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The Café Bella-Roma love-seats

A man and a woman go to Café Bella Roma either during the whole morning or whole afternoon, on just one random day every week.

The thing is, if this man and woman ever sit opposite each other, they will fall in love, and the only chance they get to do this is sitting opposite each other at the Café Bella-Roma. They choose where to sit randomly.

The Café has 3 tables, each with 2 chairs opposite each other, and never has any other patrons sitting at the tables when either the man or woman are there.

What is the average number of days it will take them to sit opposite each other and fall in love?

Suppose now that the Café has up to 6 patrons at any one time, and there is equal probability that the seats in the Café are occupied with any of 0 -6 of them.

Now what is the average number of days it will take them to sit opposite each other and fall in love?

Marcello, one of the owners, (both of whom secretly know all of this), at the beginning, Marcello wants to make an even bet that they will have met by a certain number of weeks.

What number should he bet on so that his chances of winning the bet are 50/50?"

The other owner, David, is more civic-minded and wants to know what the best thing he can do is to increase the chances of the man and the woman falling in love. His idea is to put in an extra table. Marcello reckons they should take a chair away from one of the tables. For this part, assume that still randomly 0-6 patrons are occupying the tables at any one time. There is standing room at the bar for extra's.

Which idea is the best?

edit: part c and d have been reworded for greater clarity.

Edited by Thelema

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I woke up this morning and read this little conundrum. For a moment there I tried to work it out, but my brain froze completely and I wasn't sure what the difference between two and six was, and it occurred to me that I needed some coffee, maybe a couple of cigarettes too to get this thing rolling.

Then it occurred to me, what if I was to book a flight to Rome and get to this cafe, after all, the Italians produce both the best coffee and the most beautiful women, perhaps I could get a coffee and meet this beautiful woman before the man gets to sit opposite her? I know the cafe now has up to 4 other patrons, and all of them might be female, so my question to you Thelema, the Mathematician, is, if I were to book a 24hr flight to Rome now, what are the chances I would get to meet this beautiful lady before the man does?

Knowing that the Italian day starts later than ours, we have a few hours to work this out.

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Well there's a 1/1 chance the woman will take a certain seat, on a certain timeslot. There's a 1/7 chance the guy will pick the same day, and a 1/2 chance it will be the same timeslot (1/14 running product). Then there's a 1/5 chance he will pick the only remaining seat which is opposite her (running product 1/70). So with this first set-up, love is expected to happen once every 70 days weeks.

Second and other set-ups will take more thinking ^^. Watch this space.

Edited by CβL

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Q1. i'll assume the second person to arrive never tries to sit on the lap of the first so each time the second person arrives they have a one in five crack at love IF they both attending during the same morning or afternoon, which i'm going to say is one in thirteen odds. (EDIT: AHH YES 1/14 MAKES MORE SENSE)

i don't know how to figure out how many cracks it will take on average to succeed with those odds.

Q2. is there any difference between choosing where to sit randomly, or having the choice made for you by the other seats being filled up randomly? my guess is no. unless you mean to say that there is a two in seven chance that there won't be two free seats for our potential couple. maybe this is more complex than i'd like to think, i don't actually know any of the maths behind odds.

Q3. need a way of working out the average days first.

Q4. adding a table gives the second arrival a one out of seven crack at the right chair. removing a single chair means the first arrival has a four in five chance of dodging the altar-of-infinite-loneliness, then a one in four chance that the second arrival will sit opposite.

Edited by ThunderIdeal

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the answer is: this is an ad for cafe bella roma

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Seems like there is too much coffee gazing not enough pheromones, sideways glances, need for company etc. . What if one day she wears a hot dress or is upset and crying?

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maybe david & marcello should just butt out of other peoples relationships. playing matchmaker is bound to end in tears

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thelema, if your intention was to show that only a math question can have an irrevocable answer.......

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I did a monte carlo simulation to solve this. I can't guarantee that it's correct, but I'm getting 69...for the first question anyway

EDIT: sorry, that was 69 weeks ... it's 485 days approximately

Edited by ballzac

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And I got 683 days for the second question, but of course everything from here on in is dependent on my first result being correct.

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marcello and david should take a fucking long hard look at themselves....$12.50 for a frickin BLT, just coz you cut the crusts off and stick a bloody toothpick with a flag in it!!!!

sorry. continue please

Edited by glimpse
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Well there's a 1/1 chance the woman will take a certain seat, on a certain timeslot. There's a 1/7 chance the guy will pick the same day, and a 1/2 chance it will be the same timeslot (1/14 running product). Then there's a 1/5 chance he will pick the only remaining seat which is opposite her (running product 1/70). So with this first set-up, love is expected to happen once every 70 days.

Second and other set-ups will take more thinking ^^. Watch this space.

I missed this before, but glad I had another look through the thread, because this supports my numerical result for question 1, though you made the same mistake I did :)

For question 3, if I'm understanding it correctly, we want the mode: After one million trials, the mode I found was (EDIT: well, it seems hard to believe, but I get a mode of 3 days, with a probability of 0.15%)

For question 4, assuming that there can be 0-5 or 0-8 patrons with equal probability for each part of the question, I found that if you take a chair away, the mean number of days before they meet is 731, and if you add a table (two chairs), the mean is 869. So both appear to be worse than just leaving it as is, but out of the two, taking away a chair is better.

Edited by ballzac

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Ugh I interchanged days and weeks. Amateur mistake. I'm still thinking hard on whether the number 70 is correct though (70 weeks).

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PART B
Well the woman has a 1/7 chance of arriving and finding no seat (ruining the chance of love that week, 7 as it's 0 to 6 inclusive). She has a 1/7 chance of taking the last seat (also ruining the chance). So there's a 5/7 chance she will select a seat at some point during the week and still leave at least one seat left.

Now the guy has the same chances before as finding her, which is 1/14.
But he also has the probabilities of getting the right seat, which are (1/5seats)*(1/5scenarios) + (1/4seats)*(1/5) + (1/3seats)*(1/5) + (1/2seat2)*(1/5) + (1/1seat)*(1/5) = (1/5)*(1/5 + 1/4 + 1/3 + 1/2 + 1/1)

(The 5 scenarios are from the 5/7 above, in that there are: 1) 5 empty seats, 2) 4 empty seats, 3) 3 empty seats, 2) 2 empty seats, 1) 1 empty seat). Each has equal probability (hence the 1/5 probability). It's not 1/7 as the two errant scenarios are removed by the 5/7 multiplier. It is possible to calculate this using 1/7 as the scenario multiplier, except that the possibility of success will be 0 in both of the errant scenarios, and the 5/7 multiplier would not be used in this case).

All together = (5/7) * (1/14) * (1/5)*(1/5 + 1/4 + 1/3 + 1/2 + 1/1) == (1/14) * (1/7)*(1/5 + 1/4 + 1/3 + 1/2 + 1/1 + 0 + 0) [the second way of calcing this as described above]


Which says that on average, love will happen once every 43 weeks (42.9). At first this confused me. It still kind of does. I suppose that having less empty seats kinda forces them to sit together. Maybe, unless I made a mistake. :scratchhead:


Edit due to ballzac's astute observation:
Brute force will save the day.


The guy has the same chances before as finding her, which is 1/14

Assuming they found each-other, then there are these master scenarios:

Case 6 free seats: 1/1 chance she will select an empty table. Guy has 1/5 chance of sitting across from her. P = 1/5


Case 5 free seats (1 other sitting) : 4/5 chance she will select an empty table. Guy then has 1/4 chance of sitting across from her. P = (4/5)*(1/4) = 4/20 {probably one of the few times I've typed 4/20 and meant to)


Case 4 free seats: (1/1 chance perp1 sits down, 1/5 chance perp2 will fill the table, 4/5 chance perp2 will sit at another table). => (1/5)*(1/1 chance she will take an empty table)(1/3 chance guy will sit across from her) + (4/5)(2/4 chance she will select empty table)(1/3 chance guy will sit across from her) P = (1/5)*(1/3) + (4/5)*(2/4)*(1/3) = 1/15 + 8/60


Case 3 free seats:

1/1 chance perp1 sits down.

--1/5 chance perp2 will fill the table

------1/1 chance perp3 sits down at an empty table
---------2/3 chance girl sits down at an empty table

------------1/2 chance guy sits across from girl

--4/5 chance perp2 will sit at another table.

------2/4 chance perp3 will fill a table

---------2/3 chance girl will sit at an empty table

------------1/2 chance guy sits across from girl

------2/4 chance perp3 will sit at another table

---------no chance for girl and guy to meet.

P = (1/5)(2/3)(1/2) + (4/5)(2/4)(2/3)(1/2) = 2/30 + 16/120

Case 2 free seats:

(0 = empty, 1 = filled). Bold = unique arrangement (making permutations of 2 free seats using binary rotations). Italic = a free table.

001111, 011110, 111100, 111001, 110011, 100111

010111, 101110, 011101, 111010, 110101, 101011

011011, 110110, 101101, 011011, 110110, 101101
011101, 111010, 110101, .......

15 unique arrangements. 3/15 chance she will be able to get an empty table. 1/1 chance he will then take the last remaining seat.

P = 3/15

Case 1 free seats: Woman sits down, man goes home. P = 0.


Case 0 free seats: Woman goes home, man goes home. P = 0.


P_total = (1/14) * ( (1/7)*(1/5) + (1/7)*(4/20) + (1/7)*(1/15 + 8/60) + (1/7)*(2/30 + 16/120) + (1/7)*(3/15) )

Which is 98 weeks.

Edited by CβL
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Which says that on average, love will happen once every 43 weeks (42.9). At first this confused me. It still kind of does. I suppose that having less empty seats kinda forces them to sit together. Maybe, unless I made a mistake. :scratchhead:

I think there are more scenarios than you gave. You gave the best case scenarios, that the seats are filled up in order. Consider that there may be only 3 patrons, each at a different table, and the probability of the couple meeting is zero. There are many more scenarios where the probability is lower than the best case scenario ... for each of the patron numbers. I think this explains why my numerical result is higher than your analytical one, though I wouldn't stake my life on mine being correct either.

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Ah yes, of course. Fuck. :P

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This histogram explains why the mode is so low (three days) while the average is much higher:

No other patrons:

cafe_histo.jpg

Including other patrons:

cafe_histo_with_patrons.jpg

ETA: The x-axis is the number of days until they meet. I've arbitrarily divided it into 100 bins, and this is for 10000 trials.

Edited by ballzac

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Oh, of course. The day to bet on is day 1. Each day has the same probability for them meeting, unless they've already met. So the probability has to continually drop. Hence why the mode is 1. Even after a million trials, there must be enough noise in my data to shift the mode to 3.

 

Well there's a 1/1 chance the woman will take a certain seat, on a certain timeslot. There's a 1/7 chance the guy will pick the same day, and a 1/2 chance it will be the same timeslot (1/14 running product). Then there's a 1/5 chance he will pick the only remaining seat which is opposite her (running product 1/70). So with this first set-up, love is expected to happen once every 70 days.

Second and other set-ups will take more thinking ^^. Watch this space.

In light of my comments, I think there's a problem with your analysis here. There's a 1/70 chance they will meet in week 1, but there is a (1/70)*(1-1/70) chance they will meet on day two, NOT (1/70)*(1/70), which I think your analysis relies on. This will shift the mean only slightly , and it could explain why I get 69 in my simulations, though it could also just be noise (I only ran the 1000000 trial simulation once.

(ETA: I think in practical terms it makes FA difference, so my value of 69 would just be due to not enough trials, but it's at least of analytical interest that it would be technically less than 70)

Edited by ballzac

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I don't think that we need to consider the days of the week separately. I can't quite say why, I'm too confused (after saying something over and over it loses its meaning :scratchhead: )

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I don't think that we need to consider the days of the week separately. I can't quite say why, I'm too confused (after saying something over and over it loses its meaning :scratchhead: )

We do need to consider the days separately. The question is not how often they will meet. It's a question of how long until they FIRST meet.

Consider a simplified example: What is the probability that you will get the first head on the second toss of a coin?

It's not 1/2. It is (1/2)*(1-1/2) = 1/4. There are four possibilities

HH

HT

TH

TT

In the first two cases, your first head was on the first toss, so they don't count. The only one that counts is number three. The chance that you will get the first head on the third toss is

(1/2)*(1-((1/2)*(1-1/2)+(1/2))) = 1/8

It is the same for when these two first meet. On the first day it is 1/70, but on the second day it is (1/70)*(1-1/70), and on the third day it is (1/70)*(1-((1/70)*(1-1/70)+(1/70))) The difference by the third day is only 1.4%, but it's non-zero. The histograms show this clearly.

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Such heavy head-tripping.

Damn, you'll positively scare off any potential partner as someone who is just stuck in their heads.

I think I'll go and eat my lawn now.

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The probability I have calculated, is that if there were infinite cafes, and infinite guys and girls (i.e. infinite trials of this), then the average number of weeks it would take is 70. This number includes the couples who met on the first day implicitly, and it is not needed to explicitly include them.

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Brute force will save the day.

...

Which is 98 weeks.

Awesome. I'm a big fan of brute force methods :)

That analysis looks complete to me, and we now have agreement between the monte-carlo simulation and the brute force result. :wub:

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The probability I have calculated, is that if there were infinite cafes, and infinite guys and girls (i.e. infinite trials of this), then the average number of weeks it would take is 70. This number includes the couples who met on the first day implicitly, and it is not needed to explicitly include them.

You're right. The average number of days until they meet is the probability that they will meet on day one plus the probability that they do NOT meet on day one multiplied by the average plus one:

Avg = P(meet) + P(!meet)(1+Avg)

=>Avg*(1-P(!meet)) = P(meet) + P(!meet)

=>Avg*(1 - (1-P(meet)) ) = P(meet) + 1 - P(meet)

=>Avg*P(meet) = 1

=>Avg = 1/P(meet)

=70

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Oh, of course. The day to bet on is day 1. Each day has the same probability for them meeting, unless they've already met. So the probability has to continually drop. Hence why the mode is 1. Even after a million trials, there must be enough noise in my data to shift the mode to 3.

I just realised that this is wrong, because each day in week one is equally likely because they are independent, but any day in week two is less likely than a day in week one, so if you were to bet, you would be best off picking any random day from week 1. This would explain why I didn't get 1 as the mode in my simulation.

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