Folks Of Interest A call to arms for an injured HAMBER - FREMONT, CA

(Cue Twilight Zone theme.....)

So not only were there 3 Daves yesterday,now I find out 3 of us have the same Bday of November 9

 

That's a great deal guys! Glad you guys helped a fellow HAMBer out! Stuff like this makes me proud to be back on the HAMB!
Get better soon,Maxwedge...

 

WHOA!! Thats Insane!!

 

What the Hell are the odds of something like that?!

 

good going, guys..the way it ought to be

 

You guys did good.

 

What the Hell are the odds of something like that?!

I dont know...but maybe we should pool some money and get some lotto tickets

 

Exactly!

 

Way to go guys. Love it. Get well soon Mr. Wedge! Hey...what was in the crate that was so heavy?...an engine or two? It's gotta be cool if it's for the '34!! Rick

 

fenders,Hood,Running Boards for the 34

 

What was in the crate? Was it a leg lamp? I hope it was a leg lamp.
Or perhaps some other form of major award!

 

It was in fact a leg lamp...lol!

 

Cool! Electric ***. Glad to hear it all worked out.

 

OK, you asked. Only this is the odds of having two people with the same birthday. So the odds for the three are gonna be WAY out there.


Uber geek alert -

How many people do you need before the odds are good (greater than 50%) that at least two of them share a birthday?

Answer: 23 people



Hide Solution
Solution: Let’s figure the odds that no one shares a birthday and invert that. The odds are calculated by counting all the ways that N people won’t share a birthday and dividing by the number of possible birthdays they could have.
For example, two people could have 365×365 birthday combinations. That’s the denominator. To count the numerator, imagine that the first person gets to choose their birthday. They can pick from 365 days. The second person can also pick their birthday, but can’t share a birthday with the first person. They’ve got 364 days to choose from. So the chance that two people don’t share a birthday is (365×364)/365². Subtract that from 1 and you get what you expect: that there’s a 1 in 365 chance that two people share a birthday.
For three people, the denominator is 365³ and the numerator is 365×364×363. The formula for N people is:
P(N) = [365 × 364 × · · · × (365−N+1)] / 365^N
We could have used factorials to express the numerator, but this way works better. This will be explained later.
We’re trying to solve this equation:
1 - P(N) = 0.5
I don’t know how to do that. There’s probably a way to solve equations that involve factorials, but I couldn’t find it on the Net, so I wrote a quickly Python program to try every value of N:

for N in range(1, 365 + 1):
p = 1.0

for i in range(1, N + 1):
p = p * (365 - i + 1) / 365

p = 1 - p

print N, p
</pre> This goes over 50% at 23 people. Using double precision floats, the output says “1” after 134 people, but the probability doesn’t really reach 100% until 366 people. If anyone knows how to solve this equation algebraically, please let me know.
When we first put up this page, we had the right answer but the wrong explanation. Martin Ziegler pointed that out to me (thanks!), and proved it by showing that my solution gave the wrong answer for N = 366. You should get a probability of 1, but our answer gave a probability just below that. This is my original solution, followed by a comparison of the probabilities you get from each.

If you have two people, the chance that they share a birthday is 1/365. If you have three people (A, B, and C), then you’ve got three ways (AB, BC, and AC) that birthdays could be shared. Since you want any of the three pairs to share, you can’t just add or multiply the probabilities. Instead you consider the chance that they don’t share a birthday (1 - 1/365 = 364/365) and multiply that by itself three times. If you have four people, then they could pair up in six ways, so the probability that none of those pairs shares a birthday is (364/365)⁶. The probability that any do share a birthday is 1 minus that. We want to keep increasing N, the number of people, until that probability reaches 50%.
Given N you can calculate the number of pairs with N-choose-2, meaning given N items, the number of ways to pick any two of them. It’s usually written as the letter N sitting over the number 2 with large parentheses around both. (When MathML becomes better supported these pages will become clearer. Or at least better typeset.)
The formula for A-choose-B is A! / [B! (A - B)!]. I remember that formula like this: N! is the number of permutations of N items. Imagine all the permutations of the A items we’re trying to choose from, each permutation on one line. There are A! of these lines. For each line, keep the first B items and throw away the other A - B items. You’ll be left with a lot of duplicate lines, so get rid of duplicates. For each unique line left over you threw away (A - B)! lines. Now ignore the position of the items (since we don’t care about the order of the B items we’re choosing), and get rid of duplicate lines again. For each unique line left over you threw away B! lines. That’s how you get that formula.
We have N people and want to find out how many pairs of people there are. That’s N-choose-2, or N! / (2! (N - 2)!). That simplifies to N(N - 1)/2. The probability that no two people share a birthday is:
(364/365)^{N(N - 1)/2}
So the probability that any two people do share one is:
P(N) = 1 - (364/365)^{N(N - 1)/2}
We want to find out the break-even point for P(N) = 0.50, so we plug that in and use the quadratic formula to solve for N, giving us 22.98, or 23 people.
​

Here's a modified Python program to dump both values, and this is the graph of the first 50 values of N:
<center></center>
They’re amazingly similar considering that the approaches seem so different. The wrong solution ***umes that all birthday pairs are independent, but that isn’t true. If persons A and B don’t share a birthday and B and C don’t either, then the chance that A and C share a birthday is affected by that information. (Think through the case where there are only three days in the year to choose from.)


In the formula at the top they avoided using factorials in the probability function because when N = 366 they wanted the numerator to be equal to 0, but 0! = 1, which messes up that corner case.
It’s not really a paradox, it’s just that most people expect the answer to be much higher, like 365/2 or something.



I warned you.

Vance

 

My head just literally exploded! lol...

 

you are one sick individual.......I lake that !!

 

And all these years I thought I was drinking alone?

Like "THE HANGOVER" lets go on the roof and be blood brothers, who's got the knife?

Brian

 

VANCE~ and take in to consideration of how many people have you been told what their birthday is? To have a nametag with your birthday is one thing. Like how many daves do you know compared to how many Daves p*** you by but you don't know their names. If you had a nametag with your birthday and you would say hi my birthday is _____ to open conversation you may know a few, but to meet, BS for a while and have the same hobby,

HAMB is unbelievable.

Brian

 

We all need to get together again someplace with our cars..Been trying to make that happen...

 

Well, to figure that would just be crazy. But I'll look into it.

Vance

 

make it happen....In & Out.... Morgan Hill sounds good...kinda half way

 

The Hot Rod is at Georges fixing the Weld on the steering box bracket..The Max Wedge Has 4.56 gears in it and runs 110 Octane leaded, it would cost me a small fortune to drive to Morgan Hill and back...lol

 

dude don't sweat the small stuff....details details.....

 

You have a big trunk for the gas cans.....

Name the date and place and I'll be there.

 

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LOL... I was hoping for Nation Burgers here in Fremont Tomorrow Night...

 

So when do we get to hear the Revtones "cover" of these are the Daves I know ?
Should be a billboard top 5 ! LOL

Uhh. Dave

 

I'll have to throw a huge Birthday bash for you guys!

 

^^^and you put that in print.....you know we'll hold to your word

 

No Problem!!