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Waiting for the gravy at a dinner table

Variation of 100 prisoners light problemThe Security to the Party [Part 29]Wine with dinnerThe 169 members of the Dinner ClubCutting the Cheese6, 10, 12, and 15 go out for dinner100 Prisoners' Names in Boxes - Save them all!Ernie and the Optical GyroscopesHelp! I lost my marble(s)!The “Perfect” Lineup

10

$begingroup$

You are sitting at a round dinner table with $N$ other people (i.e. $N+1$ people all together). One of the other guests is the first to pick up the gravy boat and pour some on their potatoes. It then gets passed at random to either of the adjacent people ($50%$ each way). This person, after pouring themselves some gravy, also passes it to a neighbour, and as they are so engrossed in conversation it is again equally likely to be either neighbour. This continues, with the gravy boat being passed from person to person in a random direction each time.

You don't want to be the last person to get any gravy. For you to have the smallest probability of being last, which other guest should be the first to use the gravy? And what is that probability?

I don't know the original source of this puzzle, but it has been doing the rounds for a few years and in my opinion it is destined to become a classic. See for example this Math SE question.

mathematics

share|improve this question

edited 23 hours ago

Jaap Scherphuis

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
  $endgroup$
  – MikeTheLiar
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MikeTheLiar "One of the other guests is the first to pick up the gravy" and "It then gets passed randomly to either adjacent person" and "which other guest should be first to use the gravy" is sufficient enough to clarify
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this the way one should organize seating and serving at a party?
  $endgroup$
  – Try Hard
  20 hours ago
  
  
  

  
  
  
  

add a comment | 

10

$begingroup$

You are sitting at a round dinner table with $N$ other people (i.e. $N+1$ people all together). One of the other guests is the first to pick up the gravy boat and pour some on their potatoes. It then gets passed at random to either of the adjacent people ($50%$ each way). This person, after pouring themselves some gravy, also passes it to a neighbour, and as they are so engrossed in conversation it is again equally likely to be either neighbour. This continues, with the gravy boat being passed from person to person in a random direction each time.

You don't want to be the last person to get any gravy. For you to have the smallest probability of being last, which other guest should be the first to use the gravy? And what is that probability?

I don't know the original source of this puzzle, but it has been doing the rounds for a few years and in my opinion it is destined to become a classic. See for example this Math SE question.

mathematics

share|improve this question

edited 23 hours ago

Jaap Scherphuis

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
  $endgroup$
  – MikeTheLiar
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MikeTheLiar "One of the other guests is the first to pick up the gravy" and "It then gets passed randomly to either adjacent person" and "which other guest should be first to use the gravy" is sufficient enough to clarify
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this the way one should organize seating and serving at a party?
  $endgroup$
  – Try Hard
  20 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

You are sitting at a round dinner table with $N$ other people (i.e. $N+1$ people all together). One of the other guests is the first to pick up the gravy boat and pour some on their potatoes. It then gets passed at random to either of the adjacent people ($50%$ each way). This person, after pouring themselves some gravy, also passes it to a neighbour, and as they are so engrossed in conversation it is again equally likely to be either neighbour. This continues, with the gravy boat being passed from person to person in a random direction each time.

You don't want to be the last person to get any gravy. For you to have the smallest probability of being last, which other guest should be the first to use the gravy? And what is that probability?

I don't know the original source of this puzzle, but it has been doing the rounds for a few years and in my opinion it is destined to become a classic. See for example this Math SE question.

mathematics

share|improve this question

edited 23 hours ago

Jaap Scherphuis

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

$endgroup$

share|improve this question

edited 23 hours ago

Jaap Scherphuis

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

share|improve this question

edited 23 hours ago

Jaap Scherphuis

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

asked yesterday

Jaap ScherphuisJaap Scherphuis

16k12772

2 Answers
2

active

oldest

votes

7

$begingroup$

Fascinating.

The probability of being last is $1/N$ regardless of which guest starts with the gravy.

I had to do the calculation with Markov chains to get the answer. But after getting it I thought about it.

It is a certainty that one of your neighbors will get it before you. You can choose that neighbor by sitting next to the first person, or you can wait patiently until one of your neighbors gets it.


From there, to be the last person the gravy needs to go all the way around to your other neighbor without first getting to you. There is some probability that it does this.


But, actually, everyone has the same situation. One of their neighbors will get it and then there is some probability that it goes all the way around from there to your other neighbor without coming back to you through your first neighbor.


Since everyone is in the same situation, the probability of being the last is the same for everyone and so everyone has a $1/n$ chance to be the last person getting the gravy.

share|improve this answer

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you do a lot of calculations before you arrived at that answer? :-)
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did - I used Markov chains. I'll write it up, but I'm trying to think of the clever way of getting there which must exist.
  $endgroup$
  – Dr Xorile
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is indeed a neat, calculation-free, argument for it.
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow this makes a whole lot of sense. rot13(Gur gnoyr tvirf n snyfr frafr bs cebterff jurer va ernyvgl vgf whfg yvxr lbh ner cnffvat vg qverpgyl gb n enaqbz crefba. Ab jbaqre gur ceboyrz vf cbchyne)
  $endgroup$
  – Adam
  yesterday
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose that your neighbour starts, and call the probability of you being last "x".

Then, suppose that some other person starts.

There is exactly one way for you to be the last one:

* first, the gravy comes to your neighbour

* next, it goes all around the table to your other neighbour.

As we very well know, the first happens

with probability 1

and the second happens

with probability x.

Therefore,

it doesn't matter who starts, and the probability is always the same.

By symmetry

the same applies for all the guests who didn't pick up the gravy

so the probability of it being exactly you who is last is

$frac{1}{text N}$

share|improve this answer

answered yesterday

BassBass

30.3k472186

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  can you expand on the "therefore" because I just see you restating your assumptions and then concluding something from them that I don't see following.
  $endgroup$
  – Kate Gregory
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KateGregory You get the probability of two things both happening by multiplying their individual probabilities together. In this case, that product is x, which, well, is equal to x.
  $endgroup$
  – Bass
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But you said "the chance of it reaching you last is x therefore the chance of it reaching you last is x" -- how is that proving anything? It is the "therefore" part that makes no sense.
  $endgroup$
  – Kate Gregory
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with Kate. I get what you are going for however the way you explained it on the surface can sound like you don't understand the conclusion at all. Explain how the gravy inevitably propagates around the table, not the consequence of the propagation. You say the probability of the consequence is 'X' and therefore the consequence of the probability is 'X' so where is the definition of 'X'?
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well of course you say what 'X' is at the beginning however I was referring to how you explain the consequence of event 'X' and the probability of 'X' but never quite 'X'. It is difficult to explain, when you read your answer while already knowing the solution it makes perfect sense but in reality you have a jump in your argument at "therefore" (even although there is no jump in logic) and the reader is forced to make their own conclusion at this point. I don't think it helps that everything is in little pieces, its really just a structure problem.
  $endgroup$
  – Adam
  22 hours ago
  

  
  
  
  

 | 
show 2 more comments

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7

$begingroup$

Fascinating.

The probability of being last is $1/N$ regardless of which guest starts with the gravy.

I had to do the calculation with Markov chains to get the answer. But after getting it I thought about it.

It is a certainty that one of your neighbors will get it before you. You can choose that neighbor by sitting next to the first person, or you can wait patiently until one of your neighbors gets it.


From there, to be the last person the gravy needs to go all the way around to your other neighbor without first getting to you. There is some probability that it does this.


But, actually, everyone has the same situation. One of their neighbors will get it and then there is some probability that it goes all the way around from there to your other neighbor without coming back to you through your first neighbor.


Since everyone is in the same situation, the probability of being the last is the same for everyone and so everyone has a $1/n$ chance to be the last person getting the gravy.

share|improve this answer

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you do a lot of calculations before you arrived at that answer? :-)
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did - I used Markov chains. I'll write it up, but I'm trying to think of the clever way of getting there which must exist.
  $endgroup$
  – Dr Xorile
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is indeed a neat, calculation-free, argument for it.
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow this makes a whole lot of sense. rot13(Gur gnoyr tvirf n snyfr frafr bs cebterff jurer va ernyvgl vgf whfg yvxr lbh ner cnffvat vg qverpgyl gb n enaqbz crefba. Ab jbaqre gur ceboyrz vf cbchyne)
  $endgroup$
  – Adam
  yesterday
  
  
  

  
  
  
  

add a comment | 

7

$begingroup$

Fascinating.

The probability of being last is $1/N$ regardless of which guest starts with the gravy.

I had to do the calculation with Markov chains to get the answer. But after getting it I thought about it.

It is a certainty that one of your neighbors will get it before you. You can choose that neighbor by sitting next to the first person, or you can wait patiently until one of your neighbors gets it.


From there, to be the last person the gravy needs to go all the way around to your other neighbor without first getting to you. There is some probability that it does this.


But, actually, everyone has the same situation. One of their neighbors will get it and then there is some probability that it goes all the way around from there to your other neighbor without coming back to you through your first neighbor.


Since everyone is in the same situation, the probability of being the last is the same for everyone and so everyone has a $1/n$ chance to be the last person getting the gravy.

share|improve this answer

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you do a lot of calculations before you arrived at that answer? :-)
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did - I used Markov chains. I'll write it up, but I'm trying to think of the clever way of getting there which must exist.
  $endgroup$
  – Dr Xorile
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is indeed a neat, calculation-free, argument for it.
  $endgroup$
  – Jaap Scherphuis
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow this makes a whole lot of sense. rot13(Gur gnoyr tvirf n snyfr frafr bs cebterff jurer va ernyvgl vgf whfg yvxr lbh ner cnffvat vg qverpgyl gb n enaqbz crefba. Ab jbaqre gur ceboyrz vf cbchyne)
  $endgroup$
  – Adam
  yesterday
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Fascinating.

The probability of being last is $1/N$ regardless of which guest starts with the gravy.

I had to do the calculation with Markov chains to get the answer. But after getting it I thought about it.

It is a certainty that one of your neighbors will get it before you. You can choose that neighbor by sitting next to the first person, or you can wait patiently until one of your neighbors gets it.


From there, to be the last person the gravy needs to go all the way around to your other neighbor without first getting to you. There is some probability that it does this.


But, actually, everyone has the same situation. One of their neighbors will get it and then there is some probability that it goes all the way around from there to your other neighbor without coming back to you through your first neighbor.


Since everyone is in the same situation, the probability of being the last is the same for everyone and so everyone has a $1/n$ chance to be the last person getting the gravy.

share|improve this answer

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

$endgroup$

share|improve this answer

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

edited yesterday

Adam

181119

edited yesterday

Adam

181119

answered yesterday

Dr XorileDr Xorile

13.4k22572

answered yesterday

Dr XorileDr Xorile

13.4k22572

2

$begingroup$

Suppose that your neighbour starts, and call the probability of you being last "x".

Then, suppose that some other person starts.

There is exactly one way for you to be the last one:

* first, the gravy comes to your neighbour

* next, it goes all around the table to your other neighbour.

As we very well know, the first happens

with probability 1

and the second happens

with probability x.

Therefore,

it doesn't matter who starts, and the probability is always the same.

By symmetry

the same applies for all the guests who didn't pick up the gravy

so the probability of it being exactly you who is last is

$frac{1}{text N}$

share|improve this answer

answered yesterday

BassBass

30.3k472186

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  can you expand on the "therefore" because I just see you restating your assumptions and then concluding something from them that I don't see following.
  $endgroup$
  – Kate Gregory
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KateGregory You get the probability of two things both happening by multiplying their individual probabilities together. In this case, that product is x, which, well, is equal to x.
  $endgroup$
  – Bass
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But you said "the chance of it reaching you last is x therefore the chance of it reaching you last is x" -- how is that proving anything? It is the "therefore" part that makes no sense.
  $endgroup$
  – Kate Gregory
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with Kate. I get what you are going for however the way you explained it on the surface can sound like you don't understand the conclusion at all. Explain how the gravy inevitably propagates around the table, not the consequence of the propagation. You say the probability of the consequence is 'X' and therefore the consequence of the probability is 'X' so where is the definition of 'X'?
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well of course you say what 'X' is at the beginning however I was referring to how you explain the consequence of event 'X' and the probability of 'X' but never quite 'X'. It is difficult to explain, when you read your answer while already knowing the solution it makes perfect sense but in reality you have a jump in your argument at "therefore" (even although there is no jump in logic) and the reader is forced to make their own conclusion at this point. I don't think it helps that everything is in little pieces, its really just a structure problem.
  $endgroup$
  – Adam
  22 hours ago
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

Suppose that your neighbour starts, and call the probability of you being last "x".

Then, suppose that some other person starts.

There is exactly one way for you to be the last one:

* first, the gravy comes to your neighbour

* next, it goes all around the table to your other neighbour.

As we very well know, the first happens

with probability 1

and the second happens

with probability x.

Therefore,

it doesn't matter who starts, and the probability is always the same.

By symmetry

the same applies for all the guests who didn't pick up the gravy

so the probability of it being exactly you who is last is

$frac{1}{text N}$

share|improve this answer

answered yesterday

BassBass

30.3k472186

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  can you expand on the "therefore" because I just see you restating your assumptions and then concluding something from them that I don't see following.
  $endgroup$
  – Kate Gregory
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KateGregory You get the probability of two things both happening by multiplying their individual probabilities together. In this case, that product is x, which, well, is equal to x.
  $endgroup$
  – Bass
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But you said "the chance of it reaching you last is x therefore the chance of it reaching you last is x" -- how is that proving anything? It is the "therefore" part that makes no sense.
  $endgroup$
  – Kate Gregory
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with Kate. I get what you are going for however the way you explained it on the surface can sound like you don't understand the conclusion at all. Explain how the gravy inevitably propagates around the table, not the consequence of the propagation. You say the probability of the consequence is 'X' and therefore the consequence of the probability is 'X' so where is the definition of 'X'?
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well of course you say what 'X' is at the beginning however I was referring to how you explain the consequence of event 'X' and the probability of 'X' but never quite 'X'. It is difficult to explain, when you read your answer while already knowing the solution it makes perfect sense but in reality you have a jump in your argument at "therefore" (even although there is no jump in logic) and the reader is forced to make their own conclusion at this point. I don't think it helps that everything is in little pieces, its really just a structure problem.
  $endgroup$
  – Adam
  22 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Suppose that your neighbour starts, and call the probability of you being last "x".

Then, suppose that some other person starts.

There is exactly one way for you to be the last one:

* first, the gravy comes to your neighbour

* next, it goes all around the table to your other neighbour.

As we very well know, the first happens

with probability 1

and the second happens

with probability x.

Therefore,

it doesn't matter who starts, and the probability is always the same.

By symmetry

the same applies for all the guests who didn't pick up the gravy

so the probability of it being exactly you who is last is

$frac{1}{text N}$

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$endgroup$

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answered yesterday

BassBass

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answered yesterday

BassBass

30.3k472186

answered yesterday

BassBass

30.3k472186