Probability of flipping 1 coin vs 3 to land a head given previous successes
$begingroup$
(Assuming 1/2 probability for heads or tails)
Using coin 1 I flip it say $n$ times and I have 60 heads, so probability of head:
$frac{60}{n}, n>0$
Using three coins I flip the first $m$ times and get $6$ heads, the second I flip $p$ times and get also $6$ heads, the third coin flipped $q$ times gets $7$ heads:
Coin 2: $frac{6}{m}$
Coin 3: $frac{6}{p}$
Coin 4: $frac{7}{q}$
and since they are independent probability of a head with the 3 coins is
$frac{6}{m}+frac{6}{p}+frac{7}{q}=frac{6pq+6mq+7mp}{mpq}, q,m, p>0 $
and assume : $n>q>m geq p>0$
Now I want to ask if I am to flip the coins again once, which "group" would give me the highest probability of getting a head.
Flipping only Coin 1 once or flipping all of Coins 2,3 and 4 once. Is it possible to reason without knowing $n,m,p,q$ If not what minimal information might we need to further assume.
probability
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
$endgroup$
add a comment |
$begingroup$
(Assuming 1/2 probability for heads or tails)
Using coin 1 I flip it say $n$ times and I have 60 heads, so probability of head:
$frac{60}{n}, n>0$
Using three coins I flip the first $m$ times and get $6$ heads, the second I flip $p$ times and get also $6$ heads, the third coin flipped $q$ times gets $7$ heads:
Coin 2: $frac{6}{m}$
Coin 3: $frac{6}{p}$
Coin 4: $frac{7}{q}$
and since they are independent probability of a head with the 3 coins is
$frac{6}{m}+frac{6}{p}+frac{7}{q}=frac{6pq+6mq+7mp}{mpq}, q,m, p>0 $
and assume : $n>q>m geq p>0$
Now I want to ask if I am to flip the coins again once, which "group" would give me the highest probability of getting a head.
Flipping only Coin 1 once or flipping all of Coins 2,3 and 4 once. Is it possible to reason without knowing $n,m,p,q$ If not what minimal information might we need to further assume.
probability
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
$endgroup$
1
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we'reAssuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.
$endgroup$
– zipirovich
Dec 7 '18 at 3:28
add a comment |
$begingroup$
(Assuming 1/2 probability for heads or tails)
Using coin 1 I flip it say $n$ times and I have 60 heads, so probability of head:
$frac{60}{n}, n>0$
Using three coins I flip the first $m$ times and get $6$ heads, the second I flip $p$ times and get also $6$ heads, the third coin flipped $q$ times gets $7$ heads:
Coin 2: $frac{6}{m}$
Coin 3: $frac{6}{p}$
Coin 4: $frac{7}{q}$
and since they are independent probability of a head with the 3 coins is
$frac{6}{m}+frac{6}{p}+frac{7}{q}=frac{6pq+6mq+7mp}{mpq}, q,m, p>0 $
and assume : $n>q>m geq p>0$
Now I want to ask if I am to flip the coins again once, which "group" would give me the highest probability of getting a head.
Flipping only Coin 1 once or flipping all of Coins 2,3 and 4 once. Is it possible to reason without knowing $n,m,p,q$ If not what minimal information might we need to further assume.
probability
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
$endgroup$
(Assuming 1/2 probability for heads or tails)
Using coin 1 I flip it say $n$ times and I have 60 heads, so probability of head:
$frac{60}{n}, n>0$
Using three coins I flip the first $m$ times and get $6$ heads, the second I flip $p$ times and get also $6$ heads, the third coin flipped $q$ times gets $7$ heads:
Coin 2: $frac{6}{m}$
Coin 3: $frac{6}{p}$
Coin 4: $frac{7}{q}$
and since they are independent probability of a head with the 3 coins is
$frac{6}{m}+frac{6}{p}+frac{7}{q}=frac{6pq+6mq+7mp}{mpq}, q,m, p>0 $
and assume : $n>q>m geq p>0$
Now I want to ask if I am to flip the coins again once, which "group" would give me the highest probability of getting a head.
Flipping only Coin 1 once or flipping all of Coins 2,3 and 4 once. Is it possible to reason without knowing $n,m,p,q$ If not what minimal information might we need to further assume.
probability
probability
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
asked Dec 7 '18 at 2:27
glockm15glockm15
32019
32019
1
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we'reAssuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.
$endgroup$
– zipirovich
Dec 7 '18 at 3:28
add a comment |
1
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we'reAssuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.
$endgroup$
– zipirovich
Dec 7 '18 at 3:28
1
1
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we're
Assuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.$endgroup$
– zipirovich
Dec 7 '18 at 3:28
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we're
Assuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.$endgroup$
– zipirovich
Dec 7 '18 at 3:28
add a comment |
0
active
oldest
votes
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%2f3029388%2fprobability-of-flipping-1-coin-vs-3-to-land-a-head-given-previous-successes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3029388%2fprobability-of-flipping-1-coin-vs-3-to-land-a-head-given-previous-successes%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
1
$begingroup$
If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$?
$endgroup$
– angryavian
Dec 7 '18 at 2:45
$begingroup$
@angryavian I want a head from either of the 3.
$endgroup$
– glockm15
Dec 7 '18 at 3:18
$begingroup$
Sorry, but this text doesn't make much sense. If in the very beginning you already said that we're
Assuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities.$endgroup$
– zipirovich
Dec 7 '18 at 3:28