Critical thinking
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Join Date: Aug 2011
Location: Toronto Canada
Posts: 5,148
That's how it seems to me. That probability function for B and C, totalling 2/3, collapses as soon as you know that it's not C. And it collapses to B, with a 2/3 probability. So 1/2 if he stays, 2/3 if he moves.
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Join Date: Jan 2015
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I'm assuming the king already made a choice of A, B or C and told the warden, so that's done and the choice is already made, no need to talk about 1/3 probabilities. The warden won't tell A who the choice was, but he tells him it's not C. That leaves A or B as the person pardoned. It's one or the other, 50/50 chance, so it doesn't matter if A and B switch places unless you add more to the storyline (C rats on B and tells king about the tunnel, king changes his mind, etc.). Furthermore, A and B are identical so there is no way to tell them apart. Doesn't matter if they switch places or not.
Sober Alcoholic
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Join Date: Aug 2011
Posts: 3,539
Great responses, on many different levels. I'm most interested in the reasoning behind the answers. I think it's best I keep my mouth shut at this point.
A little less than 8 and 1/12 hours to go.
A little less than 8 and 1/12 hours to go.
Trudgin
Join Date: Jul 2014
Posts: 6,348
Originally Posted by Purpleknight 
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!!
Just can't wrap my brain around it.......
So, if you started with a 1 in a 100 or 1% chance and took out 98 empty boxes, I just cannot see that the probability now isn't 1 in 2 or 50% and switching doesn't matter?!?!? How can then switching based on events that have already transpired affect the outcome of the draw????
Why is it any different than simply saying I have two boxes and one has money in it. Isn't the probability at that point 1 out of 2 regardless of switching ones choice or not???
Self recovered Self discovered
Join Date: Aug 2011
Location: Toronto Canada
Posts: 5,148
I was close but forgot how to add. The correct answer is 1/3 chance of surviving if he stays, and 2/3 chance of surviving if he changes. P + Q = 1.
Final answer.
We all agree that he has a 1/3 chance at the beginning, if he stays in A. That doesn't change. And a 2/3 chance for the other two cells. When the warden breaks the news about his buddy in B, this stuff doesn't change. It does mean however that this 2/3 chance now resides only in B.
Final answer.
We all agree that he has a 1/3 chance at the beginning, if he stays in A. That doesn't change. And a 2/3 chance for the other two cells. When the warden breaks the news about his buddy in B, this stuff doesn't change. It does mean however that this 2/3 chance now resides only in B.
Behold the power of NO
Join Date: Jan 2013
Location: WA
Posts: 7,764
Originally Posted by awuh1
Great responses, on many different levels.
I m surprised that you guys have not sent me back to the F&F forum LOL
Guest
Join Date: Sep 2014
Location: Melbourne, Australia
Posts: 1,476
Originally Posted by Carlotta
It's kind of telling though that I seem to be the only one obsessed with formulating strategies to rescue poor a or at least punish b even with the "no adding to the story rule" more than with cold probabilities.
I m surprised that you guys have not sent me back to the F&F forum LOL
I m surprised that you guys have not sent me back to the F&F forum LOL
If the king has changed the rules once, probability suggests he will do it again.
The outcome is indefinite.
Member
Join Date: Sep 2012
Location: Ireland
Posts: 25,826
Originally Posted by Flynbuy
Just can't wrap my brain around it.......
So, if you started with a 1 in a 100 or 1% chance and took out 98 empty boxes, I just cannot see that the probability now isn't 1 in 2 or 50% and switching doesn't matter?!?!? How can then switching based on events that have already transpired affect the outcome of the draw????
Why is it any different than simply saying I have two boxes and one has money in it. Isn't the probability at that point 1 out of 2 regardless of switching ones choice or not???
So, if you started with a 1 in a 100 or 1% chance and took out 98 empty boxes, I just cannot see that the probability now isn't 1 in 2 or 50% and switching doesn't matter?!?!? How can then switching based on events that have already transpired affect the outcome of the draw????
Why is it any different than simply saying I have two boxes and one has money in it. Isn't the probability at that point 1 out of 2 regardless of switching ones choice or not???
-100 boxes - 1 in 100 chance
-50 boxes - 1 in 50 chance
-2 boxes - 1 in 2 chance
That improvement in chance has to be factored into the equation when mathematicians do their fancy calculations.
Starting with only 2 boxes can't be the same as starting with 100 boxes and then switching your first choice with 1 of the final 2 boxes, it can't both be 50% because at the start there was a 1 in 100 chance rather than a 1 in 2 chance!!
So start with a 1/100, take out 98 boxes can't equal the same as simply starting with 2 boxes to begin with, which is a 1/2.
Probability surely needs to reflect the increase in probability, or the change in probability somehow with numbers, even though the options, number of boxes, number of prisoners remains the same!!
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Join Date: Nov 2012
Location: Nutmegger
Posts: 1,799
Originally Posted by Purpleknight
The other way of looking at it is that your chances have improved because many of the empty boxes have now been taken away, or 1 prisoner not being pardoned has now been taken away.
-100 boxes - 1 in 100 chance
-50 boxes - 1 in 50 chance
-2 boxes - 1 in 2 chance
That improvement in chance has to be factored into the equation when mathematicians do their fancy calculations.
Starting with only 2 boxes can't be the same as starting with 100 boxes and then switching your first choice with 1 of the final 2 boxes, it can't both be 50% because at the start there was a 1 in 100 chance rather than a 1 in 2 chance!!
So start with a 1/100, take out 98 boxes can't equal the same as simply starting with 2 boxes to begin with, which is a 1/2.
Probability surely needs to reflect the increase in probability, or the change in probability somehow with numbers, even though the options, number of boxes, number of prisoners remains the same!!
-100 boxes - 1 in 100 chance
-50 boxes - 1 in 50 chance
-2 boxes - 1 in 2 chance
That improvement in chance has to be factored into the equation when mathematicians do their fancy calculations.
Starting with only 2 boxes can't be the same as starting with 100 boxes and then switching your first choice with 1 of the final 2 boxes, it can't both be 50% because at the start there was a 1 in 100 chance rather than a 1 in 2 chance!!
So start with a 1/100, take out 98 boxes can't equal the same as simply starting with 2 boxes to begin with, which is a 1/2.
Probability surely needs to reflect the increase in probability, or the change in probability somehow with numbers, even though the options, number of boxes, number of prisoners remains the same!!
Oh, I don't know. Mind…blown as my 10 year old would say.
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