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Old 02-05-2015, 05:46 AM
  # 21 (permalink)  
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There are two gambles to consider. The original by the king, and the second by prisoner A. The first gamble had a 1 in 3 odd for winning. This does not change because prisoner A knows the outcome for C. The second gamble is whether A choose to play between cells A and B, which would be a 1 in 2 chance. However, this still falls under the first gamble which is 1 in 3. Multiplying these two chances together equals a 1 in 6 chance, versus a 1 in 3 chance under the king only gamble. Staying in cell A and not playing is the better choice.
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Old 02-05-2015, 05:55 AM
  # 22 (permalink)  
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I think it has more to do with what the warden did not tell A. The King forbade the warden to say who would go free, the warden tells A that C is condenmed. The warden can not tell A his is to be pardoned, and he hasn't said B is to go free and at the same time he hasn't told A that A is not to be freed ?
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Old 02-05-2015, 06:04 AM
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I think jazzfish is the winner!!!!!!
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Old 02-05-2015, 06:09 AM
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I'm going to say he should switch because it seems to me this would increase the probabilty of him being selected for pardon.

Though I am somewhat hung up on the king having already made the choice, but I dont think it's at all about the narrative.
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Old 02-05-2015, 06:14 AM
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Originally Posted by awuh1 View Post

The innocent prisoner in cell "A" got wind of this and privately begged the warden to tell him who would be pardoned. The warden said that the king had strictly forbidden him from saying who would go free, but confided to the prisoner in cell "A" that the prisoner in cell "C" would not be pardoned.


Assuming that the prisoner in cell "A" can somehow safely and easily switch places with the guilty prisoner in cell "B" (through the tunnel), should the prisoner in cell "A" switch? Would it increase or decrease his chances of survival, or would it make any difference?
The prisoner in cell "C" is out of the probabilities since no tunnel connects and the warden confided the cell "C" prisoner would not be pardoned.

This leaves a 50-50 chance for a pardon for both cell "A" and cell "B" and so there is no advantage to switching because it doesn't matter if the innocent prisoner in cell "A" switches with the guilty prisoner in cell "B" as the probabilities don't change either way.

Thanks for the thread awuh.
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Old 02-05-2015, 06:42 AM
  # 26 (permalink)  
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I'm staying in the tunnel!

Why is the long form of tradition 2 shorter than the short form??
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Old 02-05-2015, 07:07 AM
  # 27 (permalink)  
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"A" has 1:3 chance of living until the first man falls (it could be him). Changing cells, having inside info or doing the hokey pokey will not change his odds right now. Without a doubt, staying awake would be prudent as he is now housemates with a wacko.
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Old 02-05-2015, 07:11 AM
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Somebody said earlier that there is a hidden probability trap - we want to treat non-random information as if it were random.

I see the simple answer as hinging on a probability function that collapses when the warden tells the prisoner the bad news about his buddy in cell C. This is the similar to that phenomenon called quantum entanglement and is a good illustration of the idea that state of a certain quantity is a function of certain constraints, and the function returns a different answer as the constraints change. A Bayesian solution based on conditional probability analysis looks at it from an entirely different point of view and yields the same answer. Which is also very cool.
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Old 02-05-2015, 07:29 AM
  # 29 (permalink)  
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Wait, are we saying the cat doesn't even know until we open the box ?
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Old 02-05-2015, 08:16 AM
  # 30 (permalink)  
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Yes, and the probability of two inmates tunneling to each other's cell under the guise of escaping is uh...stupid or a plot to a gay prison porn.
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Old 02-05-2015, 08:43 AM
  # 31 (permalink)  
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oops; deleted because 24 hours aren't up.

but will just say that something like 1000 PhD's couldn't get this "right"
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Old 02-05-2015, 08:44 AM
  # 32 (permalink)  
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Ok, this is driving me crazy. But here's what I came up with. Probably wrong......

So C is out and 1 in 3 changes the odds to 1 in 2 for A and B or 1/2 = 50%
If A changes cells then he has a 1 in 4 chance ( 1/2 for event of change and 1/2 without change.) Therefore if we take both of those probabilities his overall chance is 1/4 if he changes cells or 25%.

Stay put, or hide in tunnel.......
I am on the edge of my seat!
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Old 02-05-2015, 08:47 AM
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How do you get that he then has 1 in 4 chance at this point? I don't follow that math, Flynbuy. Trying though I am terrible at math.
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Old 02-05-2015, 08:49 AM
  # 34 (permalink)  
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It never goes to a 50/50 chance. It's still the 1/3. But... he's already "selected" the first 1/3 (although he didn't really, he was jailed, but...). So, when he switches, seems that it would only increase his chance.
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Old 02-05-2015, 08:59 AM
  # 35 (permalink)  
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Before hearing that C will not be pardoned, each cell has a 33.33% chance of being pardoned. After hearing that, each of the remaining 2 cells have a 50% chance of being pardoned. I want to take a stab at it, but I'll hold off for a while.
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Old 02-05-2015, 09:02 AM
  # 36 (permalink)  
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Originally Posted by SoberJennie View Post
It never goes to a 50/50 chance. It's still the 1/3. But... he's already "selected" the first 1/3 (although he didn't really, he was jailed, but...). So, when he switches, seems that it would only increase his chance.
I agree Jennie. He has already selected the first 1/3 so that leaves a 50/50 for the remaining 2/3.. (I would imagine among the SR population we must have some card counters that could pop out this answer without blinking).


50% x 2/3=33.33

Thank god I no longer drink, I would have been working on this all night...
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Old 02-05-2015, 09:04 AM
  # 37 (permalink)  
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Nevermind. I was going to say what SJennie already said.
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Old 02-05-2015, 09:08 AM
  # 38 (permalink)  
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Originally Posted by SoberJennie View Post
How do you get that he then has 1 in 4 chance at this point? I don't follow that math, Flynbuy. Trying though I am terrible at math.
It's been 30 years since I have taken statistics, and since the OP doesn't state this is a math problem but using critical thinking, I am probably wrong ( you can figure that at 100% )

But, I believe the probability changes when we know C is out to 50 / 50.
So A can stay in his cell or leave. These are two separate risks. If he stays, I calculate 1 in 2 chance. If he leaves it's a 1 in 2 chance.

So, statistically - I am thinking ( again, it's been a really long time since class) but one should combine the probability of both events - staying or leaving to come up with an overall chance of 1 in 4.

This is why I own mattresses stores and don't teach stats!!
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Old 02-05-2015, 09:12 AM
  # 39 (permalink)  
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Jaynie, yeah, thanks... I hate numbers, lol. I think you got the math right.
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Old 02-05-2015, 10:07 AM
  # 40 (permalink)  
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In the beginning every prisoner had the same 1/3 chance of a pardon:

-A - 1/3
-B - 1/3
-C - 1/3

Once we know C will not be pardoned, the probability doesn't become 1/2 or 50%:

-Prisoner A staying where he is will remain at the original 1/3 chance
-Swapping with Prisoner B increases the chance to 2/3 - So Swapping is the way forward!!

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!!

In the same way the probability surely increases with the info that 1 prisoner won't get a pardon if he switches!!

That's my best shot, but I could be wrong!!
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