Critical thinking
Better when never is never
Join Date: Sep 2011
Location: Wisconsin near Twin Cities
Posts: 1,745
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.
quat
Join Date: Jul 2013
Location: terra (mostly)firma
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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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Join Date: Jun 2011
Location: The Deep South
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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.
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.
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?
This leaves a 5050 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.
Member
Join Date: May 2014
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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.
Somebody said earlier that there is a hidden probability trap  we want to treat nonrandom 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.
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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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!
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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Join Date: Jun 2011
Location: The Deep South
Posts: 14,636
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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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.
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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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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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!!
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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