Fun math problem!!!!1111jfklsaflkhlkaklhlsdjkf

[Spinaroonie]

You're having a dinner party. You're apparently on blood thinners or something so you've decided that you want the odds of two of your dinner guests having the same birthday (doesn't have to be year) to be 50:50.

How many guests do you have to invite?

I'll tell you later.

 

[Dixon Carter Lee]

Well, the probability is the same if you have two guests or two thousand, isn't it?

 

[Willing and Unsure]

why does this wreak of the birthday problem from my probability and stats class?

 

[Dixon Carter Lee]

"Reek".

 

[VanB]

The probability of any one guest having a birthday on a particular day is 1 in 365.25 (including leap years) and you want to have a 50% probability of 2 guests having the same birthday.

Hope you have a big budget because it's a lot of guests ( I might be wrong but pretty sure it's 181.25 so you need a midget in attendance)

 

[RosevilleCAguy]

You're having a dinner party. You're apparently on blood thinners or something so you've decided that you want the odds of two of your dinner guests having the same birthday (doesn't have to be year) to be 50:50.

How many guests do you have to invite?

I'll tell you later.

Only one if you are a twin.

 

[RosevilleCAguy]

Never mind. My drugs are wearing off.

 

[Bob_Bytchin]

You have to have 23 people in a room in order to have a 50/50 chance of having two people with the same birthday.


You get this result by using the "probability of collisions."

Mainly, as more people enter the party, the chances of two people NOT having the same birthday decreases, and the probability of them having the same birthday increases.

Once you hit 23 people, the odds are just about even.

 

[phatcat]

*sigh*

DCL is right: probability will stay the same no matter the #
However, _if_ you already had 182 guests none of whose birthdays matched, and you invited 1 more, then it would be 50:50

If I was on bloodthinners and I set out to accomplish this at the expense of ruining myself, I'd invite 149 people.

(your answer: guaranteed to be silly)

 

[seXieleXie]

Re: *sigh*

DCL is right: probability will stay the same no matter the #
However, _if_ you already had 182 guests none of whose birthdays matched, and you invited 1 more, then it would be 50:50

If I was on bloodthinners and I set out to accomplish this at the expense of ruining myself, I'd invite 149 people.

(your answer: guaranteed to be silly)

oh-so-condescending one, you're incorrect. the more guests you have at the party the more likely it is that two people will have the same birthday.

to actually do the math for this would take a long time so i'll shorten it by illustrating the principle with birth *months*

start with 2 people. first of all there are 144 possible combinations for months (12*12=144). of those 144 possible combinations 12 are the kind we want (jan + jan, feb + feb, etc) so the probability of two people being born in the same month is 12/144 or .083 (about 8%). this never changes no matter how many people are in a group, the probability of any two individuals having the same the same birth month is constant.

however, that isn't what this problem is asking. so let's look at a three person situation. with 3 people there are 1728 possible combinations (12*12*12-1728). out of those 1728 possibilities 408 are desirable (at least two people in the same month, possibly all three). if anyone is interested in how i got that number i will post that, but i suspect no one really cares

anyway so if you take 408/1728 you find the probability of at least two of the people having the same birth month is .236 or about 24%. you can see that the denominator (bottom #) has increased by a factor of 12 while the numerator (top #) has increased by a factor of 34, so obviously the probability would increase drastically.

it works the same with birth days as well.

 

[Meltzer]

damn it this was supposed to be fun. the thread said so. liar!!!!

 

[Ð¥§ƒµñ熡øñæ£]

You're having a dinner party. You're apparently on blood thinners or something so you've decided that you want the odds of two of your dinner guests having the same birthday (doesn't have to be year) to be 50:50.

How many guests do you have to invite?

I'll tell you later.

2^13466917-1 is prime

 

[Guru]

 

[Lady X]

can't we just have a party?

 

[Laurel]

damn it this was supposed to be fun. the thread said so. liar!!!!

No shit. I want my money back.

 

[SilvaTungDevil]

Math is hard


flowers are pretty

 

[Bob_Bytchin]

Sheesh. You right-brained guys. LOL...

If there are 3 people, the probability would be: 2/365 + 1/364
If there are 4 people, the probability would be: 3/365 + 2/364 + 1/363

So the problem resolves to finding x such that:
(x-1)/365 + (x-2)/364 + (x-3)/363 + (x-4)/362 + ... + 1/(366-x) >= 0.5

I could probably work that out, eventually... but instead, I'll write a quick program...

The answer is 21 people.

Do the problem again...it's 23.

 

[Spinaroonie]

Just wanted to say...

probability != odds

 

[Guru]

 

[Bob_Bytchin]

Using the formula:

(365 - n + 1)/365) where n= the number of guests at the party, you get the result of .5073 when n=23

 

[Guest]

Do the problem again...it's 23.

_________

Bob is right; 23 is the answer (good work!).

Check out this paper:

(You need Acrobat Reader to view it):


http://pumas.jpl.nasa.gov/PDF_Examples/06_07_99_1.pdf

 

[Guest]

 

[Spinaroonie]

Bob is right; 23 is the answer (good work!).

Check out this paper:

(You need Acrobat Reader to view it):


http://pumas.jpl.nasa.gov/PDF_Examples/06_07_99_1.pdf

Interesting paper.. however... if you look at the chart... You see the keyword- Probability.

 

[freakygirl]

Is there a party or not?

 

[dreamer]

You're both on the right track...and I haven't run the numbers but this may be a case of using or not using leap year...and rounding.

The basic idea without getting TOO geeky is that in a room full of people the first person you examine has the whole group to 'try' and find a match with. His odds are the number of people he asks divided by the number of days in a year...366.

NOW...you ADD his odds with everyone elses odds to get the total chance of ANY two people having the same birthday.

How do you find the odds of everyone elses chance? Imagine guy number two he can go around the room and ask everyone BUT the first guy to find a match (note we already figured the odds of guy number one and guy number two sharing a birthday when when guy number one went around the room)

SO guy number 2's odds are the number of total people in the room not counting himself AND the guy that went before him divided by the number of days in the year again...

Guy number three does the same thing but again doesn't count the people in front of him...

Repeat this until the second to last guy....he has ONE person to try...the LAST guy that hasn't asked him...then the last guy has no one to ask because he was asked by everyone...

This can be represented by:

x=the total number of people in the room.

x-1/366 (remember the guy doesn't ask himself, hence the -1)
+
x-2/366 (second guy doesn't ask himself or the guy that started this mess)
+
x-3/366
+...

now to solve for x where this total is greater than or equal to .5 (also known as 50%)

I think you cross the .5 barrier at twenty people in the room actually...hmm..