The Strange and Secret Notes of Lovers

Guy Patrick was in love with Elmira von Brut, the only daughter of Sir John and Lady Georgina von Brut, in the stately House of von Brut.

Distinguished generations of the von Brut family had passed through history, but Guy Patrick had the misfortune of being just a commoner. He was always a welcome guest, however, because of his penchant for scintillating conversation during the frequent times the Brut family had guests. Guy and Elmira kept their love secret by feigning intellectual but indifferent Platonism. But, meanwhile, Elmira regularly passed notes of love to Guy by leaving them at agreed secret drops in each of the 9 9 rooms, all of the adjacent rooms having a door in common between them.

Because of the frequent presence of guests, and also because either Sir John or Lady Georgina was invariably in one or more of the rooms, Guy would only make a sweep of 3 3 rooms at a time so as to not arouse suspicions, picking up the notes if they happened to be in any of the drops. He would sweep either 3 3 rooms in a row, or 3 3 rooms in a column, before he would try again on another night, and be satisfied with what he found, or did not find. After each night, Elmira would collect any notes left behind, and reseeded the rooms at the next opportunity.

Guy Patrick, while only a commoner, was an avid amateur scientist and mathematician, and had read the latest works of mathematicians of his age, Blaise Pascal, Christian Huygens, Abraham de Moivre, and Jacob Bernoulli whose book on probability theory, Ars Conjectandi , was just published posthumously. Flush with newfound ideas of probability theory, Guy carefully noted certain things about his successes with his 3 3 room sweeps, namely,

1) \; Whenever he made a sweep of 3 3 rooms in a row, either row 1 , 2 , 1, 2, or 3 3 , it did not matter which row he chose; half of the time he would find exactly 2 2 notes, one note in each of 2 2 rooms out of 3 3 which seemed to be at random, and the other half of the time he found nothing.

2) \; Whenever he made a sweep of 3 3 rooms in a column, either column A , B , A, B, or C C , it did not matter which column he chose; he would always find exactly 1 1 note.

Thus, the expected number of notes Guy would find in any 3 3 room sweep was always 1 1 .

Following the ideas and examples shown in the works of these men, Guy carefully reasoned that, because of the fact he never found anything other than exactly 1 1 note when making a sweep of 3 3 rooms in a column, he proposed that what might be happening is that Elmira was always either leaving behind only 2 2 notes, one each in a room, and only the rooms that were in a row, which was chosen at random, or she left behind nothing at all. He made himself a sketch of all 10 10 possibilities as follows:

He hypothesized that each of the first 9 9 cases was occurring with equal probability, while the last case where Elmira left behind nothing was occurring with a probability not necessarily equal to any of the other 9 9 cases. He then did the math on this, figuring out what the probability of the last case of her leaving behind nothing had to be, in order to get a result that matched his experience and observation, which he did successfully.

He knew that Elmira was an extraordinary creature, as men who are in love often say about those they loved, but he was just enough of a mathematician to realize that what was happening was truly extraordinary. He couldn’t quite put his finger on it, but he felt as if he was glimpsing into a future when someday physicists would understand and explain this to him, but right now he found it hard to make any sense out of it, even though the mathematics of it was working out just fine.

What was the probability of Elmira leaving behind nothing? Give answer in decimal form.


Note: This problem is asking for the probability of case 10 10 , assuming that

1) Guy Patrick is indeed finding the notes in the manner he has been experiencing, without either him or Elmira knowing ahead of time how, when, and where the notes would be left and be [or not be] found (other than the location of the secret drops);

2) the probabilities of the first 9 9 cases are all the same, so that once the probability of case 10 10 is computed, the probability of each of the first 9 9 cases immediately follows, given that the total of the probabilities of all 10 10 cases is 1 , 1,

...in order to deliver results that Guy Patrick has been observing, whether or not such results are "possible" in "real life."


The answer is -0.5.

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1 solution

Michael Mendrin
Feb 5, 2017

This graphic shows what are the probabilities of each case occurring. If Guy makes a sweep of, say, Row 1, then the probability of finding 2 2 notes is

1 6 + 1 6 + 1 6 = 1 2 \dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{1}{2}

while the probability of finding nothing is

1 6 + 1 6 + 1 6 + 1 6 + 1 6 + 1 6 1 2 = 1 2 \dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6} -\dfrac{1}{2}=\dfrac{1}{2}

as he had found. Similarly, if Guy makes a sweep of, say, Column A, then the probability of finding 1 1 note is

1 6 + 1 6 + 1 6 + 1 6 + 1 6 + 1 6 = 1 \dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6} =1

while the probability of finding nothing is

1 6 + 1 6 + 1 6 1 2 = 0 \dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{2}=0

as he had also found. Hence, the math works out, and predicts results that agrees with his personal experiences in collecting the notes. So, the probability of Elmira leaving behind nothing is 1 2 = 0.5 -\dfrac{1}{2}=-0.5

Moderator note:

There is possible practical value for negative probabilities beyond the world of quantum physics. The have also been theorized as a tool in finance, like in this article. They remain controversial and the financial literature often treats them as "nonsense" or a failure in the model, but they may aid in certain calculations.

As hinted in your question, negative probability has taken foot in modern physics. I'm interested in this connection, can you tell me more about it?

Julian Poon - 4 years, 4 months ago

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It was first seriously considered by physicists during the "dawn of quantum physics" in the early 20th century, who were coming to the realization that ordinary probabilities were inadequate in explaining quantum behavior. Negative probability today is not a standard feature in quantum theory, it served as a kind of a stepping stone towards quantum probability much like how Bohr's simple model of electron orbits paved the way for quantum atomic orbitals.

Here's one paper by Richard Feynman, one of the principal authors of quantum field theory, on what he had to say about Negative Probability , arguing for its use as a matter of utility. Theoretical physicists are notorious for using things "as a matter of utility", i.e., "if it works, why not use it?"


Michael Mendrin - 4 years, 4 months ago

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Very interesting. I did the question and was uncertain because I got a negative probability but didn't find anything wrong with my analysis. I put the negative half anyways to see and got it right.. Thank you for this input, sir.

Anirudh Chandramouli - 4 years, 4 months ago

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@Anirudh Chandramouli Theoretical physics is often like that. It's like what Sherlock Holmes said, "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth"

Michael Mendrin - 4 years, 4 months ago

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@Michael Mendrin But I find it hard to interpret

As in if I had a probability of zero,I would interpret it as an event that is not going to happen and similarly a probability of 1 would tell me that the event will definitely happen

So, what does a negative probability yet me??

Rohith M.Athreya - 4 years, 4 months ago

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@Rohith M.Athreya In ordinary classical world of hamburgers and solar systems, it doesn't have meaning. But it would be wrong to think that there are no observable physics that doesn't involve non-standard probability. As a matter of fact, quantum mechanics "generates" a "real probability" (i.e., a number between 0 and 1) for a particular event to happen, but the mathematical machinery of quantum mechanics behind it heavily involves non-standard probability. So far, nobody has been able to give a satisfying "intuitive explanation" of how or why this is so. Like with this story, the mathematics works just fine, leaving Guy and the rest of us baffled.

Negative probabilities appear to be suppressive, much like how quantum interference behaves. It negatively influences the probabilities of other outcomes. In this story, for example, it makes impossible for Guy to walk away empty-handed after a 3 room sweep of a column. Quantum interference is a profound fact of quantum reality, a great many ordinary physical behavior that we see depends on this kind of "probability meddling".

Michael Mendrin - 4 years, 4 months ago

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@Michael Mendrin As Feynman said in the paper u attached, the intermediate steps could be rearranged to end up with all positive probabilities but just would complicate the analysis but always yield a positive probability in the end

Can we not suitably rearrange or at least establish that we can do so.

Also, he said that that the system may be applying an additional constraint which may lead to the physical indeterminable nature of such events. What then is such a limitation on the system that has been established in this problem

I just want to try understand from a more experienced person than myself as that paper (first few pages) couldn't make full sense to me

Rohith M.Athreya - 4 years, 4 months ago

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@Rohith M.Athreya As already noted above, quantum physics doesn't use "negative probabilities", because it's using something even harder to make sense out of, involving complex quantities from which a "real" probability is generated to predict a particular event. Nevertheless, by first playing around with the concept of negative probabilities, one can better appreciate the conceptual challenges of quantum physics.

Nobody has ever been able to "do quantum physics" using only ordinary probabilities. That is the reason why physicists speak of the divide between the classical and the quantum. Things behave one way in the classical world, and quite another way in the quantum world.

Michael Mendrin - 4 years, 4 months ago

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@Michael Mendrin Also, suppose you were Guy, and you didn't want to use negative probabilities. Is there another way to explain it using ordinary probabilities? Go give that a try.

Michael Mendrin - 4 years, 4 months ago

This is a fascinating read. Do you know anymore texts that teaches us about Negative Probability ?

Pi Han Goh - 4 years, 3 months ago

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@Pi Han Goh Unfortunately, there is not a lot of literature discussing negative probabilities because, as I have noted elsewhere here, it's been superseded by quantum probability which is based on complex quantities. Nevertheless, Wikipedia does have an article on it as a starting point Negative Probability Interestingly enough, it has found use in mathematical finance. Calvin, are you listening?

Michael Mendrin - 4 years, 3 months ago

@Pi Han Goh Here's a good, interesting short [blog] paper on negative probability. A Neighborhood of Infinity . Here, it specifically mentions Feynman's suggestion that an alternative formulation of quantum probabilities is possible using negative probabilities. So it's possible, it's just that it's not easily workable. The takeaway from this is that when it comes to quantum phenomenon, there isn't any essential difference between the concept of negative probabilities and the use of complex quantities in quantum probability. It's very common to have alternative mathematical formulations for the same things in physics, and which one to use is a matter of choice. "Will that be Hamiltonian or Lagrangian, sir?"

Michael Mendrin - 4 years, 3 months ago

As an algebraic problem this has a solution but it's a real world problem here. And in this world there is one note in each column, or three in total. Which contradicts the evidence of from Guy's sweep of the rows. My conclusion is that he could not analyse the data correctly.

Malcolm Rich - 4 years, 3 months ago

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As I've already asked in other comments here, if you were Guy Patrick, and this was happening to you, what's another way to assign probabilities to make this work? Let's assume this is actually happening instead of arguing that it cannot possibly, because this is the exactly the sort of thing theoretical physicists have to deal with all the time.

Michael Mendrin - 4 years, 3 months ago

I don't completely agree with the solution presented by Michael Mendrin. I don't quite understand how Guy could only find 2 notes half of the time and no notes the other half, whenever sweeping a row. Reading some of the discussion led me to think that Guy got it wrong in interpreting the problem, and those probabilities would only make sense to me if Guy knew in advance that no notes would be found in 1 of 3 row or column.

Rodrigo Machado - 4 years, 3 months ago

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Rodrigo, as I have asked others here, suppose you were Guy Patrick, and this was happening to you. How would you work out the probability numbers? I would be very interested in any alternative probabilistic model of how this could even happen. "Can't happen" isn't an interesting argument. There's thousands of books on standard probability theory that all say this can't happen. Antimatter at one time was another one of those "can't happen" things.

Michael Mendrin - 4 years, 3 months ago

To Challenge Master (and others). I was very careful with the wording of the problem, which is, IF Guy Patrick were to actually have such an experience in picking up the notes, and IF he has hypothesized the 10 10 possible scenarios, 9 9 of them occurring with equal probability and the remaining one with another, what would the "other" probability have to be in order to deliver such a result? There is not a single mention whether this could even happen in the "real" world. In other words, this problem is not supposed to be an debate about whether or not this can even happen in the real world. The problem is supposed to be finding that probability of the case where Elmira has left no notes.

Michael Mendrin - 4 years, 3 months ago

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Indeed you were careful. If you're worried about the note, that was intended to expand on what you said about physicists.

Jason Dyer Staff - 4 years, 3 months ago

How is the probability of finding note in row 1= 1/6?

Aditya Naik - 4 years, 3 months ago

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If we imagine this scenario as "conventional", then this doesn't make any sense, given there are three rows and she'd be equally likely to pick any row.

Essentially, what's happening is even while it is given "without either he or Elmira knowing ahead of time how and when and where the notes would be left" there's an apparent reversal of cause and effect; the probability of finding notes when checking a row is fixed at 1/2 even given any of the 3 rows are equally likely. Since each row has a 1/2 chance of success, then, the probability of an individual case from that row happening is 1/2 / 3 = 1/6. This ends up with a 1/2 + 1/2 + 1/2 = 1.5 overall probability, which isn't possible, hence the extra 10th case which is assigned a negative probability of -0.5 to balance it back to 1.

If this hurts your brain, that's because this is a contextualization of what actually happens in quantum physics.

Jason Dyer Staff - 4 years, 3 months ago

Hi Challenge Master & Others, I have found the column wise sweep probabilities and row wise sweep probabilities.

Column wise:

Case A: dots found in two 1st sweeps.

Case B: dots found in 1st and 3rd sweep.

Case C: dots found in 2nd and 3rd sweep.

No of ways 1st sweep column is chosen: 1/3

No of ways a dot is there: 1/3

=> No of ways a dot is there in the first column: (1/3)*(1/3) = 1/9

No of ways 2nd sweep column is chosen and a dot is there: 1/2 .. [ a dot is present in the same row only.. so either this 2nd sweep column or not in the 2nd sweep column.

=>=> Probability of getting the dots in 1st 2 sweeps= (1/9)*(1/2) = 1/18 ....... (case A) = (case B)

No of ways 1st sweep column is chosen: 1/3

No of ways a dot id not there: 1

=> no of ways that the dot is not there in the first column: (1/3)*1 = 1/3

No of ways any one of the remaining two columns chosen: 1/2

No of ways that Guy gets a dot there: 1/3 => the other dot is in the last column, same row.

=>=> Probability that Guy gets the dots in 2nd two sweeps: (1/3) (1/2) (1/3) = 1/18 ....... (case C)

adding up: probability that Guy gets the two dots by column sweeps: 1/18 + 1/18 + 1/18 = 1/6

Row wise sweep: No of ways a row is chosen: 1/3 => Peobability of getting the two dots with a single row wise sweep.

These two results are different.

from column wise sweep result: the probability of finding nothing: 1-1/6 = 5/6

from row wise sweep result: the probability of finding nothing: 1-1/3 = 2/3

if we consider the identicality of the two dots in row wise sweep: we get the probability as: (1/3)*(1/2)=1/6

if we consider this single line present in the problem statement: "and the other half of the time he found nothing. "

=> the probability of finding nothing at all: 1/2=+0.5

there is no 9*p+q=1 kind of scenario as far as i can understand. Can anybody explain rightly?? if sweeps are considered.

I have also tried to identify the probabilities of each configuration without thinking of the sweeps. it gives me 3/10 probability of finding the first dot and 2/7 possibility of finding the second dot out of all favourable and non-favourable events. => probability of finding two dots: (3/10)*(2/7) = 3/35

9 of such cases: 9*(3/35) = 27/35 probability of finding nothing: 1-(27/35) = 8/35 = 0.2285..

i found a correction: Probability of finding the dots is:

3/7*2/7 = 21/49 (a dot is in col 1);

and (1/4)*(1/4) = 1/16 (no dot in 1st column .. first dot in anywhere in the second column or not there. second dot is anywhere in the 3rd column or not there);

=>(21/49) + (1/16) = 385/784 => 1- (385/784) = +0.5089..

Please please tell me where am I wrong.

Ananya Aaniya - 4 years, 3 months ago

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"Row wise sweep: No of ways a row is chosen: 1/3 => Peobability of getting the two dots with a single row wise sweep."

This bit is wrong. The probability of getting two dots with a single row wise sweep is given in the problem scenario as 1/2:

"half of the time he would find exactly 2 notes, one note in each of rooms out of 3 which seemed to be at random, and the other half of the time he found nothing."

It might help to also look at my response to Aditya Naik below. This is a troubling problem to think about because it's a real-life contextualization of a scenario that can happen in quantum physics.

Jason Dyer Staff - 4 years, 3 months ago

Aside from the unrealizable negative probability, it seems to me this problem is based on a false assumption. Since in all 9 cases where any note was left, one of three columns is always empty. Even if Elmira never failed to leave a note, random choice of a column should once in a while yield no note. Since there is only a finite number of samples, there is no justification to assume that the next sample can't be a null result. If you start with a false assumption, you can "prove" anything. Quantum probabilities only apply in the quantum domain. In the real world of low speeds, classical rules apply.

As an aside, how do you assign a probability of 1/6 to each of 9 equally likely states in the first place?

Tom Capizzi - 4 years, 3 months ago

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As I've already asked elsewhere here in comments, suppose you were Guy Patrick, and this was happening to you. What mathematical model would you propose would "explain" such strange results? I'm more interested in seeing what can be done mathematically rather than hearing arguments "on realistic grounds" why this can't possibly ever be happening. Do you have an alternative probability model that would explain or predict such results?

Michael Mendrin - 4 years, 3 months ago

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