Mixing
Is it possible to assign equal probabilities to individuals and groups of individuals?
$begingroup$
A popular music festival with a fixed number of available tickets uses a system (algorithm?) that randomly picks from a group of registered individuals who is able to purchase a ticket. The group of registered individuals is considerably larger than the number of available tickets. Nothing confusing so far.
However, the festival organization provides the possibility for registered individuals to become part of a group (up to 20 individuals). The idea is that the system either picks everybody or nobody from a group. The festival organization claims that the probability of being picked by the system is equal for individuals and for groups. I wonder whether this is possible or not?
probability statistics algorithms
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
$endgroup$
|
show 3 more comments
$begingroup$
A popular music festival with a fixed number of available tickets uses a system (algorithm?) that randomly picks from a group of registered individuals who is able to purchase a ticket. The group of registered individuals is considerably larger than the number of available tickets. Nothing confusing so far.
However, the festival organization provides the possibility for registered individuals to become part of a group (up to 20 individuals). The idea is that the system either picks everybody or nobody from a group. The festival organization claims that the probability of being picked by the system is equal for individuals and for groups. I wonder whether this is possible or not?
probability statistics algorithms
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
$endgroup$
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45
|
show 3 more comments
$begingroup$
A popular music festival with a fixed number of available tickets uses a system (algorithm?) that randomly picks from a group of registered individuals who is able to purchase a ticket. The group of registered individuals is considerably larger than the number of available tickets. Nothing confusing so far.
However, the festival organization provides the possibility for registered individuals to become part of a group (up to 20 individuals). The idea is that the system either picks everybody or nobody from a group. The festival organization claims that the probability of being picked by the system is equal for individuals and for groups. I wonder whether this is possible or not?
probability statistics algorithms
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
$endgroup$
A popular music festival with a fixed number of available tickets uses a system (algorithm?) that randomly picks from a group of registered individuals who is able to purchase a ticket. The group of registered individuals is considerably larger than the number of available tickets. Nothing confusing so far.
However, the festival organization provides the possibility for registered individuals to become part of a group (up to 20 individuals). The idea is that the system either picks everybody or nobody from a group. The festival organization claims that the probability of being picked by the system is equal for individuals and for groups. I wonder whether this is possible or not?
probability statistics algorithms
probability statistics algorithms
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
edited Jan 22 at 10:22
Quinten Zuurbier
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
asked Jan 22 at 10:16
Quinten ZuurbierQuinten Zuurbier
133
133
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45
|
show 3 more comments
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It is possible if groups are counted as an individual. So if the system forms a list: person1, person2, ..., personN, group1, group2, ..., groupM and then chooses randomly from the list until all the tickets are gone. If the number of available tickets is way larger than 20, the probability to be picked either as an individual or a group member are practically the same.
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
$endgroup$
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3082965%2fis-it-possible-to-assign-equal-probabilities-to-individuals-and-groups-of-indivi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is possible if groups are counted as an individual. So if the system forms a list: person1, person2, ..., personN, group1, group2, ..., groupM and then chooses randomly from the list until all the tickets are gone. If the number of available tickets is way larger than 20, the probability to be picked either as an individual or a group member are practically the same.
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
$endgroup$
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
add a comment |
$begingroup$
It is possible if groups are counted as an individual. So if the system forms a list: person1, person2, ..., personN, group1, group2, ..., groupM and then chooses randomly from the list until all the tickets are gone. If the number of available tickets is way larger than 20, the probability to be picked either as an individual or a group member are practically the same.
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
$endgroup$
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
add a comment |
$begingroup$
It is possible if groups are counted as an individual. So if the system forms a list: person1, person2, ..., personN, group1, group2, ..., groupM and then chooses randomly from the list until all the tickets are gone. If the number of available tickets is way larger than 20, the probability to be picked either as an individual or a group member are practically the same.
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
$endgroup$
It is possible if groups are counted as an individual. So if the system forms a list: person1, person2, ..., personN, group1, group2, ..., groupM and then chooses randomly from the list until all the tickets are gone. If the number of available tickets is way larger than 20, the probability to be picked either as an individual or a group member are practically the same.
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
answered Jan 22 at 10:51
Vasily MitchVasily Mitch
2,3841311
2,3841311
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
add a comment |
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
$begingroup$
I agree with this answer but think it could be improved with an explicit bound (based on the total number of tickets), and a proof of that bound.
$endgroup$
– Mees de Vries
Jan 22 at 10:52
add a comment |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3082965%2fis-it-possible-to-assign-equal-probabilities-to-individuals-and-groups-of-indivi%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
$begingroup$
Consider a simple case in which there are four tickets for six people. If these people are not in group, they each have a probability of $frac{4}{6} = frac{2}{3}$ of being selected. Now put them together in groups of two. We can select two groups to assign the tickets, with every group having a probability of $frac{2}{3}$ of being selected. Can you see how this applies to larger groups as well?
$endgroup$
– jvdhooft
Jan 22 at 10:25
$begingroup$
@jvdhooft What if you have a single person, a group of 2 and a group of 3 with 4 tickets? Then either the single and the double go, or the single and the triple go. The single is sure to get a ticket. Nobody else is.
$endgroup$
– paw88789
Jan 22 at 10:29
$begingroup$
@jvdhooft Thx, I see how that applies to larger groups as well. I think an important assumption here is that the system views a group as an individual?
$endgroup$
– Quinten Zuurbier
Jan 22 at 10:42
$begingroup$
@paw88789 I assumed a fixed group size, but upon reading the question again ("the possibility to become part of a group") you are right that it does not apply when variable group sizes are allowed.
$endgroup$
– jvdhooft
Jan 22 at 10:43
$begingroup$
@QuintenZuurbier The important part is that variable group sizes are possible, which, as paw88789 pointed out above, can result in some individuals having a higher probability of getting a ticket than others.
$endgroup$
– jvdhooft
Jan 22 at 10:45