Another Problem for Fun

[Debater Slayer]

In a room of how many people is there a 50% chance that at least two of them share the same birthday?

This is known as the birthday paradox, but if you know, calculate, or google the answer, please only post it behind spoiler tags!

 

[Polymath257]

A much harder problem: how many people are required to have a 50% chance that *three* people have the same birthday?

 

[ChristineM]

Ah ha, i know this one, it went the rounds of our studio some years ago.
And i didn't believe it then
So I'll leave it there

 

[Polymath257]

In a room of how many people is there a 50% chance that at least two of them share the same birthday?

This is known as the birthday paradox, but if you know, calculate, or google the answer, please only post it behind spoiler tags!

A classic problem with a surprising answer.

Another surprise: if you double the number of people, the probability goes up to 95%.

 

[Debater Slayer]

A much harder problem: how many people are required to have a 50% chance that *three* people have the same birthday?

This is absolutely nasty, but after thinking about it, this is how I would approach it:

Spoiler

Frame the problem using a Poisson distribution where k = 0 and lambda = the probability that at least three people share the same birthday. To find lambda, use nC3/(365)^2. This is because the probability of two people sharing the same birthday is 1/365. Adding a third person, this becomes 1/365 x 1/365 = 1/(365)^2.

Finally, the above calculates the probability that none of the outcomes will be the one we want, so, to find our desired outcome, we will calculate 1 - the above.

 

[Left Coast]

In a room of how many people is there a 50% chance that at least two of them share the same birthday?

This is known as the birthday paradox, but if you know, calculate, or google the answer, please only post it behind spoiler tags!

If this is fun for you, I have no clue what they would do with you in hell.

 

[Debater Slayer]

If this is fun for you, I have no clue what they would do with you in hell.

They would make me read sociology papers.

I deeply enjoy math, but I have considerable difficulties with certain other subjects.

 

[Secret Chief]

I don't think I can cope with this relentless fun.

 

[Shadow Wolf]

I used to know this one. But now I don't so I'll just say 0% chance if it is actually my birthday because no one is stealing my thunder.

 

[Debater Slayer]

I don't think I can cope with this relentless fun.

Per my calculations, there's a 1% chance you won't be able to handle this much fun.

 

[Alien826]

I looked it up, confirmed my recollection that it is a surprisingly small number, and failed to follow the math.

Something to remember when thinking about it is that it deliberately ignores things like leap years, and the fact that births are not uniform through the year.

 

[The Hammer]

In a room of how many people is there a 50% chance that at least two of them share the same birthday?

This is known as the birthday paradox, but if you know, calculate, or google the answer, please only post it behind spoiler tags!

Spoiler: Spoils be here

Isn't it like 25

 

[Debater Slayer]

Spoiler: Spoils be here

Isn't it like 25

Spoiler

23.

 

[ChristineM]

This is absolutely nasty, but after thinking about it, this is how I would approach it:

Spoiler

Frame the problem using a Poisson distribution where k = 0 and lambda = the probability that at least three people share the same birthday. To find lambda, use nC3/(365)^2. This is because the probability of two people sharing the same birthday is 1/365. Adding a third person, this becomes 1/365 x 1/365 = 1/(365)^2.

Finally, the above calculates the probability that none of the outcomes will be the one we want, so, to find our desired outcome, we will calculate 1 - the above.

Spoiler: French

Poisson is fish in french, and it makes more sense

 

[The Hammer]

Spoiler

23.

Oof. I knew it was small.

 

[It Aint Necessarily So]

In a room of how many people is there a 50% chance that at least two of them share the same birthday?

Spoiler

If there are two people in the room, A and B, the odds that they do NOT share the same birthday is 364/365 (I'm ignoring February 29th)
If there are three, A,B, and C, we have three pairs to consider, AB, AC, and BC, so the odds that no two share a birthday becomes (364/365)^3
If there are four, we have AB, AC, AD, BC, BD, and CD = six pairings, so the odds are (364/365)^6 that they do not share a birthday.
Since the exponents are 1, 3, and 6, generalizing, the odds than N people have no pair with identical birthdays becomes (364/365)^n, where n = N+ (N-1) + (N-2) ... 2 + 1

So, the question is, for what n is (364/365)^n closest to 0.5, and what N corresponds to that n. Here's where it gets shaky for me. I think the next step needs to be
log 0.5 = n(log (364/356))
-0.30102999566 = n(0.00965138567)
so n = -31.1903394966, but that seems like nonsense, so my solution would be to plug in values for n and find the one closest to 0.5

(364/365)^20= 0.94660846509
(364/365)^50 = 0.87181825733
(364/365)^100= 0.76006707381
(364/365)^200= 0.57770195669
(364/365)^250=0.50365111314
(364/365)^253= 0.49952284596

So n is about 252-253. N must be 1+2+3 ... + N until we get close to 252
For N=10, n is 1+2+3+4+5+6+7+8+9+10=55
N=15, n= 55+11+12+13+14+15= 120
N=20, n= 120+16+17+18+19+20 = 210
N=23, n=210+21+22= 253

I've seen the answer and know that it is correct, but I can't derive it mathematically.

 

[Debater Slayer]

Spoiler

If there are two people in the room, A and B, the odds that they do NOT share the same birthday is 364/365 (I'm ignoring February 29th)
If there are three, A,B, and C, we have three pairs to consider, AB, AC, and BC, so the odds that no two share a birthday becomes (364/365)^3
If there are four, we have AB, AC, AD, BC, BD, and CD = six pairings, so the odds are (364/365)^6 that they do not share a birthday.
Since the exponents are 1, 3, and 6, generalizing, the odds than N people have no pair with identical birthdays becomes (364/365)^n, where n = N+ (N-1) + (N-2) ... 2 + 1

So, the question is, for what n is (364/365)^n closest to 0.5, and what N corresponds to that n. Here's where it gets shaky for me. I think the next step needs to be
log 0.5 = n(log (364/356))
-0.30102999566 = n(0.00965138567)
so n = -31.1903394966, but that seems like nonsense, so my solution would be to plug in values for n and find the one closest to 0.5

(364/365)^20= 0.94660846509
(364/365)^50 = 0.87181825733
(364/365)^100= 0.76006707381
(364/365)^200= 0.57770195669
(364/365)^250=0.50365111314
(364/365)^253= 0.49952284596

So n is about 252-253. N must be 1+2+3 ... + N until we get close to 252
For N=10, n is 1+2+3+4+5+6+7+8+9+10=55
N=15, n= 55+11+12+13+14+15= 120
N=20, n= 120+16+17+18+19+20 = 210
N=23, n=210+21+22= 253

I've seen the answer and know that it is correct, but I can't derive it mathematically.

I somehow missed this earlier. Apologies.

Spoiler

Your method of equating log(1/2) to nlog(364/365) is indeed correct, but your post has a typo where 365 is instead 356. That's why n = -31.19 instead of the correct 252.652. Since we're talking about combinations of pairs here, we can't have 252.652 combinations. We need whole numbers, so we'll round it up to 253.

Now you have found the correct exponent. The way to derive it mathematically is to examine what it denotes: 253 is the number of combinations obtained from comparing every person in the room to the others. A different way to word this is thus: how many people produce 253 combinations when we compare the birthday of every person in the room with all of the others'?

Since we compare two people at a time, we know that in nCr, our r = 2. We need to find the n, which is the number of people in the room. You have already correctly calculated nC2 to be 253, so we can directly plug 253 into the formula for nC2 and find n:

nC2 = 253 = (n!)/2!(n - 2)!. Solving for n, this gives n = 23. So the number of people in the room is 23.