 Posers and Puzzles

1. 21 Oct '05 22:05
A company has called a meeting, in which there are enough people so that there is a chance higher than 50% of two birthdays at the same date. What is the minimal amount of people there must be at this meeting?

The answer I got when asked by my friend is quite surprising!
2. 21 Oct '05 23:48
Originally posted by fetofs
A company has called a meeting, in which there are enough people so that there is a chance higher than 50% of two birthdays at the same date. What is the minimal amount of people there must be at this meeting?

The answer I got when asked by my friend is quite surprising!
I believe the answer is 23 people.
3. 22 Oct '05 00:37
Your belief is correct. Can you explain the reasoning?
4. 22 Oct '05 11:04
Originally posted by fetofs
Your belief is correct. Can you explain the reasoning?
Well I acctually think its even less..
It's because of the great number of pairs you can examine.

23*22=506 pairs.

for the exact calculations of probability I can't remember just now, seem to have some trouble remembering my math today, to much on my mind no doubt...

but it should be something along these lines:
We are examinig a true of false question.
Any pair we examine can either have the same birthday or they don't.

This means that to get a 50% chance we must examine 365 pairs,
since given person 1 we need 356 possible person 2 have the even shot at finding a match.
this means we accually need only 20 people (19*20=380)
5. 22 Oct '05 11:14
the answer n is determined as 1 - the probability that all n birthdays are different. There are 366 possible dates, hence

1 - 366/366 *365/366* ...*(366-n+1)/366 >= 0.5
which holds for n>=23
6. 22 Oct '05 11:18
Originally posted by Mephisto2
the answer n is determined as 1 - the probability that all n birthdays are different. There are 366 possible dates, hence

1 - 366/366 *365/366* ...*(366-n+1)/366 >= 0.5
which holds for n>=23
Yes... Now I remember :-) Thanks!
However. Can you please tell me where I went wrong in my
logic. It still seem sound to me :-( (I'm a knucklehead)
7. 22 Oct '05 11:35
Originally posted by chasparos
Yes... Now I remember :-) Thanks!
However. Can you please tell me where I went wrong in my
logic. It still seem sound to me :-( (I'm a knucklehead)
I tried really hard, but I am not even capable of reading and understanding what you wrote. Pairs of what? Dates or people? I see that 19*20=380 is the possible combination of pairs formed out of 20 people, but how does that relate to >=50% as being compared to 365 (should be 366 really)? The fact that 18*19=342 which is less than 365 seems to lead you to the conclusion that 20 is the turning point. But of what, and how does that relate to 50% and not 40% or 70%?
8. 22 Oct '05 13:243 edits
Originally posted by Mephisto2
the answer n is determined as 1 - the probability that all n birthdays are different. There are 366 possible dates, hence

1 - 366/366 *365/366* ...*(366-n+1)/366 >= 0.5
which holds for n>=23
I'm sure you meant less than or equal 0.5, because it's the probability of diferent dates.
9. 22 Oct '05 13:28
Originally posted by fetofs
I'm sure you meant less than or equal 0.5, because it's the probability of diferent dates.
It is '1 - the probability of different dates' that has to exceed 0.5.
10. 22 Oct '05 18:281 edit
Originally posted by Mephisto2
I tried really hard, but I am not even capable of reading and understanding what you wrote. Pairs of what? Dates or people? I see that 19*20=380 is the possible combination of pairs formed out of 20 people, but how does that relate to >=50% as being compared to 365 (should be 366 really)? The fact that 18*19=342 which is less than 365 seems to lead you to ...[text shortened]... n that 20 is the turning point. But of what, and how does that relate to 50% and not 40% or 70%?
I'll see if I can explain. Frist of all:
I know you are right. The math is unassilable. What I want to know is where my way of thinking fails.

Now the explanation:
The total number of combinations of two days is 366*366 (=133956)
Of these 366 are the same day twice. (366*366-(366*365))

This means one pair in 366 is a pair of matching dates.
If you have 23 people with unknown birthdays you have a total of 23*22 diffrent pairs (=506 pairs) which should be more than enough to hit one that is two people with the same birthday.

I have made an error: acctually (23*22)/2 is the number of unique pairs. Since the order in which you choose two persons doesn't matter. this would push the answer I would expect to find with this line of Thought up to 28 (28*27/2=378).
This does not change my question. Why does this not work?

11. 22 Oct '05 19:02
Originally posted by chasparos
I'll see if I can explain. Frist of all:
I know you are right. The math is unassilable. What I want to know is where my way of thinking fails.

Now the explanation:
The total number of combinations of two days is 366*366 (=133956)
Of these 366 are the same day twice. (366*366-(366*365))

This means one pair in 366 is a pair of matching dates.
If y ...[text shortened]... ought up to 28 (28*27/2=378).
This does not change my question. Why does this not work?

I still find it very hard to follow the reasoning. Perhaps the problem is that you omit the possibilities of 3, 4, ... matching dates?
12. 22 Oct '05 21:58
Originally posted by Mephisto2
I still find it very hard to follow the reasoning. Perhaps the problem is that you omit the possibilities of 3, 4, ... matching dates?
I'll move this to a much simpler problem..

When rolling two six sided dice you have to roll 18 times to
have a 50% chance to get a twelve. (right?)

Now assume the year has six days then there is a total of 21
( (5*6/2)+6 )
ways to combine two days since the combination 1 and 2 is the same as 2 and 1.

Out of these 21 combinations six are the same day twice (11 22 33 44 55 66) this means you would need to pick 21/6=3,5 to have a 50% chance to get at least one?

Back to orignal numbers:
If the dice has 366 sides. You would have to roll them 66978 times to have 50% chance of getting 732. (right?)

There are 366 results with the same day twice out of 67161 ((366*365)/2 +366) you need then 67161 /366 =183,5 to get 50%
and now magicaly I'm back at (19*20)/2=190

That is 20 people to get 50%.
And As I have said I KNOW this is wrong. But why?
Please I'm going Mad 🙂 And since I'm only examining pairs of dates 3 or more matching dates is not an issue.
13. 23 Oct '05 16:432 edits
14. 23 Oct '05 16:441 edit
Let's see if I can sort this out:

When rolling two six sided dice you have to roll 18 times to
have a 50% chance to get a twelve. (right?)

No. The probability of rolling a 12 is 1/36, which means the probability of not rolling a 12 is 35/36. The probability of not rolling a 12 in n trials is (35/36)^n, so the probability of rolling at least one 12 in n trials is 1-(35/36)^n. To get this above 50%, we need at least 25 rolls.

Now assume the year has six days then there is a total of 21
( (5*6/2)+6 ) ways to combine two days since the combination 1 and 2 is the same as 2 and 1.

Yes, this is the same as saying the number of ways to choose 2 days out of 6 where order does no matter is to calculate 6c2 (6 choose 2) and add 6 more combinations to cover all matched pairs (11, 22, etc...). However, it's not relevant to the question.

Out of these 21 combinations six are the same day twice (11 22 33 44 55 66) this means you would need to pick 21/6=3,5 to have a 50% chance to get at least one?

The probability of picking one pair out of 21 combinations is 6/21, the probability of not picking a pair is 15/21, so the probability of picking at least one pair in n trials is 1-(15/21)^n. To get this above 50% we need n to be at least 3.

Back to orignal numbers:
If the dice has 366 sides. You would have to roll them 66978 times to have 50% chance of getting 732. (right?)

No. 732 is the maximum roll with two 366-sided dice. The probability of rolling a 732 is 1/366^2, the probability of not rolling a 732 is 1-1/366^2, and the probability of rolling at least one 732 in n trials is 1-(1-1/366^2)^n. To get this above 50%, we need n to be at least 92851.

There are 366 results with the same day twice out of 67161 ((366*365)/2 +366) you need then 67161 /366 =183,5 to get 50%
and now magicaly I'm back at (19*20)/2=190

First, using the same procedure as above, the number of trials required to bring the probability of picking at least one of the 366 pairs out of 67161 possibilities to above 50% is 127. But I think I'm seeing the problem with your reasoning more clearly now.

That is 20 people to get 50%.
And As I have said I KNOW this is wrong. But why?
Please I'm going Mad 🙂 And since I'm only examining pairs of dates 3 or more matching dates is not an issue.

You are trying to find the probability of picking two random people and seeing if they have the same birthday, then doing that again and again until you find them, just like rolling dice. That is not the question we are trying to answer. We are trying to find the probability that any two people will have the have birthday, and we compare all birthday dates simultaneously (that is, no specific pair is required to have the same birthday, any pair or more will do). This is another application of the Pigeon Hole principle. Imagine a calendar; as people mark their birthdays on the calendar, there are fewer spots available for the next person to plot a clean day because more and more spots are taken. This makes it more and more likely that two people will have the same birthday as the number of people increases, until you get to 367 people at which point it is guaranteed that two people's birthdays will coincide.

I believe you estimation technique of dividing the number of possibilities by the number of desired outcomes to find the number of trials required for 50% success was misleading you, as it's just plain wrong but it's resoponsible for giving you numbers closer to the real answer then you would otherwise have.

Does that help?
15. 23 Oct '05 17:32
Originally posted by PBE6

I believe you estimation technique of dividing the number of possibilities by the number of desired outcomes to find the number of trials required for 50% success was misleading you, as it's just plain wrong but it's resoponsible for giving you numbers closer to the real answer then you would otherwise have.

Does that help?[/b]
Yes it does! 🙂 Thank you.

I was (incorrectly) certain that the method would yield the same result as I thought I was describing what the math was doing.