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Another Maths Problem

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zebo | 08:24 Tue 14th Apr 2015 | ChatterBank
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http://www.dailymail.co.uk/sciencetech/article-3037266/The-maths-problem-set-Singapore-teenagers-left-people-world-stumped.html

I know it's the DM but it's certainly an interesting problem. At the moment, I have considerably less hair on my head!
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"Of course, perhaps the more important question is whether Cheryl deserves a birthday gift at all after putting us through all of this."

Ha! Sadly a joke made several times already, but amusing all the same.
That's just given me a headache!
it's in the Guardian too so between them they must have got the answer right (they did get the same answer). I couldn't get it, though it's straightforward enough when you see the reasoning.
Seems simple enough.

It was obvious Albert does not know the birthday since he has only been told the month and no month has a single day in it. So that need not be stated.

He thinks Bernard also does not know which imples he has not been told a month with a unique day in it. This rules out May (19th), June (18th) leaving only July & August.

Bernard knows the day so we can rule out any common days in July & August (14th) and then go with whatever month has just one day left for Bernard to seize on (July 16th ?).

At which point all know.

That would be my guess.
it seems there is another answer, though not the one intended

http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/apr/15/why-the-cheryl-birthday-problem-turned-into-the-maths-version-of-thatdress

This is getting a bit too subtle for me
I suppose a semi-valid alternative solution was always going to come up. The way I had read the problem originally included some level of probability running in the background. You can essentially construct a sort of probability tree. The logic then demonstrates that all but one branch of this tree in fact vanishes, leaving only July 16th.

But then probability problems require assumptions about the scenario that can be open to interpretation. Hence, for example, the classic "if one child of a two-child family is a boy, what is the probability that the other is also a boy?", a problem with the answer 1/3 or 1/2 depending on the interpretation of the question. Roughly speaking it runs that, if you have not seen the boy, then the probability that the other is also a boy is 1/3, but if you know anything about that boy then the probability that the other is also a boy is 1/2 instead.

There's a similar change here, although I haven't constructed the full tree for the "second" solution. I think here it's introducing something that was not stated in the problem, and so is invalid, but I can understand how people got to it.
I'd need to double check that alternative but it seems to me, at present, that Bernard could not claim to know the answer. All that has occurred is the removal of days 18th & 19th and the month of June. There may only be the one 17th left, but I do not presently see how he is able to rule out the 14s, 15, or 16s to be left with a unique date.

I'll check that again if I have time at some point.
Oh that one Jim. Yes, been brought up here somewhere. The problem is supposed to be stated in a way to avoid ambiguity. It is only when someone tries to tell it having only remembered the gist of it, that problems arise.

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