Expected number of trials before I get two out of three types
I am a biologist trying to quantify the likelihood of observed genetic phenomena… I’ve tried my best to formalize my question in accepted terms as follows:
Given a stack of n cards, each colored red (x cards) , blue (y cards), or green (z cards) where x + y + z = n, what is the expected number of draws, without replacement, before you draw both a red and a blue card? It doesn’t matter whether the red or blue card comes first. Preferably, a solution would be defined in general terms, but for demonstration’s sake, one could assume a stack of 10 cards, where x = 3, y = 5, z = 2.
Intuitively, I figured this could be solved using multivariate hypergeometric distributions. However, because we are looking for the expected minimal number of draws before two successes occur (both a red and a blue card), this approach appeared to require some tricky combinatorics.
I have also looked, without success, into adopting solutions from other problems. My sense is that between the following two posts there is a solution, but one I have not been able to intuit.
Expected number of card draws to get all 4 suits
Expected number of trials before I get one of each type
Any direction would be greatly appreciated -- thanks so much!
probability conditional-probability
asked Nov 23 at 19:07
hemen
62
add a comment |
I am a biologist trying to quantify the likelihood of observed genetic phenomena… I’ve tried my best to formalize my question in accepted terms as follows:
Given a stack of n cards, each colored red (x cards) , blue (y cards), or green (z cards) where x + y + z = n, what is the expected number of draws, without replacement, before you draw both a red and a blue card? It doesn’t matter whether the red or blue card comes first. Preferably, a solution would be defined in general terms, but for demonstration’s sake, one could assume a stack of 10 cards, where x = 3, y = 5, z = 2.
Intuitively, I figured this could be solved using multivariate hypergeometric distributions. However, because we are looking for the expected minimal number of draws before two successes occur (both a red and a blue card), this approach appeared to require some tricky combinatorics.
I have also looked, without success, into adopting solutions from other problems. My sense is that between the following two posts there is a solution, but one I have not been able to intuit.
Expected number of card draws to get all 4 suits
Expected number of trials before I get one of each type
Any direction would be greatly appreciated -- thanks so much!
probability conditional-probability
asked Nov 23 at 19:07
hemen
62
add a comment |
I am a biologist trying to quantify the likelihood of observed genetic phenomena… I’ve tried my best to formalize my question in accepted terms as follows:
Given a stack of n cards, each colored red (x cards) , blue (y cards), or green (z cards) where x + y + z = n, what is the expected number of draws, without replacement, before you draw both a red and a blue card? It doesn’t matter whether the red or blue card comes first. Preferably, a solution would be defined in general terms, but for demonstration’s sake, one could assume a stack of 10 cards, where x = 3, y = 5, z = 2.
Intuitively, I figured this could be solved using multivariate hypergeometric distributions. However, because we are looking for the expected minimal number of draws before two successes occur (both a red and a blue card), this approach appeared to require some tricky combinatorics.
I have also looked, without success, into adopting solutions from other problems. My sense is that between the following two posts there is a solution, but one I have not been able to intuit.
Expected number of card draws to get all 4 suits
Expected number of trials before I get one of each type
Any direction would be greatly appreciated -- thanks so much!
probability conditional-probability
asked Nov 23 at 19:07
hemen
62
I am a biologist trying to quantify the likelihood of observed genetic phenomena… I’ve tried my best to formalize my question in accepted terms as follows:
Given a stack of n cards, each colored red (x cards) , blue (y cards), or green (z cards) where x + y + z = n, what is the expected number of draws, without replacement, before you draw both a red and a blue card? It doesn’t matter whether the red or blue card comes first. Preferably, a solution would be defined in general terms, but for demonstration’s sake, one could assume a stack of 10 cards, where x = 3, y = 5, z = 2.
Intuitively, I figured this could be solved using multivariate hypergeometric distributions. However, because we are looking for the expected minimal number of draws before two successes occur (both a red and a blue card), this approach appeared to require some tricky combinatorics.
I have also looked, without success, into adopting solutions from other problems. My sense is that between the following two posts there is a solution, but one I have not been able to intuit.
Expected number of card draws to get all 4 suits
Expected number of trials before I get one of each type
Any direction would be greatly appreciated -- thanks so much!
probability conditional-probability
probability conditional-probability
asked Nov 23 at 19:07
hemen
62
asked Nov 23 at 19:07
hemen
62
asked Nov 23 at 19:07
hemen
62
asked Nov 23 at 19:07
hemen
62
asked Nov 23 at 19:07
hemen
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $yleq x.$ Then apply the inclusion exclusion approach in the first answer you linked.
answered Nov 23 at 19:59
kodlu
3,380716
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%2f3010717%2fexpected-number-of-trials-before-i-get-two-out-of-three-types%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
This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $yleq x.$ Then apply the inclusion exclusion approach in the first answer you linked.
answered Nov 23 at 19:59
kodlu
3,380716
add a comment |
This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $yleq x.$ Then apply the inclusion exclusion approach in the first answer you linked.
answered Nov 23 at 19:59
kodlu
3,380716
add a comment |
This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $yleq x.$ Then apply the inclusion exclusion approach in the first answer you linked.
answered Nov 23 at 19:59
kodlu
3,380716
This may help. Clearly the maximum number of draws is $z+x+1$ assuming with no loss of generality that $yleq x.$ Then apply the inclusion exclusion approach in the first answer you linked.
answered Nov 23 at 19:59
kodlu
3,380716
answered Nov 23 at 19:59
kodlu
3,380716
answered Nov 23 at 19:59
kodlu
3,380716
answered Nov 23 at 19:59
kodlu
3,380716
3,380716
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3010717%2fexpected-number-of-trials-before-i-get-two-out-of-three-types%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
