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Old 02-05-2015, 01:40 PM
  # 61 (permalink)  
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Awaiting the dawn sat three prisoners wary,
A trio of brigands named Tom, Dick and Mary.
Sunrise would signal the death knell of two;
Just one would survive, the question was who.

Young Mary sat thinking and finally spoke.
To the jailer she said, “You may think this a joke,
But it seems that my odds of surviving till tea
Are clearly enough just one out of three.

But one of my cohorts must certainly go,
Without question, that’s something I already know.
Telling the name of one who is lost
Can’t possibly help me. What could it cost?”

The shriveled old jailer himself was no dummy.
He thought, “But why not?” and pointed to Tommy.
“Now it’s just Dick and me!” Mary chortled with glee,
“One in two are my chances, and not one in three!”

Imagine the jailer’s chagrin, that old elf.
She’d tricked him. Or had she? Decide for yourself.

— Richard E. Bedient
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Old 02-05-2015, 01:46 PM
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If Mary claims to be pregnant, under the law she will be granted a stay of execution therefore by default Dick will be executed.
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Old 02-05-2015, 01:50 PM
  # 63 (permalink)  
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1 pea, 3 shells. The event that would knock "C" out of the game has not occurred, therefore has no bearing on the 3 living today but only 1 tomorrow. The odds obviously change when one is removed but not until. Switching shells by diversion never changes the peas' status--it will always be under one of 3 shells. The odds for A are the same. I still think A and B are making the best of a tense situation by comforting each other.
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Old 02-05-2015, 01:57 PM
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Originally Posted by Carlotta View Post
If Mary claims to be pregnant, under the law she will be granted a stay of execution therefore by default Dick will be executed.
Mary's survival would therefore be 100% . . . and Dick is in a pickle!!
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Old 02-05-2015, 01:57 PM
  # 65 (permalink)  
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so based on that little ditty, if he stays his chances will always be one in 3, but if he changes his chances are 1 in 2...

based on nothing else - just numbers - staying where he is he has a better chance. By relocating, the whole formula is then started over, he has a 50/50 chance. But it actually does not matter. The starting variables have changed.

however, even though it started at 1/3 regardless of the start it is now 1/2. it doesn't matter.

say a football team has won 9 of 12 games with 6 games left. what are their chances of ending up over .500? 6 weeks later they are now 9 and 8 with one last game to play. What are their chances of ending up over .500 now?

They are not the same as they were six weeks ago. Same thing for the guy in the cell. Not the same as yesterday folks - a "game" has already been played since then.


annataboy - I disagree, because it is already understood C is no longer in the game, that 'shell' has been removed from the game. You are looking at it from an uninformed observer's viewpoint. You have to look at it from the informed participant's viewpoint - from this viewpoint, there are only two shells.
It's like the stupid briefcase came, the more there are eliminated the better the chances because each round starts with a new number.

Last edited by LBrain; 02-05-2015 at 02:04 PM. Reason: saw annataboy post
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Old 02-05-2015, 02:40 PM
  # 66 (permalink)  
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Okay, a game has been played alright but I'm not sure we agree on what that is.
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Old 02-05-2015, 03:06 PM
  # 67 (permalink)  
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I will suggest that while these puzzles and riddles are entertaining exercises in deductive reasoning, they really don't have much to do with critical thinking.
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Old 02-05-2015, 04:36 PM
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I very much enjoyed the fun of this thread and the competitive banter. Thanks awuh1.
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Old 02-05-2015, 04:46 PM
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Originally Posted by anattaboy View Post
I very much enjoyed the fun of this thread and the competitive banter. Thanks awuh1.
So did I. I generally don't hang out in the Secular Connections section but this was fun and I cannot wait to hear the verdict and the reactions to it (I cheated and looked it up LOL)

Ps: What is it with almost no one thanking anyone in this corner of SR? LOL
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Old 02-05-2015, 04:52 PM
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If it is A & C that are not pardoned, would the warden say it is C that is not pardoned and not say that A is not pardoned? In other words- if it is A & C, is there a 100% chance of the warden saying it is C that is not pardoned? Assuming that is true: 1 * .33 = .3333

If B & C are not pardoned, I assume there is a 50/50 chance of the warden saying C is not pardoned. .5 * .33 = .1667

It has been over 20 years since I last took statistics. I thought those two numbers should equal 1. Hmmm...

If I was A, I'd switch. (But I agree with Carlotta about both A & C telling the truth about B! I want justice!)
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Old 02-05-2015, 04:53 PM
  # 71 (permalink)  
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It was a fun puzzle to think about today, thanks Awuh.
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Old 02-05-2015, 04:55 PM
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Originally Posted by Carlotta View Post
Ps: What is it with almost no one thanking anyone in this corner of SR? LOL
You do have on that Codie hat today, Carlotta
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Old 02-05-2015, 05:07 PM
  # 73 (permalink)  
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Hope I can stay awake to hear the answer.
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Old 02-05-2015, 05:13 PM
  # 74 (permalink)  
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awuh1, thanks so much for starting this. I've just read all the posts, but will not reveal what my decision is until your answer is written. Isn't that what you asked? That written, in some ways it reminds me of the Monty Hall Let's Make a Deal Decision process, but this involves more randomness so it's probably a cousin to that question.
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Old 02-05-2015, 05:36 PM
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I probably should have divided by .5 instead of 1. But why is the probability of C equal to .5? (Oh! I get it!)

P (C) = 0*.33 + .5*.33 + 1*.33 = .5 (probability that warden says C is not pardoned)

P (B+C | C) = (.5 * .33) / .5 = .33

P (A+C | C) = (1 * .33) / .5 = .67

Thanks awuh1!!! This was so much fun to think about today.
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Old 02-05-2015, 05:49 PM
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We don't say thanks so much here cuz the Badassery just shines so brightly no words can convey.....
I'm with you brynn--I'm a morning poster and early to bed. I think 24 will be up when I'm R.E.M. -ing
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Old 02-05-2015, 06:03 PM
  # 77 (permalink)  
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anatta....we gotta try and stay awake for this one! It's been driving me nuts all day!

And I was tempted to look it up too, Carlotta! Ha!
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Old 02-05-2015, 06:11 PM
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Originally Posted by Purpleknight View Post

In probability terms if it was 100 boxes with 1 box having a $1 million cheque, if you chose 1 box at the beginning, then 98 boxes containing nothing were removed, would you swap? yes because there was more probability in the beginning, 99 out of 100 boxes that you chose a box with no cheque in it, this makes it more statistical, using people in the problem blinds the numbers I think!!
Purpleknight- Thank you for this example.
I assume in this example, the 98 boxes that were removed were known not to contain the check, right? That is why it makes sense to switch. I may be wrong, but I think if 98 boxes were randomly removed (and no one knows if one of them contained the check), then it makes no statistical difference if you switch or stay with your original box. Make sense? (I could be wrong. It has been many years since I studied stats.)


I'm assuming that in awuh1's brain teaser, it matters if the warden randomly chooses who to say is not pardoned.
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Old 02-05-2015, 06:45 PM
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Originally Posted by awuh1 View Post
Awaiting the dawn sat three prisoners wary,
Are clearly enough just one out of three.


Telling the name of one who is lost
The shriveled old jailer himself was no dummy.
He thought, “But why not?” and pointed to Tommy.
“Now it’s just Dick and me!” Mary chortled with glee,
“One in two are my chances, and not one in three!”

Imagine the jailer’s chagrin, that old elf.
She’d tricked him. Or had she? Decide for yourself.

— Richard E. Bedient
Here Mary's chances of dying RISE to 1 in 2 chances….she should be weeping.
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Old 02-05-2015, 07:32 PM
  # 80 (permalink)  
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Women are statistically far less likely to get executed. if she is Caucasian and plays the pregnancy card she ll be off the hook but crying can't hurt either
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