 0.33 or 0.0027. Wagwhahfwhahwhffahwhwhfgagaaaaaa

can someone who’s actually smart come save me please

 

Yikes. Um, let me find that book I had. It will explain things much better than I will.

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So, I think I figured this out.

It's easiest if you think of it as dice rolls, as in "how many 365-sided dice do I need to roll so that the probability of at least one die coming up a given value is between 98-99%?"

I started by trying to find the formula that tells us, "given x n-sided dice, what is the probability of at least one die coming up a given value?" Probability is (# of desired outcomes) / (# of total outcomes). Total outcomes for dice is actually really simple; it's just n^x. Desired incomes, after a bit of math, turned out to be the function D(x) = (n - 1)(D(x - 1)) + n^(x -1), where D(1) = 1. Huzzah! The formula we want turns out to be (n - 1)(D(x - 1)) + n^(x -1) / n^x.

So, plug in the numbers, and we need to solve 0.98 < 364(D(x -1) + 365^(x - 1) / 365^x < 0.99

That's all for now, cos I have a class soon, but I shall return!

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1 hour ago, FriarFritz said:

So, I think I figured this out.

It's easiest if you think of it as dice rolls, as in "how many 365-sided dice do I need to roll so that the probability of at least one die coming up a given value is between 98-99%?"

I started by trying to find the formula that tells us, "given x n-sided dice, what is the probability of at least one die coming up a given value?" Probability is (# of desired outcomes) / (# of total outcomes). Total outcomes for dice is actually really simple; it's just n^x. Desired incomes, after a bit of math, turned out to be the function D(x) = (n - 1)(D(x - 1)) + n^(x -1), where D(1) = 1. Huzzah! The formula we want turns out to be (n - 1)(D(x - 1)) + n^(x -1) / n^x.

So, plug in the numbers, and we need to solve 0.98 < 364(D(x -1) + 365^(x - 1) / 365^x < 0.99

That's all for now, cos I have a class soon, but I shall return!

That is awesome. Good job!

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I have returned, and created a better formula!

See, I was having trouble figuring out how to find D(x) without just running the sequence (which could get pretty nasty), so I started looking for patterns with low-sided dice (2, 3...), and it turns out you can also find D(x) by this formula: D(x) = n^x - (n - 1)^x

So our equation becomes .98 < (365^x - 364^x) / 365^x

I think I can solve now.

I am wrong. The numbers are too big for Desmos or my calculator. I must find the formula for (a - 1)^x / a^x

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3 minutes ago, FriarFritz said:

I have returned, and created a better formula!

See, I was having trouble figuring out how to find D(x) without just running the sequence (which could get pretty nasty), so I started looking for patterns with low-sided dice (2, 3...), and it turns out you can also find D(x) by this formula: D(x) = n^x - (n - 1)^x

So our equation becomes .98 < (365^x - 364^x) / 365^x

I think I can solve now.

I am wrong. The numbers are too big for Desmos or my calculator. I must find the formula for (a - 1)^x / a^x

Try doing it on Google, that's what I do.

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