Waiting for the gravy at a dinner tableThe Security to the Party [Part 29]Wine with dinnerThe 169 members of...
Is my plan for fixing my water heater leak bad?
Where is this triangular-shaped space station from?
Closure of presentable objects under finite limits
Is there a German word for “analytics”?
Linear regression when Y is bounded and discrete
Whom do I have to contact for a ticket refund in case of denied boarding (in the EU)?
How to mitigate "bandwagon attacking" from players?
"Murder!" The knight said
Use comma instead of & in table
What is the difference between throw e and throw new Exception(e)?
How to count words in a line
What is better: yes / no radio, or simple checkbox?
What if I store 10TB on azure servers and then keep the vm powered off?
Equivalent to "source" in OpenBSD?
I can't die. Who am I?
How can I be pwned if I'm not registered on that site?
How to properly claim credit for peer review?
How can I find an Adventure or Adventure Path I need that meets certain criteria?
When was drinking water recognized as crucial in marathon running?
Why is working on the same position for more than 15 years not a red flag?
I encountered my boss during an on-site interview at another company. Should I bring it up when seeing him next time?
Skis versus snow shoes - when to choose which for travelling the backcountry?
Compare four integers, return word based on maximum
Is there a better way to make addon working on both blender 2.80 and 2.79?
Waiting for the gravy at a dinner table
The Security to the Party [Part 29]Wine with dinnerThe 169 members of the Dinner ClubHoles in the TableRemoving chips from the tableSolve this math table6, 10, 12, and 15 go out for dinnerUncle's mathematics puzzles after dinnerHelp! I lost my marble(s)!Ernie and the Case of the Singing Sisters
$begingroup$
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.
mathematics
asked 3 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.
mathematics
asked 3 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
50 mins 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.
mathematics
asked 3 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.
mathematics
mathematics
asked 3 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 3 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
edited 3 hours ago
Jaap Scherphuis
asked 3 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 3 hours ago
Jaap ScherphuisJaap Scherphuis
16k12771
asked 3 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
50 mins 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
50 mins ago
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
50 mins ago
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
50 mins 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 1 hour ago
Adam
141119
answered 2 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
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour 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 1 hour 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
1 hour 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
38 mins 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
29 mins 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
15 mins 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
5 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "559"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80243%2fwaiting-for-the-gravy-at-a-dinner-table%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 1 hour ago
Adam
141119
answered 2 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
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour 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 1 hour ago
Adam
141119
answered 2 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
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour 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 1 hour ago
Adam
141119
answered 2 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 1 hour ago
Adam
141119
answered 2 hours ago
Dr XorileDr Xorile
13.3k22572
edited 1 hour ago
Adam
141119
edited 1 hour ago
Adam
141119
edited 1 hour ago
Adam
141119
141119
answered 2 hours ago
Dr XorileDr Xorile
13.3k22572
answered 2 hours ago
Dr XorileDr Xorile
13.3k22572
answered 2 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
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour ago
add a comment |
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour ago
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
2 hours ago
$begingroup$
Did you do a lot of calculations before you arrived at that answer? :-)
$endgroup$
– Jaap Scherphuis
2 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
2 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
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 hours ago
$begingroup$
There is indeed a neat, calculation-free, argument for it.
$endgroup$
– Jaap Scherphuis
2 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
1 hour 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
1 hour 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 1 hour 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
1 hour 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
38 mins 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
29 mins 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
15 mins 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
5 mins 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 1 hour 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
1 hour 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
38 mins 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
29 mins 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
15 mins 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
5 mins 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 1 hour 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 1 hour ago
BassBass
30.3k472186
answered 1 hour ago
BassBass
30.3k472186
answered 1 hour ago
BassBass
30.3k472186
answered 1 hour 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
1 hour 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
38 mins 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
29 mins 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
15 mins 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
5 mins ago
add a comment |
$begingroup$
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
1 hour 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
38 mins 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
29 mins 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
15 mins 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
5 mins 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
1 hour 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
1 hour 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
38 mins 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
38 mins 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
29 mins 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
29 mins 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
15 mins 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
15 mins 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
5 mins 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
5 mins ago
add a comment |
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80243%2fwaiting-for-the-gravy-at-a-dinner-table%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
To clarify, for whatever reason, I'm not allowed to just pick up the gravy boat before anyone else does?
$endgroup$
– MikeTheLiar
50 mins ago