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Old 02-05-2015, 07:37 PM
  # 81 (permalink)  
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awright - I'm going to go outside and make a fire in the thing, me and the wife are gonna sit in the hot tub, it's 10 degrees outside and a full moon (pretty full) not a cloud in the sky... It's our Friday night - we don't get many nights together.

So I'll be checking back later to find out how dumb I am.
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Old 02-05-2015, 07:59 PM
  # 82 (permalink)  
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This is the first of three posts. The second has "the answer".

Thank you all for responding. As a group you have a great sense of humor. I laughed out loud many times while reading. I particularly liked Flynbys idea of staying in the tunnel. Clearly a case of thinking outside the box.

I must admit that when I first encountered this problem I got the answer wrong. Not only did I get the answer wrong but I was completely certain that I had it correct. (As some of you may have pieced together from some of my previous SR posts, this is not an altogether uncommon event.)

Several folks got the correct answer. This was far more than I had predicted. My suspicion is that some of them had help via Google, but that's unimportant really. The important thing is to examine your own thinking. I think the wisdom from the poem above "decide for yourself" points in that direction.

I believe that more than 95% of the population who first encounter this problem fail to get the correct answer. My guess is that most of those people believe (as I did) that the chances for A go from 33% to 50% and that it does not matter if A stays or switches cells with B. That's just my estimate, but I bet I'm fairly close in that estimate.

You might be interested to know that many professional mathematicians have gotten the answer wrong while at the very same time ridiculing those with the correct answer. Also, getting the correct answer does not seem to be related to intelligence. BTW, A famous mathematician named Paul Erdos got it wrong initially! Took him a while to find a way to understand it.

Next, the "answer".
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Old 02-05-2015, 08:08 PM
  # 83 (permalink)  
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This problem is known as the Monte Hall Problem. In this "Let's Make a Deal" scenario, a contestant picks one of three doors which might hide the "prize". The host Monte Hall knows the door that holds the prize, so after the contestant chooses a door he opens one of the other doors and shows that it does not have the prize. This leaves only two doors. He then asks the contestant if they wish to change their original choice and pick the remaining door.

This is the same problem as the prisoner example from the OP. So why didn't I just use the Monte Hall example above? Because I know you alcoholic drug using sorts. You want to feel good now! You don't want to think too hard. If I would have called it the Monte Hall problem 80% of you would have Googled it and gotten the answer. I wanted you to THINK. OK enough ranting lol.

The best way (for me) to understand the best strategy (to stay or switch cells) is to do a decision tree. I'll use the Monte Hall example with Door #1 Door #2 and Door #3.

Plot the outcomes for the two different decisions (stay or switch) given that the prize "belongs to" door "A". Lets call this real world #1



This is where the prize is behind door number 2. You get the same win/loss ratio depending on whether you choose to stay with your first pick or switch. It's always better to switch because you will win 2 out of three times if you do.

There are also many youtube videos that explain this in a variety of ways. Just enter "Monte Hall problem". I found the above decision tree to be the best explanation. It's directly applicable to the prisoner example. It's twice as good for him to switch cells with B, as he will have a 2/3 chance of freedom as opposed to only 1/3 if he stays in cell A.

One more post to come.
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Old 02-05-2015, 08:18 PM
  # 84 (permalink)  
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I haven't read the responses, but the correct answer is to switch cells. Staying in the orginal cell gives Prisoner A a 1/3 chance of being pardoned while switching gives him a 2/3 chance of being pardoned.

This puzzle is a variation of the Monty Hall Problem. It's a counter-intuitive problem. Personally, I've always had problems wrapping my head around it, but the math is clear IMO. For those who would like to explore the issue further, just Google:

Monty Hall Problem
Monty Hall Problem explained.

And this site gives a decent explanation as to why switching is the best option Monty Hall Problem Explained With Worked Solutions
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Old 02-05-2015, 08:19 PM
  # 85 (permalink)  
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The Monty Hall Problem and Card Counting

I was sort of kidding about card counting( honestly)…but I just plugged in Monty Hall problem/card counting and look what I found (by Jove!) It makes sense that this method is employed because the odds continue to change.
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Old 02-05-2015, 08:20 PM
  # 86 (permalink)  
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So why go to all the bother of presenting this problem? My intention was to get people to look their own thinking. IMO HOW we come to know/believe things is quite important. We all need to make decisions, but IMO we also need to be flexible enough to change our views of things when new information (as well as information in novel forms) becomes available.

For me, the Monte Hall problem really drove this home.

Incidentally I also believe that (to a much larger extent than is known) we are biologically hard wired to think in certain ways. Occasionally (albeit rarely) this interferes with our interpreting reality correctly. It's my hunch that this is a big part of our problem in understanding quantum mechanics. My thoughts went in much the same direction as those that freshstart expressed in this regard.

None of the verbal explanations that I have read anywhere (in this thread or online) have convinced me that either choice to switch or stay is preferable. I had to do my own decision tree and compare the results in order to be convinced. It was as simple as counting the wins and losses.

For me a problem still remains in how to succinctly and verbally express the mechanism by which the increased probability of a positive outcome is conveyed to only one of the remaining choices. This defies me. Can anyone do this?

So to sum up I'll just repeat the wisdom of a bumper sticker I spotted a few years ago. It said "Don't believe everything you think". I've taken that to heart.

I'm glad so many of you had fun with the thread.
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Old 02-05-2015, 08:40 PM
  # 87 (permalink)  
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I was looking at the table and it's wrong. lol In the "stay with" table at the top of the picture, the column with the 1's should say "original choice" (or the number of the original choice) instead of 1. Sorry.
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Old 02-05-2015, 09:15 PM
  # 88 (permalink)  
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didn't do the fire thing but when our hair froze it was time to come in...

okay - let me just call bullS on this one. I totally agree with the monty hall explanation - never heard of it by the way. Sure it works if you get three tries at it.
But in the scenario where there is only one choice - one attempt - it is 50/50. The guy can't come back next week and try to win a pardon is he chooses incorrectly.
It's totally random in this case. In a one and done situation, the law of averages means squat.
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Old 02-05-2015, 09:30 PM
  # 89 (permalink)  
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Just read this whole thing, missed my shot at an answer. Agree with LBrain. It is not like LMAD. The prison cell was chosen for him. So was the Kings decision.

You could look at it this way. Before the guard told him C, there was also a 2/3 chance A or C was chosen too, exactly like B or C. Eliminating C, which was beyond their control, still leaves A or B at 2/3, which then becomes 50/50.

Has to do with the wording. If A had gotten to choose which cell to be in before the guard told him, then the guard told him, the Monty Python thing applies. Just my take. I'm sure they'll be a lot of debate on this.
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Old 02-05-2015, 09:36 PM
  # 90 (permalink)  
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I say he should switch to b. before he knew anything he had a 66% chance of dying. the guard told him c was one of the ones to be put to death. his chance of being put to death is still 66% based on the assumption the guard would not have told him if he was the one to be put to death. so I say if he switches to to b his chances of being put to death drops to 33%.
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Old 02-05-2015, 09:50 PM
  # 91 (permalink)  
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silentrun, Very nice observation about the assumption that the Warden would not tell A if he was going to be executed. This is necessary in order to make the Monte Hall example compatible with the prisoner example.
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Old 02-05-2015, 09:54 PM
  # 92 (permalink)  
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oh **** you gave the answer
I think you sell us short thinking we wouldn't get it tho.

It is easy to see if instead of 3 cells there where 100 if only 1 was to survive and they started killing people but me and one other person I would assume that person had a 99% chance of surviving
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Old 02-05-2015, 09:58 PM
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The Monty Hall problem and the prisoner problem have different parameters and probabilities in their setups. Since "C" was for sure not pardoned, then only "A" or "B" had any probability of pardon, and so since only "A" or "B" but not both, it then becomes 50-50.

I too call BS.

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Old 02-05-2015, 10:01 PM
  # 94 (permalink)  
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Thanks for posting the answer!

I think it is significant that in the Monty Hall problem, Monty knows where the prize is and will not open the door to the prize.
The Monty Hall page (explanation)
The Monty Hall page (to play the game online)

This is what convinced me that switching was a good idea-
If prisoners A & B are not pardoned, there is a 0% chance that the warden will say that prisoner C will not be pardoned.
If prisoners B & C are not pardoned, there is a 50% chance that the warden will say that prisoner C will not be pardoned. (and a 50% chance that the warden will say that prisoner B will not be pardoned)
If prisoners A & C are not pardoned, there is a 100% chance that the warden will say that prisoner C will not be pardoned. (Assumption: The warden will not tell prisoner A that he will not be pardoned.)
The above three combinations have an equal chance of happening (1/3 chance for each option). Although we know that the first combination did not happen (A & B are not both not pardoned), since C is not pardoned.

I used Bayes theorem, which gives the correct answer (although I might be using Bayes theorem incorrectly).
Bayes' theorem - Wikipedia, the free encyclopedia (Look at the "Introductory Example" section)


How did other people come to an answer???
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Old 02-05-2015, 10:18 PM
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Originally Posted by RobbyRobot View Post
The Monty Hall problem and the prisoner problem have different parameters and probabilities in their setups. Since "C" was for sure not pardoned, then only "A" or "B" had any probability of pardon, and so since only "A" or "B" but not both, it then becomes 50-50.

I too call BS.

I tend to agree. If prisoner a has 2/3 chance of making it by switching then prisoner b has also 2/3 chance of survival by switching since the only difference between them two is that one is evil (but the warden does not know which one) so they are exactly on the same footing and the same story could have been written about b and according to your computations, b by going to cell a would also have had a 2/3 chance of making it.

2/3 plus 2/3 does NOT equal one.

Anyway, I still think he should have done b in
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Old 02-05-2015, 10:30 PM
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Actually oak, that makes sense. So he has:

Staying in A - 50% * 50% = 25% chance of pardon
Switching to B- 50% * 50% = 25% chance of pardon
Switching to B(alt. Scenario)- 50% * 100% = 50% chance of pardon

So the way I see it, switching to B actually gives you a 75% chance of pardon, under that assumption about the guard telling him. I originally assumed the guard would tell him. Anyway I was wrong, need to admit it. Better to switch to B. Anyway, makes my head spin. Good stuff
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Old 02-05-2015, 10:33 PM
  # 97 (permalink)  
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Originally Posted by RobbyRobot View Post
The Monty Hall problem and the prisoner problem have different parameters and probabilities in their setups. Since "C" was for sure not pardoned, then only "A" or "B" had any probability of pardon, and so since only "A" or "B" but not both, it then becomes 50-50.

I too call BS.


Think of it this way Robbie. The status of prisoner C is unknown until the Warden tells prisoner A that C will not be pardoned. This is exactly like not knowing what is behind door number 3 until Monte points out that nothing is there. In both cases there are only 2 options remaining (A and B or 1 and 2).

Did you work a decision tree?
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Old 02-05-2015, 10:55 PM
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Geez, what I wrote b4 is wrong too, not finished, disregard. I just wrote it down and get it now, feel stupid. 66% chance if he switches to B. Good puzzle Awuh.
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Old 02-05-2015, 11:12 PM
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The story presented here isn't the Monty Hall problem, nor is it the Three Prisoner Problem. The answer remains, it doesn't matter if he switches or not.

Moe, Larry and Curly are in boxes A, B, or C. The king pardons the person in either box A, B or C.

Case 1: Moe, Larry and Curly in A, B, C. King pardons A, Moe lives.
Case 2: Moe, Larry and Curly in A, B, C. King pardons B, Larry lives.
Case 3: Moe, Larry and Curly in A, B, C. King pardons C, but this cannot happen.
Case 4: Larry, Moe and Curly in A, B, C. King pardons A, Larry lives.
Case 5: Larry, Moe and Curly in A, B, C. King pardons B, Moe lives.
Case 6: Larry, Moe and Curly in A, B, C. King pardons C, but this cannot happen.

There are two ways for Larry to live, and two ways for Moe to live.
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Old 02-05-2015, 11:17 PM
  # 100 (permalink)  
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Mirage, it's not about being stupid. Really. PhD statisticians get this wrong!
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