Wikipedia:Reference desk/Archives/Mathematics/2009 March 23#Birthdays Part 2

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= March 23 =

Birthdays Part 2

This is (sort of) a follow-up to my question above. Let us assume that each day of the year is equally likely as any other to be a randomly selected person's birthday. Thus, all of the days of the year have an equal probability of 1/365. (Again, let's "pretend" that February 29 does not exist as a birthday.) If I randomly selected 365 people and, say, placed them in a room ... shouldn't everyone in that room have a different (unique) birthday? In other words, should all 365 days of the year be represented ... or no? Something tells me "no" ... but, why not? Also, if the answer is indeed "no" ... how many people would it, in fact, take to fill a room such that we would insure all 365 birthdays are represented? Thanks. (Joseph A. Spadaro (talk) 04:32, 23 March 2009 (UTC))

: The problem of how many people are needed before probably two have the same birthday is the Birthday problem. The problem of how many people are needed before all 365 days appear is the Coupon collector's problem. McKay (talk) 04:38, 23 March 2009 (UTC)

::And the direct answer to the second question is that if you have no information about actual birthdays, you could put the entire population of the world in your room [to implement this, I recommend an inversion transformation :-)] and you still couldn't be sure that all be sure that all 365 birthdays are represented. For all you know, it's possible that a February 16 birthday, like North Dakota, does not really exists -- by chance, nobody alive has that birthday. Sure, the probability of that is about (364/365)^6,750,000,000, which is a pretty small number, but you wanted certainty. --Anonymous, 20:25 UTC, March 24, 2009.

:And just to have an idea, it's quite unlikely to get all 365 days among the 365 people: around 1 over 10155 --84.220.118.44 (talk) 09:01, 23 March 2009 (UTC)

::It hinges on the fact that once you have n people in a room, assuming they have different birthdays, then you have only a (365-n)/365 chance that the next person has a different birthday. As n increases, and assuming that the probabilities are independent, the probability of all people having a different birthday decreases significantly. At around 20 you have a 50% probability of two people with the same birthday. -mattbuck (Talk) 14:03, 23 March 2009 (UTC)

vertex connectivity of a graph

Hello. I want to prove that the following construction of a graph G with n vertices and e edges gives us a graph with maximum possible vertex connectivity which is \left\lfloor\frac{2e}{n}\right\rfloor. First construct a regular graph on n vertices where each vertex has degree \left\lfloor\frac{2e}{n}\right\rfloor. Now add the remaining e-\frac{n}{2}\left\lfloor\frac{2e}{n}\right\rfloor edges arbitrarily. The resultant graph has vertex connectivity \left\lfloor\frac{2e}{n}\right\rfloor.

I can understand that such a graph would have a vertex v of degree \left\lfloor\frac{2e}{n}\right\rfloor and removing the adjacent vertices of v would thus disconnect the graph but I want to prove that removing fewer vertices can never disconnect G. How can I do that please?--Shahab (talk) 16:24, 23 March 2009 (UTC)

:As stated, this procedure obviously fails. For example, with n=e=6, one such G is a pair of triangles, which is not even connected. I'm not sure yet if requiring connectivity of the \left\lfloor\frac{2e}{n}\right\rfloor-regular graph gets you anywhere. Algebraist 16:31, 23 March 2009 (UTC)

::Update: it doesn't. There are 3-regular connected graphs which are not even 2-connected. Algebraist 16:43, 23 March 2009 (UTC)

:There's another problem, too. n and \left\lfloor\frac{2e}{n}\right\rfloor may both be odd, in which case there is no regular graph with that degree. A suggestion for a different construction: form a clique of the desired connectivity, connect every other vertex to each clique vertex, and use up any leftover edges by connecting pairs of non-clique vertices. —David Eppstein (talk) 17:03, 23 March 2009 (UTC)

::OK. I see this construction is faulty. I was reading from [http://books.google.com/books?id=Yr2pJA950iAC&printsec=frontcover&dq=Graph+Theory+with+Applications+to+Engineering+and&ei=M9LHSbDJJ5qGkATgrbCBDg#PPA78,M1 this book(Pg 78)]. David, why will removing the vertices in the clique disconnect the graph? Non-clique vertices may be forming a path.--Shahab (talk) 18:20, 23 March 2009 (UTC)

:I don't understand the implicit claim in the second sentence of the original post. The complete graph on n=k vertices and e=k(k-1)/2 edges is k-vertex connected and (k-1)-edge connected, right? However, k is quite a bit larger than floor( 2e/n ) =floor( k/2 ) = k-1, right? In particular, K4 has n=4, e=6, k=4 > floor(2e/n) = 3. Is it asking for a graph that somehow manages to NOT be highly connected? JackSchmidt (talk) 18:02, 24 March 2009 (UTC)

::2e/n=k-1 in that case. Where I come from (Bollobas's text book agrees with me), the complete graph on k vertices by definition has vertex-connectivity k-1. This slightly perverse definition ensures that the connectivity is always at most the minimal degree, which is of course at most floor(2e/n). Algebraist 18:11, 24 March 2009 (UTC)

:::Combinatorics is significantly harder for people who cannot handle counting more than 2 thing or doing simple arithmetic. Ok, with your two corrections to my understanding, I think it makes sense. Then David Eppstein's argument shows how to keep the connectivity high while filling in the rest of the nodes and edges. Ok, all good. The answer to the OP's second question is then simply "because it can't have more than the maximum connectivity," right? Someone else should answer it, as I find most of combinatorics baffling. JackSchmidt (talk) 19:07, 24 March 2009 (UTC)