The Birthday Problem/Paradox for Dummies

[Dilb]

I think Just Thinking is confused by your simplified fraction. 365! is, per Wolfram Alpha,

25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

so the numerator is somewhat more awkward in a naive calculation.

 

[W.D.Clinger]

Don't the number of zeros at the end of a power-of-ten factorial have a long string of nines at the end?

1000000! has 249998 zeros at the end.
1000001! has 249998 zeros at the end.
1000002! has 249998 zeros at the end.
1000003! has 249998 zeros at the end.
1000004! has 249998 zeros at the end.
1000005! has 249999 zeros at the end.
...
999999999999! has 249999999985 zeros at the end.
1000000000000! has 249999999997 zeros at the end.
1000000000001! has 249999999997 zeros at the end.
...
(10^50)! has 24999999999999999999999999999999999999999999999989 zeros at the end.
...

 

[W.D.Clinger]

I think Just Thinking is confused by your simplified fraction. 365! is, per Wolfram Alpha,

25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Yep, that's what I got.

so the numerator is somewhat more awkward in a naive calculation.

But the naive calculation is a lot more awkward than reducing the fraction to lowest terms at each step of the calculation, especially when you don't have pencil and paper handy so you have to do all the calculating in your head.

 

[Dilb]

? You can multiply 22 three-digit numbers together in your head?

 

[W.D.Clinger]

? You can multiply 22 three-digit numbers together in your head?

No, I'm not that good. I use Karatsuba's algorithm.
(As implemented on a computer.)
http://www.larcenists.org/

 

[UnrepentantSinner]

Way back when I was at university studying math, a professor mentioned this. There were about 40 students, and he said he was willing to bet that at least 2 of us were sharing a birthday. So we pointed to the twins in the front row: "Yeah, it's those two".

 

[lionking]

I absolutely love this problem. I often raise with very well paid management people, and I haven't found anyone who knew the answer. I've even had some tell me I'm obviously wrong. When I ask people how many pairs can you form from four people I haven't had anyone say 6. Maths is fantastic.

 

[Lothian]

Okay, making as much fun of my lack of math comprehension as you choose, can anyone explain the probability of the birthday paradox in terms that someone who's math-challenged can understand.

Forget Wiki. I have a hangover and it's far too thick(or I am).

Is the birthday paradox

'The older a woman gets the larger the difference beween her age and the number of birthdays she has had.'

Or is that the birthday rule?

 

[Lothian]

You can multiply 22 three-digit numbers together in your head?

I can, as long as one of them is 000.

 

[Foolmewunz]

bump for Nobby Nobbs.

 

[Skeptic Ginger]

Thanks, guys. I sat and did some of the math, and it actually checks out once I got an understanding of the "series" (if that's the right term) I was looking at. And it does come out to somewhere just over fifty persons to come out to a 99% chance that you'll find two people matching.

Now comes the difficulty of explaining it to someone. I trusted the fact that it worked, but I was making a dog's breakfast of the explanation.

The way I look at it (and the math people can tell me it's wrong, I might be but it is how I see the logic) is each half of the match, person who matches the other person, contributes to the odds, rather than the pair being the subject of the odds.

 

[Skeptic Ginger]

...Maths is fantastic.

And probability can be downright counter-intuitive.