wolfgang
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Everything posted by wolfgang
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If B>C and A>B (as in 1a),so A=30,B=20 and C=10....you chosse (No) for the 1st question to make him a truth teller.
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Thanks...see maurice approach...he was right.
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you can concider them as one truth teller and two random(actually you don`t need them). I said it doesn`t matter wether you can arrenge them ,10..20..30,,,or 30..20..10 all you can do is to ask the truth teller " do you have 20$?" if YES,then put him in the middle. if NO,ask him the second question"has the man in the middle the 20$?" if YES,leave them as they are, if NO,the put the third man in the middle. We can also solve it with Mathematical equations( I`ll leave it to my friend Araver )
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with more cases,we get less %, I reached to the 7th order,and I`ve got only 55%
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there are 10 orders with 4 to 1 out of 31 odds which are about 32.258% and 20 orders with 3 to 2 out of 31 odds making about 64.516% and by joining both of them you will have only 10 odds to be right!
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I meant 6 out of 8 will be right...which makes 75%
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I understand...Sorry firstly I thought you mean to move who has 30,but now it is clear...good job though!!
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Is it allowed to only one person to change his place? If yes,,It will be very easy.
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See the order 10,,30,,20 will give you 2 times(no) also if you were asking the 10. and the same with 30,,10,,20 and 30,,20,,10 if you were asking the 30
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by sending them within a planed system,they will act like Morse signals,they`ll bring the informations even if they didn`say anything !
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If they were: 10,,20,,30 so by moving 30 to the middle,will get 10,,30,,20 and you will get (no) answer. and by moving himself to the middle,will get either 20,,10,,30 OR 10,,30,,20 in both cases you get(no) again.
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Thank you Dear araver... You know...If I`ve said each one has a different underware color( e.g.), so no one can solve it !! but I said ( money ),,and fixed amount...a+b+c=60... what can I say more?
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It makes no differemce..it is the same so or so.
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Yes they do..
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You know Dear Araver...intellegent people(like you),some times complicates a very simple situation. Let me give you a hint...what about asking anyone of them a (yes)question?.
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To ask them about(the next question),will not force them to tell the truth,they will remain totaly "at random" to the first question whatever it is.
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I didn`t refer to the details,cuase this topic is well known to all members here. I concider it as very easy one,you need maximal 5 min. to solve it..
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Sorry...but I didn`t say they are standing in this arrengement... there are six possiblities in this case as you know... sorry again..and thanks
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I have to add,each of them knows who is who.
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I hope so
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Three men, each one has money in his pocket. The first one has 10$, the second one has 20$, and the third one has 30$. Each one of them upon being asked the first( yes or no) question ,will randomly choose to be a truth teller or a liar. And as he did he will be fixed to that persona and will not change it( i.e.If he tells the truth,will always tell the truth, if he lied,will always lie). You are to ask only three ( yes or no ) questions.to any one of them at a time,or the same question to different persons..its up to you,in order to arrenge them in a row from the one with 10$ then 20$ then 30$. so your goal is to arrenge them as follows:- 10$....20$....30$
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Yes...you are...
