CaptainTux Posted October 20, 2006 Report Share Posted October 20, 2006 First letter of every word. Ya forgot to ask what time. ATtaCkwedDawn Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 20, 2006 Author Report Share Posted October 20, 2006 That is correct. Well done. Quote Link to comment Share on other sites More sharing options...
CaptainTux Posted October 20, 2006 Report Share Posted October 20, 2006 Just to be a smart alec I almost said becuase it was #2 and all the answers I have gotten right have been option #2. Quote Link to comment Share on other sites More sharing options...
babygirl Posted October 20, 2006 Report Share Posted October 20, 2006 First letter of every word. Ya forgot to ask what time. ATtaCkwedDawni dont see the word wednesday in the first letter of the words Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 20, 2006 Author Report Share Posted October 20, 2006 'AL: The trucks arrive, Charles knows when every day. Do answer when necessary.Attack Wed. Dawn :) Quote Link to comment Share on other sites More sharing options...
CaptainTux Posted October 20, 2006 Report Share Posted October 20, 2006 Confession: Once a semester I lecture at a local college regarding cryptography, social engineering, and exploiting weaknesses in computer systems. Much of that requires pattern recognition and logic exercises have never hurt. Quote Link to comment Share on other sites More sharing options...
babygirl Posted October 20, 2006 Report Share Posted October 20, 2006 'AL: The trucks arrive, Charles knows when every day. Do answer when necessary.Attack Wed. Dawn :)oh...okay i see, thanx Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 20, 2006 Author Report Share Posted October 20, 2006 That sounds great Tux. What a fun lecture. :) Quote Link to comment Share on other sites More sharing options...
CaptainTux Posted October 20, 2006 Report Share Posted October 20, 2006 It is. About a week before the presentation I compromise the security of as many students as possible and they are unaware until the day of the lecture when I ask them to pull my presentation from their hard drive and tell them what their new password is. I do this by pretending to be a headhunter offering a juicy internship to each student on the phone and after a 15 minute phone call I usually have enough info to get into their system. In a class of 20, I average about 3 successful invasions. Quote Link to comment Share on other sites More sharing options...
pushka Posted October 21, 2006 Report Share Posted October 21, 2006 Dr. T. I thought I was really clever when I got the answers right in post 16, but the other ones stumped me... Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 21, 2006 Author Report Share Posted October 21, 2006 Pushka, You are clever. Just because you can't get them all it doesn't say that you are not. That kind of all or nothing thinking usually doesn't pan out. Look at other evidence. :) Dr. T Quote Link to comment Share on other sites More sharing options...
Lindy Posted October 21, 2006 Report Share Posted October 21, 2006 Don't feel bad Pushka... I just read this thread.... I had fun TRYING to answer the questions.... I did get some right... then when I read M's answers I either felt.... ALRIGHT! or DUH!.....lol all in fun, all in fun. Thanks Dr T for the smiles :) Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 21, 2006 Author Report Share Posted October 21, 2006 My pleasure Lindy. :) It brings me smiles too. Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 21, 2006 Author Report Share Posted October 21, 2006 Warning: This one might be tricky. Tom, Joe and Harry are in prison. One of them has been randomly selected to die in the morning, and the other two will be set free. Their guard knows which one will die, but none of the prisoners does. The guard is under strict instructions not to divulge the identity of the doomed man. Tom is desperate for any information beyond the fact that his odds of death are one in three. He begs the guard to throw him an informational bone. Finally, to shut him up, the guard agrees to reveal only the following: the name of one of Tom's fellow prisoners who will be set free rather than killed. The guard then says that Joe will be set free. After receiving this information from the guard, what is the most accurate calculation Tom can make of the probability that he is the doomed man? 1 None of the other options is correct. 2 One out of two. 3 One out of three. 4 Two out of three. Edit for change in name that was not allowed Quote Link to comment Share on other sites More sharing options...
miztrniceguy Posted October 21, 2006 Report Share Posted October 21, 2006 1 in 3 Quote Link to comment Share on other sites More sharing options...
Maureen Posted October 21, 2006 Report Share Posted October 21, 2006 Don't feel bad Pushka... I just read this thread.... I had fun TRYING to answer the questions.... I did get some right... then when I read M's answers I either felt.... ALRIGHT! or DUH!.....lol all in fun, all in fun. Thanks Dr T for the smiles :)I have to agree with Dr. T! I think everyone has their niche when it comes to solving problems. I'm avoiding anything that has to do with calculating the "odds", no thanks; but the word game I liked. The definitions made it easier and with 2 of the questions dictionary.com's Thesaurus came in handy. My philosophy is, do the ones you're interested in and be all amazed by the answers you didn't know. M. Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 21, 2006 Author Report Share Posted October 21, 2006 Well done Miztrniceguy! An analogous illustration might help facilitate understanding of this problem. Assume that there are 100 children and 100 boxes. Each box has a different one of the children's names on it. I pick one at random to put a gold star inside; inside all the rest I put a silver star. I then seal all the boxes. Now suppose that Tom is one of the children. Because I selected the box at random, as far as he knows there is a 1% chance that he received the gold star in his box. That is, if we did this whole "experiment" 100 times, in approximately 1 run of the experiment Tom would get the gold star and the other 99 boxes would contain silver stars; and in approximately 99 runs of the experiment Tom would get a silver star, 98 other boxes would contain a silver star, and 1 other box would contain a gold star. Note that in all 100 runs, at least 98 boxes other than Tom's box contain silver stars; that will be true no matter where the gold star is. Of course, I know which box contains the gold star. I can therefore always generate a list of 98 names (always excluding Tom's name) that correspond to boxes containing silver stars. I can do this if Tom's box contains the gold star and I can do this even if Tom's box does not contain the gold star. Suppose that I compile such a list for Tom's benefit, having assured him in advance that the list will contain precisely 98 names of persons other than Tom whose boxes contain silver stars; in other words, that no matter which box contains the gold star the list will reduce the candidate gold-star boxes to two, Tom's and one other. HAVE TOM'S ODDS OF RECEIVING THE GOLD STAR ROCKETED FROM 1% TO 50%? CONSIDER THAT IF WE "RUN" THE "EXPERIMENT" 100 TIMES, WE ONLY EXPECT TOM TO RECEIVE THE GOLD STAR ONCE, AND YET I CAN PRODUCE A LIST OF 98 NON-TOM SILVER-STAR NAMES IN _EVERY SINGLE RUN_. WILL GENERATION OF A 98 NON-TOM SILVER-STAR-NAME LIST CAUSE TOM TO RECEIVE THE GOLD STAR FIFTY PERCENT OF THE TIME? OBVIOUSLY NOT; I CAN GENERATE SUCH A LIST WITHOUT AFFECTING THE ODDS. TOM STILL ONLY HAS A 1% CHANCE OF RECEIVING THE GOLD STAR ON ANY PARTICULAR "RUN." The Prisoner Problem is similar. Change the number of boxes from 100 to 3, and the gold star to a death warrant, and the logic is identical across the two situations. The information Tom was given had no effect on his own odds. Note that although Tom's odds remain 1 in 3, Harry's odds of being the one to be killed can now be rated at 2 in 3. This is because of the conditions the guard imposed: he said he was generating a one-person non-Tom list of people not being killed; he didn't promise to make a non-Harry list, so the new information that, of persons who are not Tom, ###### is safe doubles the probability that Harry is going down. *** Here is ANOTHER way of explaining it. You can set this up and try it at home. In fact, please REALLY DO try this at home before you write to me. Pick three friends to help you, and you be the jailer. Pick one of your friends to be "Tom." Put your friends' names in a hat, and pick one of them at random: that one will be "killed." Tell Tom the name of one of the other two friends who will not be killed. Ask Tom what his odds of being killed are. He will probably say 50%. Reveal to him who will actually be killed. Write down the name of the person to be killed. Do this 30 times. It may be that Tom always thinks he has a 50-50 chance after the candidates for being killed are narrowed down to two people, but in fact, at the end of the day, you will have written Tom's name down only approximately 10 times (1/3 of 30). And of the approximately 15 times that you said "Harry will not be killed" you will have written down "###### was killed" approximately 10 times (2/3 of 15), and vice versa. All get killed approximately the same number of times over the long haul, but only the calculation of odds for the non-Tom prisoners change as a result of the guard's information. *** [iF YOU GOT THIS QUESTION WRONG, AND HAVE TROUBLE UNDERSTANDING HOW YOUR ANSWER COULD POSSIBLY BE WRONG, YOU ARE IN GOOD COMPANY: THIS PROBLEM IS WELL STUDIED BY PSYCHOLOGISTS, LOGICIANS, AND GAME THEORISTS, AND IT IS WELL DOCUMENTED THAT EVEN VERY WELL-EDUCATED PEOPLE GET THIS PROBLEM WRONG MORE OFTEN THAN NOT, AND HAVE TROUBLE UNDERSTANDING THE CORRECT ANSWER. IT IS HIGHLY COUNTER-INTUITIVE. This problem has been exhaustively analyzed by scholars. If you think that you have uncovered an error so far undetected by any of the many academics who have bult careers trying to understand how humans reason over problems like this one, then please tell them why they are wrong while you are correct. HOWARD MARGOLIS has several books addressing this problem, but there are others. once again, well done Miztrniceguy! :) Quote Link to comment Share on other sites More sharing options...
miztrniceguy Posted October 22, 2006 Report Share Posted October 22, 2006 thank you Quote Link to comment Share on other sites More sharing options...
pushka Posted October 22, 2006 Report Share Posted October 22, 2006 Hi Dr. T. My post with the face was tongue in cheek. I'm really enjoying trying to solve these puzzles too...I do enjoy the word ones more, because I naturally was better at English than Maths at school, and sometimes have a blind spot when I see number puzzles. I was tempted to say 1 in 3 for the answer to the last puzzle, but went for 1 in 2 as you state most people probably do...it is confusing, but I understood your explanation of why that was the wrong answer. I actually enjoyed probability when I was re-sitting my Maths GCSE in my 30's! Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 22, 2006 Author Report Share Posted October 22, 2006 I'm happy to hear it Pushka. What is a GCSE? I'm guessing some kind of exam but I have no idea. Thanks, Dr. T Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 22, 2006 Author Report Share Posted October 22, 2006 What is 1/2 of 1/4 of 2/9 of 3/7 of 84? Quote Link to comment Share on other sites More sharing options...
Princess3dward Posted October 22, 2006 Report Share Posted October 22, 2006 iunno. Why does it matter? Quote Link to comment Share on other sites More sharing options...
pushka Posted October 22, 2006 Report Share Posted October 22, 2006 Dr. T. GCSE's are the exams that all 16 year old students take as they leave school. In the UK they used to have O'Levels and CSE's at that age, followed by A Levels taken if you wished to carry on with further education, these translated as: Ordinary Level, Advanced Level and Certificate of Secondary Education. Sometime in the 1980's it was decided to combine the O'Level and CSE's into 1 exam: GCSE, but each pupil would be assessed at to what level of that exam they would be aiming to reach, A-B, C-D, E-F, Ungraded, according to their abilities and would be taught in classes in which the difficulty of the work being taught had been adjusted accordingly. Hope that reads okay!! Quote Link to comment Share on other sites More sharing options...
Princess3dward Posted October 22, 2006 Report Share Posted October 22, 2006 Dr. T. GCSE's are the exams that all 16 year old students take as they leave school. I have never heard of these... and I am almost 17!!! I know there was a literacy test two years ago in grade 10...Is that the same thing? Quote Link to comment Share on other sites More sharing options...
Dr T Posted October 23, 2006 Author Report Share Posted October 23, 2006 Thanks Push Quote Link to comment Share on other sites More sharing options...
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