Probability Quiz Question.

What is the probability that it is in the right pocket? If - instead - the lab assistant reveals the left-hand pocket to be empty and he only ever reveals the contents of empty pockets at random, what are the chances then?

I think it's a variant on the Monty Hall problem where you should always switch if revealed an empty box out of three.

1) 33%
2) 50%

I think. I answered quickly on purpose so as to fall into the trap though

Could the cat be male (A tom... stoppard)

The above sounds correct to me.

To me, the probability that it's in the box (50%) doesn't change, so 50% to both questions.

Well done on getting the cat joke so quickly Lori, and I also thought of the Monty Hall problem but no, that is not the answer given.

Monty Hall doesnt reveal random doors though, its always a loser. For me its 50/50 in both cases still.

Its badly worded he only ever reveals the contents of empty pockets at random

No reason to say empty. Should be just contents of pockets. If he only reveals empty pockets its Monty Hall

Brendanuk and 1.01 Layer I came up with the same answer as you. 50% chance or 50-50, call it what you will.  But that is not the answer given either.

The reason I get 33% for the first one

There's four places the key can be. Pocket 1, Pocket 2, left half of box, right half of box.

If you cross one out, it becomes one in three

*but if the assistant reveals an empty pocket, it doesn't seem to matter.

Yes Lori and that is the heart of the logic used in the printed answer.  But I think that the answer is flawed.

Printed answer coming. I'm a slow typist so bear with me.

For the second one

Left pocket 25%: Assistant shows Right pocket, key is in coat
Right pocket 25%: Assistant shows Left pocket, key is in coat
Box 50%: assitant shows either pocket 50% of time. Key is in box

Observations:

25% of time, assistant shows right pocket, key is in coat
25% of time, assistant shows right pocket,key is in box
25% of time, assistant shows left pocket, key is in box
25% of time, assitant shows left pocket, key is in coat

50% of time he shows right pocket. half of those times it's in box
50% of the time he shows left pocket. half of those times it's in box

It is however kinda strangely worded, so I'm still willing to be wrong.

"Imagine there are 4 pockets: 2 real ones belonging to Erwin and 2 "belonging to" the box. Before looking the chances were 1/4.  After peeking in 1 pocket, the chances of the key being in the other are 1/3. In the second scenario the chances become 1/2 after the lab assistant's revelation."  Says the Guardian.

Interesting they did it a similar way, it's gonna look like I read the piece now, i can assure you I don't have time for papers!

The other way of doing it is to see that there's now a 25% chance of it being in the coat and a 50% chance of it being in the box.

As that's only 75% you have to scale it up to 100%

The way I see it is this.  I am happpy to grant the box 2 imaginary pockets to counterbalance the 2 real pockets of the lab coat.  This maintains the symmetry between box and coat and gives the chance of the key being in the right hand pocket of the coat at 1/4.  So far so good.  But when you look in the left hand pocket and find it empty then the question becomes "Where is the key? In the right hand pocket or in the box?".  The box loses its "need" for 2 imaginary pockets and the choice is  just 50-50 from there.

I can feel a letter coming on.

Whats the difference between q1 and q2? I am not seeing it

I suppose the answer is that a theoretical mathematician can come up with any answer they want....
Suppose there were 8 places in the box and 12 places in the lab coat...

Is this just badly worded Monty Hall?

If you focus on the pocket rather than the box, what chance is it that it's in the pocket?

What I mean, is that you're saying "okay, it's still 50% that it's in the box, so it's 50% it's in the pocket", which is entirely understandable.

However, if you've got two pockets in the coat and it's 25% that it's in one or the other, you just eliminated 25% by looking in the pocket, so you "should" make it that it's only 25% to be in the coat!

The options are as follows:

L R Box






If you look in the left pocket and are sad, there are three other options left, one of which is a happy.

As I see it Brendan, I think it is.

my reading of it was:

"The lab tech always shows you an empty pocket, no matter whether the key is in the bx or the coat"

The final way to get your head around it would be to get a coin and an area to represent the box and an area to represent the coat.

Toss the coin to see whether it's in box or coat, toss again if it's in the coat to see which pocket it's in.

Then "look" in the left pocket and see how often it's in the box when you don't find it in the left pocket.

Yes it is badly worded but where I think the reasoning is wrong is that after looking in the left pocket of the coat and finding it empty we have not lost a 25% chance of finding the key but a 50% chance.  As soon as the left pocket is found to be empty then the need for the 2 imaginary pockets in the box disappears. Effectively you are losing two 25% chances simultaneously.

forget the box entirely, however you want to picture it. I'm happy to say that there's a 50% chance of the box originally. Let's say there's a 50% chance it's not in the coat.


What I want to get you to picture is what happens to the coat?

There's a 50% chance it's in the coat originally. When you look in the left pocket - without thinking about the box - what chance would you make it that the key is still in the coat.

If you say 50% I'll be a bit confused, because surely looking in one of the pockets lowers the chance that it's in there.

I get the alternative way of looking at the problem but if there are 3 places the key could be, then the opening statement (..is either in his lab coat (probability 1/2) or in the box with the (probability 1/2)) is not only misleading but incorrect.

Wouldn't be the first time the Guardian were guilty of this

ok cool, I understood wrong and then put up wrong answer to my wrong understanding

Its 33% if sample space is 2 pockets and 1 box. left pocket is empty, 1 in 3 chance its in the other.
When lab guy does it, sample space is 1 pocket and 1 box.  1 in 2 chance

I'm not trying to win an argument here.  Just trying to understand.  The opening statement remains correct. The key is in either the lab coat or in the box. Stopping right there the chances of the key being in the coat or the box are 50-50.  Can we agree on that?  The second part of the question says that the key can be either in the left pocket or the right. Once we discover that the key is not in the left hand pocket then the question reverts to the original question, "Is the key in the coat or in the box?" It can be in one of only two places - either it is in the right hand pocket of the coat or it is in the box.  So- it is a 50-50 shot.

Stopping right there the chances of the key being in the coat or the box are 50-50.  Can we agree on that?

Definitely. FWIW the reason I'm trying a lot of different ideas is that I'm not arguing either. Just come across a bit factual/tetchy sometimes as I'm not very tactful.


The second part of the question says that the key can be either in the left pocket or the right.

Ok


Once we discover that the key is not in the left hand pocket then the question reverts to the original question, "Is the key in the coat or in the box?"

That's where I'd disagree.

You're ignoring (accidentally) the line

If in the lab coat, it is equally likely to be in the left as the right pocket.

This makes the original question

The key is 25% likely to be in the left pocket, 25% likely to be in the right pocket and 50% likely to be in the box.

Remove the coat from the equation and go with:

The key is 25% likely to be in the safe, 25% likely to be on the table and 50% likely to be in the box, see if that helps.

In fact let's go the whole way.

"The key is 25% likely to be in the safe, 25% likely to be on the table and 50% likely to be in the box.  He checks the table and the key is not there. What's the chances that it's in the safe?"

Going even further:

"There's a 25% chance it's on the table, 25% chance it's in the safe, 25% chance it's under the mat, and 25% chance it's in the cat.

He checks the table and it's not there. What's the chances it's under the mat"

What you need is pre-posterior probability analysis (not meant to be rude).  Once you've discovered an outcome what are the probabilities of the possible causes.
What caused the key to not be in the pocket? It was either in the other pocket or the box. What are the probabilities of these two causes?

Hi Lori.  In your last post you ask us to imagine that the key has a 25% chance of being in the safe,a 25% chance of it being on the table and a 50% chance of being in the box.  This cannot be.  If the key can  be in any one of three locations then the chances of it being in any one of them must be 33% in each.

Not at all. It could be in the lab, down the drain , or in his car. These have all been assigned a 0% chance though, not a 16% one.

Hi Sonic Jerry.  That seems to be playing into my way of thinking.  Eliminate the left hand pocket and there are 2 other possible outcomes.  Either the key is in the right hand pocket or it is in the box. The probability in each case is 50%. Surely?

You're assuming that all places are equally likely, there's no reason that should be the case.

You could have a puzzle where there's a 95% chance it's in the coat and 5% chance it's in the box, based on how forgetful the professor is

Imagine the key could be in any one of 3 boxes. 100% probability it is in one of them. 50% for box 1 (the real box) 25% for box 2 (one pocket) and 25% for box 3(the other pocket).
Now you open box 2 (left-hand pocket)and the key is not in it.
The probability of the key being in one of the two remaining boxes remains at 100%.
So what is the probability of 1) the key being in box one (original probability 50%) and 2) the key being in box 3 (original probability 25%)

Sonic J, and Lori, yes I am assuming that all locations of the key are equally likely. It would be impossible to proceed otherwise. The answer to the question is phrased to suggest that there are 4 possible locations for the key when in reality there are only 3.  The left pocket, the right pocket and the box.  All have a probability of 33% of being correct.  Eliminate one of those outcomes (the lft pocket) and only 2 possible outcomes remain.  Right hand pocket or the box.  Probability 50% in each case.

Anyway I'm off to the pub now.  It's been fun.  I shall look in later to see if this is still smouldering.

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

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)

Ah yes Lori. It is random selection.

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 [;)]

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%.

Ach, should read "subtract C2 and you are left with C1=B".  But you got my meaning.

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

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.

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.

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..

I have just looked up the Monty Hall problem on Wikipedia and it has not helped my certainty at all.

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.

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.

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.

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.