Mind # Question

[sxk1693]

This is a confusing question for me, but probably not for the geniuses on this forum:

A person enters a party that has 20 other people. What is the probability that one of the party members shares the same birthday with the newcomer?

Along the same lines,

What if the person entered another party with 380 party members. What is the probability that one of the party members shares the same birthday with that one person.

On these problems, ignore leap years(meaning no Feb.29 as a birthday) and assume a year has 365 days.

Note: I do not know the answer to these questions, so let the debate begin.
By the way, I tried looking this up on the internet, but there was nothing there, so I decided to post it on Chit Chat and Stuff. I hope you guys do not think of this as wasting forum space. HAVE FUN!! (WARNING: this problem might cause severe brain damage, especially the 380 party members one. It did for me.)

[sxk1693]

Why is "blo-wing" censored?

[darksaber21]

My coach last year told me that the first one is close to 50% (I think the exact percentage is like around 41.1%, found that out on my own days later). but the second one makes me think the probability is much higher, with more people that is.

Assuming that the likelihood of one person to have a birthday on one day is the same as another (meaning all 365 days are equally probable), in a party with 380, I would say the probability is like 99.99999%, although theoretically should be 100% (considering there are more people than days, haha). But since when are probabilities in situations like this ever 100.

That's just what I think.

[sxk1693]

[darksaber21]

That is true. Well, let's look at it from the other side, and determine the probability of someone not having the same birthday.

Let's say there is two people, including yourself in the room, right? Well, the probability of the other person not having the same birthday is [unparseable or potentially dangerous latex formula], meaning the probability of the other person having the same birthday being [unparseable or potentially dangerous latex formula]. But let's say there is three people in the room. The probability would be [unparseable or potentially dangerous latex formula], but the probability for someone to have the same birthday is not [unparseable or potentially dangerous latex formula], because the sums in that case would not add up to one. I don't know where the other missing part have went, but since we are only looking at two possibilities, the two probabilities should just add up to one.

So, we rely on the probability of someone not having the same birthday to determine the probability of someone having the same birthday, mostly since its better to deal with the negatives than the positives, probably.

Will that work?

[101dalmatians]

Let us all think... Is there going to be a question like thison any test - in what chance? IDK but i know not very high a chance...
If it does occur, then both questions have a lesser chance of occuring on the same test...
So, I conclude that for these kinds of problems, my skipping strategy works the best!
What do you all think?
I mean if I tried this during a math test, I would be wasting my time!!!

[Zubair_Marediya]

Hi guys! I think I may have a possible solution. First off, @darksaber21, to answer your last post, add up the following numbers: (1/365)^2 + 2(1/365)(364/365) + (364/365)^2. It should add up to one. You were right about both of the other members having the same birthday, which is accounted for with the (1/365)^2 part, and about them not having the same birthday, which is accounted for with the (364/365)^2 part. The other missing part occurs when "member A" has the same birthday while "member B" does not. There are two different ways to set this situation up so 2(1/365)(364/365) accounts for this.

To build upon the problem above, if one simply wanted to calculate the probability of at least one member having the same birthday, they could simply do the following: 1 - (364/365)^2. Subtracting away the probability of no members having the same birthday gives the probability of at least one member having the same birthday.

In conclusion, the setup that we should extract from this is the following: 1 - (364/365)^(Number of members). So if there are twenty other members, you find that 1 - (364/365)^20 will give you about a 0.0534 probability of finding at least one other member with the same birthday. Similarly, if there are 380 other members, then you find that 1 - (364/365)^380 gives you about a 0.647 probability of finding at least one other member with the same birthday.

To test this, I wrote a function in a software that I have called MATLAB. In the function, there are two inputs, the number of simulations and the number of members. The function makes a row vector with a random numbers ranging from 1 to 365 for the number of simulations that wish to be ran. The length of this row vector is (Number of members + 1). The first number in this row represents the birthday of the newcomer, and the rest of the numbers represent the birthdays of the members. Here's an example if you are confused: If there are 5 members at the party, then the function will make an array like [4 75 32 98 277 123]. The "4" represents the birthday of the newcomer and the following five numbers represent the birthdays of the members. The function runs through all the simulations and outputs the results of finding at least one other member with the same birthday.

For the 20 member problem, running the code 10 times with 10000 simulations gave the following results: 0.0552, 0.0542, 0.0533, 0.0529, 0.0543, 0.0597, 0.0521, 0.0532, 0.0524, 0.0510. You can see that these numbers are pretty close to the expected 0.0534 probability.

For the 380 member problem, running the code 10 times with 10000 simulations gave the following results: 0.6485, 0.6523, 0.6511, 0.6485, 0.6454, 0.6466, 0.6408, 0.6407, 0.6382, 0.6473. You can see that these numbers are pretty close to the expected 0.647 probability.

I hope this is correct and I hope it helps!

P.S. Sorry about not using latex. I was just too lazy to figure it out right now!

[sxk1693]

Wow, I think that works. That is a really cool way to test for the probability. I should have thought about using Java. Like I said, I do not know the answer, but this seems correct unless someone else wants to argue otherwise.

[darksaber21]

Yeah, is it hypothetically correct to me, and to this article that I found .

It covers what we just talked about and more, and extremely in depth too. That is why wikipedia is awesome with information such as that.

[sxk1693]

Yes! I think I figured out an alternative solution that is along the lines of Zubair's solution.

We can use the binomial probability thing that I just learned a few days ago for math uil.

[unparseable or potentially dangerous latex formula] represents the probability of a situation happening where a represents the probability of a situation happening and b represents the probability of the situation not happening. N represents the number of people in the party in this case.

[unparseable or potentially dangerous latex formula] in this case is [unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula] is [unparseable or potentially dangerous latex formula]

So, [unparseable or potentially dangerous latex formula] can be represented as [unparseable or potentially dangerous latex formula]

The binomial expansion would yield:
[unparseable or potentially dangerous latex formula]

The last term represents the probability of no one having a birthday that coincides with the birthday of the newcomer.
Therefore, like Zubair said, the probability that at least one person will have the same birthday is:

[unparseable or potentially dangerous latex formula] where n represents the number of people at the party already.
This is just another approach to get the same expression. You guys found the same expression another way. Anyways thanks for hurting your brains over this problem.

[sxk1693]

Does anyone know why my latex is so small? I'm new to this.

[darksaber21]

Either use "\displaystyle" right after the first "tex," or use "\dfrac" as oppose to "\frac" for fractions. "\displaystyle" typically makes the fractions, limits, and other stuff work a lot better.

And, use "\left" and "\right" with parenthesis or brackets so they will adjust size to fit what they are going around.


All without quotes of course.

[Zubair_Marediya]

No problem! I just wanted to clear one thing up. The solution that I wrote calculates the probability of at least one party member having the same birthday as the newcomer. The Wikipedia page calculates the probability of any of the people in a group having a birthday that matches any of the others. I wasn't really sure which situation you were asking about, so I answered the one I assumed you meant. Either way, it seems that the solution to both are in your hands!