RobbyRobot

Thanks awuh, I guess we're just not going to agree. What you offered with the prisoners is not the same as the Monty Hall problem. The question posed to the reader by you was asked AFTER the warden confides the info about "C". Since the question is asked last, the status of the problem at THAT TIME is "C" is not pardoned and only "A" and "B" have any probabilities of being pardoned. The timing of your question is important. Its a 50-50 chance when the question is posed and switching cells doesn't change the odds.

In any case, thanks for the thread. No, I totally didn't work a decision tree as one is not required.

Mirage74

Thx awuh, I can see how that can happen, bc the key is the assumption that the guard wouldnt tell him if A(himself) isn't pardoned. So:

If A is pardoned (50% chance), guard says B or C isn't pardoned (2 scenarios, 50% each)

If B is Pardoned (50% chance), guard says A or C isn't pardoned. (2 scenarios, 100% outcome for 1,We know he won't say A isnt pardoned)

So that leaves 3 scenarios, bc the guard actually said C, 2 of them B as pardoned.

Funny, I'm still not convinced. Doesn't seem exactly like LMAD or Monty Python. I'm still leaning towards doesn't matter. This is bending my mind, ha ha. Anyway, I'll come back to this later.

awuh1

The timing is exactly the same Robbie. Prisoner A has the option to switch with B after he is informed that C will not be pardoned. This is done in the same way as the game show contestant is given the option to change to door 2 after door 3 is shown not to have anything there. Same point in the story. Same effect.

RobbyRobot

In the Monty Hall problem the question is asked to the active contestant if he wants to switch. In the prisoner problem the question is asked only to the reader ie me. This makes all the difference when making out the odds and so its really a matter of perspective (mine) which makes a difference in the parameters when answering the question.

Mirage74

Actually, and I need to verify this, in LMAD, when u pick a door out of 3, you have a 2/3 chance of being wrong. Once it's revealed to you one the second door is empty, the (erroneous) assumption is the 3rd door remains at 66% so you should switch to that door. I believe though, the probabilities revert back to 50-50, shifting similarly to how probabilities of winning hands shift once community cards are revealed in poker. The fact is, the 3rd door began at 33% and the shifts to 50% once the 2nd door is revealed empty, just like your door. I believe this is what's known as fool's gold, causing a contestant to switch to the 3rd door believing he's increased his odds, when really they didn't change at all.

Anyway, that's my final answer. Thx again for this, I like this stuff, plus it helped keep me sober tonight.

RobbyRobot

And actually, "A" and "B" can freely switch by way of the connecting tunnel before the warden confides with "A" and so the warden doesn't even know who he is actually confiding in, as the two prisoners have a probability of 50-50 changing before the warden even gets to talking with "A"... or was it prisoner "B" after all??

awuh1

Robbie, I'm sorry but I had to ask myself if you are being serious. Think about it. This is not about questions or who is asked those questions. Its about probabilities and about choices. If you are prisoner A, in that specified situation, you have choices, and probable outcomes are involved in those choices. It's equally true that if you are a game show contestant, in that defined situation, that you also have choices and those choices have exactly the same probabilities. This is true regardless of your "perspective" or even what you had for breakfast this morning. Your "perspective" is irrelevant because it is not part of the stated example. The question has not been asked of you, it is being asked about individuals in the specific examples given. Neither your perspective nor mine changes those percentages.

Regarding the prisoners changing places before the question was posed to the warden. This again attempts to change the basic parameters of the situation specified in the story. I addressed this on the first page of this thread. You seem to be sort of argumentative Robbie. Perhaps someone else will address some of your future points in a way that is more satisfactory to you

PurpleKnight

Yeap the 98 boxes removed are known not to contain the cheque, so in the beginning the box you chose was from a 1/100 chance probability, and now you have a 1/2 chance, so switching increases you're probability of picking the cheque!!

Mirage74

Nevermind

anattaboy

Thanks awuh...again. The biggest bonus for me was having my 13 yr. old step D home yesterday (snow day) and she actually put her phone down for 5 whole minutes and we discussed this problem (I swear I did not influence her in any way). This was and continues to be a moment. I'm still grokking and she has school shortly but this may get her out of bed on time and we can have some more Brain-to-Brain interaction. Huge bonding going on here. Thanks.

silentrun

You are getting distracted by details that have nothing to do with the problem. The survivor was chosen at random so guilt or innocence has nothing to do with it. The warden knows who will survive so him revealing the ones to die is not random. Having "information" on a cell not chosen at random does not alter the math from 66% die to 50% die. It's just information.

jaynie04

!

Carlotta

Then prisoner C did not die in vain
Seriously this is priceless.

Wholesome

I had my kid trying to figure it out with me too! lol

Wholesome

I still don't get it, but maybe I'm brain damaged. To me both A and B have a 2/3 chance, A doesn't know the outcome of either just that C is certain death.

LBrain

awuh, you totally ignored my response...

my point is that --- I still have not been proved wrong,

in this scenario if the warden told him B was not to be pardoned, he has no choice but to remain, C is NEVER an option in this scenario

And please explain to me how the probability is different when there is only ONE chance at it.

In PK's example of the 100, it also does not apply because YOUR box is chosen AFTER 98 of the boxes are removed - so 50/50

In this case of the prisoners, you choose AFTER you know there are only two choices. And as I've said, this is a one time guess. Theory does not come into play here. Statistics do not come into play.

The prisoner does have the opportunity to "change" his guess he only gets to choose one time after the odds are already revealed.

If they were choosing between steak and eggs or mush every morning where 2 of 3 were mush from 3 covered plates, and one was revealed after he chose, the MH thing would definitely work. He would be better off changing his choice every day. I agree with the premise. But if he were given just two plates to choose from there is no better chance to get the steak and eggs.

LBrain

correction to above: prisoner does NOT have the opportunity to change his guess

ubntubnt

maybe the dialogue is the point

dwtbd

Thx Awuh enjoyed the posting, kept me thinking all day.
I wanted it to be a riddle with an answer that would provide A with a way to definitely secure his survival. Granted the OP does ask about the probability of the prisoners chances of being pardoned and if switching could affect the probability of the outcome, which would entail a answer from the statistical perspective. I think I would have been invested emotionally different with the problem if the outcome involved a contestant either receiving a prize or not based on the statistical probabilities of switching .
So the answers and arguments "against" the math are interesting if viewed from that perspective, yes ?
I get the decision tree and table of results, the counter intuitive ness of probabilities still don't sit comfortably, lol, but I understand how the math is generated. No matter the probabilities there was only one outcome, and it's heart warming that most everyone seems to want to root for A
I would have an easier time of being comfortable with A getting not the steak and eggs, and tell him to quit whining about the mush, just pick luckier on the morrow, but I really hope he doesn't have to use statistical analysis to survive to every morrow. Because ultimately probabilities be damned he is going to get what he gets.

PurpleKnight

Ah but my point is if you choose 1 box from 100 so 99 chances to not pick the cheque and 1 chance to pick the cheque.

or

have 2 boxes that you know 1 has a cheque and 1 has not, by switching your box at the end you are dealing with a higher probability that you are picking the original box with the cheque inside.

The idea that starting with 100 boxes narrowed down to 2 by removing 98 boxes that are known not to contain the cheque, and then switching, is different from simply starting with 2 boxes.

Again it's the increase in probability that needs to be reflected, rather than both scenarios having simply the same 2 boxes to decide between.

I'd love to be able to do the fancy calculations and equations to work it out mathematically!!