Waiting for the gravy at a dinner table
$begingroup$
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
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
$endgroup$
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
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
$endgroup$
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
@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
8 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
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
$endgroup$
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
mathematics
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
edited 1 hour ago
Jaap Scherphuis
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 12 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
16k12771
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
@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
8 hours ago
add a comment |
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
@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
8 hours ago
$begingroup$
@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
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
edited 9 hours ago
Adam
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
$endgroup$
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
add a comment |
$begingroup$
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
$frac{1}{text N}$
answered 9 hours ago
BassBass
30.3k472186
$endgroup$
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
1
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 hours ago
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited 9 hours ago
Adam
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
$endgroup$
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
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.
edited 9 hours ago
Adam
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
$endgroup$
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
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.
edited 9 hours ago
Adam
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
$endgroup$
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.
edited 9 hours ago
Adam
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
edited 9 hours ago
Adam
141119
edited 9 hours ago
Adam
141119
edited 9 hours ago
Adam
141119
141119
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
answered 10 hours ago
Dr XorileDr Xorile
13.3k22572
13.3k22572
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
add a comment |
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
add a comment |
$begingroup$
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
$frac{1}{text N}$
answered 9 hours ago
BassBass
30.3k472186
$endgroup$
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
1
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 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
$frac{1}{text N}$
answered 9 hours ago
BassBass
30.3k472186
$endgroup$
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
1
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 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
$frac{1}{text N}$
answered 9 hours ago
BassBass
30.3k472186
$endgroup$
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
$frac{1}{text N}$
answered 9 hours ago
BassBass
30.3k472186
answered 9 hours ago
BassBass
30.3k472186
answered 9 hours ago
BassBass
30.3k472186
answered 9 hours ago
BassBass
30.3k472186
30.3k472186
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
1
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 hours ago
|
show 2 more comments
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
1
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
@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
9 hours ago
1
1
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
I think what you're saying is that it doesn't matter who starts, because it can never reach you without first reaching one or the other of your neighbours, at which point it is exactly as though your neighbor started.
$endgroup$
– Kate Gregory
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
|
show 2 more comments
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$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
9 hours ago
$begingroup$
@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
8 hours ago