6Foot2 Posted April 15, 2015 Posted April 15, 2015 The puzzle: http://i.snag.gy/j7b6q.jpg The solution:
featured_spectre Posted April 15, 2015 Posted April 15, 2015 Took me about an hour...July 16th is what I got. Am I right?
6Foot2 Posted April 15, 2015 Author Posted April 15, 2015 Took me about an hour...July 16th is what I got. Am I right? You are correct. The linked BBC video explains the logic behind the solution.
Popular Post Domino Posted April 16, 2015 Popular Post Posted April 16, 2015 Beat Cheryl with a stick till she tells you. 5
LosOjos Posted April 16, 2015 Posted April 16, 2015 Actually they're wrong; her birthday is June 30th. 1
Arreks Posted April 16, 2015 Posted April 16, 2015 Actually they're wrong; her birthday is June 30th. Beat me to it..
tech_guy Posted April 16, 2015 Posted April 16, 2015 My wife sat me down, slapped me, refused to let me play my guitar or drink JD on Monday night until I understood it. She has a Mensa level IQ whereas I have the mindset of a hamster.
woodham Posted April 16, 2015 Posted April 16, 2015 according to our sims cheryl's birthday is non of them and she's really old
unixman_again Posted April 16, 2015 Posted April 16, 2015 Me neither. I asked the computer to help but it just sat there looking stupid - no change then!
MkII Posted April 16, 2015 Posted April 16, 2015 I don't get the radio 4 one. Just doesn't make sense to me - anyone help?? There were three prisoners, one of whom was blind. They were offered their freedom if they could succeed in the following game. Their jailer produced three white hats and two red hats and, in the dark, placed a hat on each prisoner. The prisoners were then taken into the light where, except for the blind man, they could see one another (but they could not see the hat on their own head). The game was for any prisoner to state correctly the colour of the hat he himself was wearing. The jailer asked one of the men who could see if he knew, and the man said 'no'. Then the jailer asked the other man who could see if he knew, and his answer was 'no'. The blind man at this point correctly stated the colour hat he was wearing, winning the game and freedom for all three. What colour hat was he wearing, and how did he know? Surely, the two sighted prisoners could see either two white hats or a white and a red hat. If they'd seen 2 red hats they would have known the colour of their own hat (white). The blind prisoner could conclude that either he was the one with the red hat, or the white hat. How would he know which???
Seb1780 Posted April 16, 2015 Posted April 16, 2015 I don't understand the explanation..... Step 1 – Albert says “I don’t know the birthday, but I know Bernard doesn’t know it either.” Bernard has been told the day of the birthday, if that day was the 18th or the 19th then he would know the month too as these dates only appear once. For Albert to state categorically that he knows Bernard doesn’t know the birthday then Albert, who knows the month, must know it is NOT in May or June. Step 2 – Bernard then says, “I didn’t know at first, but now I do know.” Bernard knows the date of the birthday and now has only to pick the month between July and August. Because he states that he now knows what the birthday date is we can rule out the 14th as this appears in both July and August and if it was the 14th he could not state he knew the birthday. Step 3 – Albert then says, “Now I also know Cheryl’s birthday.” Albert now knows the birthday is one of July 16th, August 15th or August 17th. He does not know the day of the birthday, but because he states categorically that he now knows then it must be July because if it was August he would still have two dates to choose from. So it is July 16th
Seb1780 Posted April 16, 2015 Posted April 16, 2015 Surely, the two sighted prisoners could see either two white hats or a white and a red hat. If they'd seen 2 red hats they would have known the colour of their own hat (white). The blind prisoner could conclude that either he was the one with the red hat, or the white hat. How would he know which??? There are only six combinations of hats, three of which have the blind man wearing a red hat. Two of these combination would involve the second red hat being worn by one of the other two prisoners, meaning that one of them would have seen two red hats and declared himself white-hatted. In the third scenario where the blind man is red-hatted the first prisoner looks and sees one red hat and one white hat and is unable to declare. Knowing that the first prisoner could not declare, the second prisoner, seeing the red hat on blind man knows he has a white hat on, otherwise the first man would have declared. In the scenario in the question, the only time the two sighted men would not be able to declare is if the blind man is wearing a white hat! 1
MkII Posted April 16, 2015 Posted April 16, 2015 (edited) Got it! Thanks No wait... There are seven combinations of hats. rr(w), rw®, rw(w), ww(w), ww®, wr(w), wr®. (blind prisoner in brackets) How can the sighted prisoners observing ww® & ww(w) have any influence on the blind mans conclusion? They can't declare two whites = themselves as r or w. I think the bbc may have messed up the puzzle - the originals are different: http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle Edited April 16, 2015 by MkII
Seb1780 Posted April 16, 2015 Posted April 16, 2015 There are seven combinations of hats. rr(w), rw®, rw(w), ww(w), ww®, wr(w), wr®. (blind prisoner in brackets) Only three of which have the blind man in a red hat. If the blind man was wearing a red hat then either one of the sighted prisoners would see two red hats or the second sighted prisoner would use the logic above. If neither sighted prisoner can draw a conclusion the blind man must be wearing a white hat.
Arthur Posted April 16, 2015 Posted April 16, 2015 Took me about an hour...July 16th is what I got. Am I right? You are. https://abustamam.github.io/blog/2015/04/16/solving-the-singaporean-math-problem
MkII Posted April 17, 2015 Posted April 17, 2015 (edited) Only three of which have the blind man in a red hat. If the blind man was wearing a red hat then either one of the sighted prisoners would see two red hats or the second sighted prisoner would use the logic above. If neither sighted prisoner can draw a conclusion the blind man must be wearing a white hat. Not true. If neither sighted prisoner can draw a conclusion the blind man cannot know what colour hat he is wearing. OR ~ the two people that the sighted prisoners can see are not both wearing red hats (the only scenario where anyone can know anything) 1. rr(w) 2. rw(w) 3. ww(w) 4. ww® 5. wr(w) All the above are inconclusive to the 2 sighted prisoners yes? (*) = blind man In 4 the blind prisoner can be wearing a red hat. Edited April 17, 2015 by MkII
tech_guy Posted April 17, 2015 Posted April 17, 2015 Chantelle has six children under the age of ten. Bryony - 9 Ebonii - 7 Connor - 6 Scarlett - 5 Boston - 3 Elija - 1 Chantelle keeps getting asking by her children who their daddy is. Chantelle knows who the father of her three eldest children is. Chantelle thinks she knows who fathered Scarlett & Boston. Elija was born on the 24th December 2013. 9 months previously, Chantelle had a one-night stand with Tom on the 3rd of April, Dick on the 6th, Harry on the 9th, and Bob on the 12th Who is Elija's daddy? Scroll down for the answer if you can't work it out. Answer: Chantelle's a slapper. 1
Seb1780 Posted April 17, 2015 Posted April 17, 2015 1. rr(w) 2. rw(w) 3. ww(w) 4. ww® 5. wr(w) All the above are inconclusive to the 2 sighted prisoners yes? (*) = blind man Correct; almost! In 4 the blind prisoner can be wearing a red hat. When this is the case the second prisoner can make the argument thus:- "I can see the blind man is wearing a red hat I know the first prisoner could see the blind man is wearing a red hat Therefore, if I too was wearing a red hat the first prisoner would have known he was wearing a white hat and would have declared, so I must be wearing a white hat." Thus ruling out scenario 4 above and leaving only scenarios where the blind man is wearing a white hat! 1
MkII Posted April 17, 2015 Posted April 17, 2015 Yes I see it now. It hangs on prisoner 2 (S2), who will know that if the blind prisoner (BP) is wearing a red hat, that then his hat will be white. (Because his sighted mate (S1) just said he couldn't decide, so that means that S1 could see rw or ww. As S2 can see that BP is wearing a white hat, he can't say what colour his own hat is. Therefore BP knows without looking that his own hat is white! This is going to seem so obvious in retrospect! 1
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