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Geek Culture / Newsletter Puzzle - May 2014

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BatVink
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Posted: 1st May 2014 00:17
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Here is the thread to discuss the puzzle from the May edition of the newsletter. If you're reading this before or early on May 1st, you're too soon. The newsletter will be here shortly.

As with last time here's the simple rules:

1. If you have an answer, put it in code tags so you don't spoil it for everyone else.

2. This puzzle thread is for people who want to exercise their brains, not find the answer via Google.

3. This month you're going to have to include some explanation on how you got to the answer

4. I don't claim any credit for the puzzle, I'm simply passing on a fun(?) challenge. I apologise if you have already seen this one.

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easter bunny
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Posted: 1st May 2014 02:03 Edited at: 5th May 2014 12:33
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So, here's my answer:



Reason:




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Quirky1
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Posted: 1st May 2014 16:55
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Matty H
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Posted: 1st May 2014 17:15
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I think I know the answer but not from working it out I have performed the experiment to see if it was true at a family gathering.

It never worked out that time, you can't trust randomness, although we got two birthdays separated only by one day.

semi-spoiler


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French gui
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Posted: 1st May 2014 18:17
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May I ask if the date of the party is a leap year? O_o
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Green Gandalf
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Posted: 1st May 2014 23:28
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Good question.



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BatVink
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Posted: 2nd May 2014 11:07
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@Quirky1, yes that's the way to do it and is the right answer.
Perhaps somebody will come up with the answer, and a smarter way to work it out. I admit, I'm not smart enough to come up with a neater method.

@French GUI, try it with both. Once you have the answer for both, you'll be able to work out if it matters or not.

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Battoad
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Posted: 2nd May 2014 12:05 Edited at: 2nd May 2014 12:54
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I have a different answer.




Answer


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Green Gandalf
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Posted: 2nd May 2014 12:26 Edited at: 2nd May 2014 12:28
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Quote: "yes that's the way to do it"


Is it? I couldn't get the answer he gave using the method as described. His initial approach seems to be correct (as is his numerical answer) but I couldn't quite put his pieces together to get the answer given. Perhaps the steps given in the code box were over-simplified a bit? [I used to teach this particular example to undergraduate maths students in their introductory probability course.]



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BatVink
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Posted: 2nd May 2014 13:32 Edited at: 2nd May 2014 13:33
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Quote: "Is it? I couldn't get the answer he gave using the method as described. His initial approach seems to be correct (as is his numerical answer) but I couldn't quite put his pieces together to get the answer given. Perhaps the steps given in the code box were over-simplified a bit? "


GG you are right, when I look at the explanation from Quirky1, it has been oversimplified...but the answer is correct.

Quote: "[I used to teach this particular example to undergraduate maths students in their introductory probability course.]"

Maybe you should be our resident newsletter puzzle-setter

@Juney, I'm afraid you are incorrect. And I think you are incorrect for the same reasons explained in the newsletter editorial (probability of is versus is not. If you take your calculation further (more people) do you end up with more than 100% probability very quickly?

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Battoad
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Posted: 2nd May 2014 15:47
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Quote: "@Juney, I'm afraid you are incorrect."


Fair comment but it depends on how the question is interpreted.

You are basing your answer on the condition that no one else has a birthday the same as anyone else, i.e conditional probability.
In reality this may not be the case so the probability should be unconditional therefore providing, lets say an alternative answer.

Nice puzzle though, although maybe not fully qualified.

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BatVink
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Posted: 2nd May 2014 15:56
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Quote: "You are basing your answer on the condition that no one else has a birthday the same as anyone else"


This is the mutually exclusive condition of the answer. If this is untrue, then at least 2 of the party-goers will share the same birthday, which is the answer we are looking for.

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Battoad
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Posted: 2nd May 2014 16:23
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Quote: "This is the mutually exclusive condition of the answer. If this is untrue, then at least 2 of the party-goers will share the same birthday"


You are initially using conditional probability to ensure that no-one has the same birthday as anyone else but then reversing this to find someone with the same birthday. But by reversing you are actually finding a condition where everyone shares a birthday.

But as you say, you have the answer you are looking for.

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BatVink
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Posted: 2nd May 2014 16:34 Edited at: 2nd May 2014 16:36
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The reverse of nobody sharing a birthday is that at least two people share a birthday. This also includes everybody sharing a birthday, as well 2,3,4,5 etc people sharing a birthday.

I have to confess that when I was given the answer, I had to sit down for a good 30 or 40 minutes to understand it fully.

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Green Gandalf
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Posted: 2nd May 2014 17:02
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Quote: "I have to confess that when I was given the answer, I had to sit down for a good 30 or 40 minutes to understand it fully."


Sounds like you are getting something out of this task so perhaps you should continue setting these.



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French gui
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Posted: 4th May 2014 18:32
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So I gave up and have a Google search. No remorse as I couldn't have found it by myself.
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nonZero
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Posted: 4th May 2014 23:42
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So lemme get this straight:
Probability of birthday collision must be 50%, how many people?



You're a bad man!
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easter bunny
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Posted: 5th May 2014 12:35
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So here's my answer and reason. Apparently it's incorrect. Why?
It makes sense to me...



Reason:




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BatVink
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Posted: 5th May 2014 13:15 Edited at: 5th May 2014 13:16
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@NonZero, unfortunately no, here is a hint/spoiler for your answer:



@Easter Bunny, here's a tip/spoiler for your solution:


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nonZero
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Posted: 11th May 2014 11:06
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Okay, I think the problem is I'm being too esoteric in my general approach to solving this. Well that and I barely remember anything I learned in school.

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BatVink
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Posted: 11th May 2014 19:21
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A helping hand for anyone that wants it.

Tip 1:


Tip 2:


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nonZero
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Posted: 13th May 2014 10:52
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Okay, figured out this


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BatVink
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Posted: 13th May 2014 15:59
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@nonZero (if you want another tip)


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Mr StixBC
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Posted: 30th May 2014 06:44
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So I think I understand....
[sol'n:

2 people - 1 and 2 - 1 potential match = 364/365 chance they DONT
3 people - 1 and 2, 1 and 3, 2 and 3 - 3 potential mathces = 362/365 chance they DONT
4 people - 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4 - 6 potential matches = 359/365 chance they DONT
5 people - 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, 3 and 4, 3 and 5 - 9 pot. matches = 354/365 chance they DONT

so n people means (n-1)! matches or (n-1)!/365 chance they DONT

50% = 365/2 = 182.5 or 183(talking whole people)

so for 19 people - (19-1)!/365 chance = 171/365 = 46.8% they DONT or

100% - 46.8% = 53.2% chance they DO!

19 is the answer, not 20]
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