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Q1 Initially
P(left)=0.25 P(right)=0.25 P(box)=0.5 left is eliminated and the probabilities for the remaining change to : P(right)=0.25/0.75 P(box)=0.5/0.75 Q2 Is it possible to reveal the contents of empty pockets at random in this case? Lori, and the paper apparently solve this in a way that suggests anything but random selection, as the lab assistant always shows the contents of an empty pocket. There is no random selection process as there is only one option to choose from. If there were three pockets, then one of the two empty ones could be picked at random. If I have one coin and I ask you to pick a coin at random, you only have one choice. Can you call that a random selection? Another way of putting this is saying the key is either in one of the pockets or the box. Now I tell you that one of the pockets is empty, how does that change the probability of the key being in one of the pockets? Well, it doesn't because we already knew one of the pockets was empty. No new information, the probability doesn't change. P(key is in one of the pockets)=0.5 P(key is in one of the pockets given that one of the pockets is empty)=0.5 |
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I took it to mean random if there were a choice (So in the instances where it's in the box, he can randomly pick a pocket)
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Ah yes Lori. It is random selection.
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I should change my answer to Q2 slightly:
P(key is in one of the pockets)=0.5 P(key is in one of the pockets given that at least one of the pockets is empty)=0.5 If exactly one pocket is empty, obviously the key is in the other pocket [;)] |
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Good grief is this still running! Hello to you The Investor. I can only partly agree with your analysis. The part that I agree with is the conclusion that you draw, namely that P(key is in one of the pockets) =0.5. I think that this is beyond dispute. Is it not?
However I do dispute the method you have used to get there. Q1 did not state the following: P(left)=0.25 P(right)=0.25 P(box)=0.5 Q1 merely stated that: C(coat) =0.5 B(box)=)0.5 The secondary part of the question led to pockets R and L of the lab coat as options. The thought experiment (if I can call it that) suggested that if C(coat) had two parts to it then in order to be properly weighted with the contents of the box (B) then the box should contain two imaginary parts, namely B1 and B2. This approach neatly balances B and C. Where The Investor and others seem to fall into the same trap as the question setter is in then granting B1 and B2 actual credence. They are not separate entities. They do not exist. There is only one box. Subtract C1 and you are left with C2=B. Subtract C2 and you are lft with C2=B. Again, the only solution is that the probability of the key being in either of only two remaining locations is 50%. |
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Ach, should read "subtract C2 and you are left with C1=B". But you got my meaning.
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Topnag, you say:
Q1 merely stated that: C(coat) =0.5 B(box)=)0.5 But it states more than this, see underlined: "The key to the box in which Erwin has locked his cat, Stoppard, is either in his lab coat (probability 1/2) or in the box with the (probability 1/2). If in the lab coat, it is equally likely to be in the left as the right pocket." Given that the initial probability of the key being in the lab coat is 0.5, and that it is equally likely to be in either pocket the probability of the key being in the left pocket is 0.25 and the right pocket is 0.25 |
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The Investor,
Let me cut straight to your last sentence. "Given that the initial probability of the key being in the lab coat is 0.5" (I am with you here) "and that it is equally likely to be in either pocket" (no problem so far), "the probability of the key being in the left pocket is 0.25 and the right pocket is 0.25." Here we differ. In this one sentence it is tacitly acknowledged that there are three possible locations for the key. If the key is in either coat or box then there is a 0.5 chance of the key being found in either of two places. Do we agree so far? But when the options are expanded to include left hand pocket of coat, right hand pocket of coat or in the box then the options are increased to three possible locations. Therefore it follows that the probability of the key being found in any one of those locations must be 33% in each case. Assuming equal likelihood of each possibility of course. |
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No because it states that the combined probability of the key being in either coat pocket is 0.5.
Using your logic, if we expanded the possible locations to 100 (the box and a lab coat with 99 pockets), there would be a 1% of the key being in the box. But it doesn't matter how many pockets there are, you've already been told there is a 50% chance of the key being in the box. |
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Back to the original question which goes on to state "you look in the left pocket and it is empty." The question asked is this...."what is the probability of the key being in the right pocket?" So there are only two possible locations for the key. The right hand pocket of the coat or in the box. Two possibilities. Two outcomes. 50% chance.
Am I going mad here? I am going to email the quiz setter. I will let you know what he has to say. If anything.. |
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I have just looked up the Monty Hall problem on Wikipedia and it has not helped my certainty at all.
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It is similar to the monty hall problem (it was probably the inspiration for this problem).
Try to figure it out topnag [;)] If you don't understand the solution, you could try doing a simulation. At least that will show you what the correct answer should be! That will at least put your mind at rest about that. |
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Believe me I am trying to figure it out! OK here is another way of looking at it... Suppose that the answer printed in the paper is correct and the probability of the key being in the right pocket really is 1/3 as stated.
For this to be the case there would have to be one correct solution to the question, for instance the key is to be found in the right pocket of the lab coat. There would also have to be two incorrect solutions to the problem. The key is not to be found in the box or....or where....? Not somewhere else in the box either? No, this is an either/or question. A 50-50 chance. |
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If in the lab coat, it is equally likely to be in the left as the right pocket.
This is new information and so changes the initial probability. |
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topnag1, you are wrong because you choose to ignore the fact that this experiment would end 25 % of the times when you find the key in the first pocket you are looking.These 25 % of failed attemys has to be included in the probablities calculation.
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