What is the expected percentage of targets hit when n arrows are shot to n targets?
$begingroup$
n arrows are shot to n targets and each arrow will hit one of the targets, each target having the same probabilty (1/n) to be hit by the arrow, independent of the previous shots.
We divide the n targets into the ones hit (by 1 or more arrows) and the ones not hit. What is the expected percentage of the targets hit?
It is easy to see that it is more than 50% and less than 75%. Is the golden ratio 61.8% the expected percentage? (in the limit for growing n)
Additional question: If one considers only the hit targets, what is the expected percentage of them to be hit by exactly 1 arrow?
probability
$endgroup$
add a comment |
$begingroup$
n arrows are shot to n targets and each arrow will hit one of the targets, each target having the same probabilty (1/n) to be hit by the arrow, independent of the previous shots.
We divide the n targets into the ones hit (by 1 or more arrows) and the ones not hit. What is the expected percentage of the targets hit?
It is easy to see that it is more than 50% and less than 75%. Is the golden ratio 61.8% the expected percentage? (in the limit for growing n)
Additional question: If one considers only the hit targets, what is the expected percentage of them to be hit by exactly 1 arrow?
probability
$endgroup$
add a comment |
$begingroup$
n arrows are shot to n targets and each arrow will hit one of the targets, each target having the same probabilty (1/n) to be hit by the arrow, independent of the previous shots.
We divide the n targets into the ones hit (by 1 or more arrows) and the ones not hit. What is the expected percentage of the targets hit?
It is easy to see that it is more than 50% and less than 75%. Is the golden ratio 61.8% the expected percentage? (in the limit for growing n)
Additional question: If one considers only the hit targets, what is the expected percentage of them to be hit by exactly 1 arrow?
probability
$endgroup$
n arrows are shot to n targets and each arrow will hit one of the targets, each target having the same probabilty (1/n) to be hit by the arrow, independent of the previous shots.
We divide the n targets into the ones hit (by 1 or more arrows) and the ones not hit. What is the expected percentage of the targets hit?
It is easy to see that it is more than 50% and less than 75%. Is the golden ratio 61.8% the expected percentage? (in the limit for growing n)
Additional question: If one considers only the hit targets, what is the expected percentage of them to be hit by exactly 1 arrow?
probability
probability
asked Dec 6 '18 at 19:39
user623654
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$
After $n$ tries, the probability that a target in not hit is $(1-p)^n$.
The probability that a target is hit is therefore
$$P_h = 1 - (1-p)^n = 1 - left(1 - frac{1}{n}right)^n$$
This correspond to the pourcentage of hit targets.
As $n to infty$, this probability converge to $1-e^{-1}$, about 63.2%.
The probabilty that one target is hit by exactly one arrow is equal to
$$np(1-p)^{n-1} = left(1-frac{1}{n}right)^{n-1}$$
If we consider the hit targets only, the ratio is equal to
$$frac{left(1-frac{1}{n}right)^{n-1}}{1 - left(1 - frac{1}{n}right)^n} $$
As $n to infty$, this probability converges to
$$frac{e^{-1}}{1-e^{-1}} = frac{1}{e-1} $$
About 58%
answered Dec 6 '18 at 20:42
DamienDamien
58214
$endgroup$
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
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%2f3028943%2fwhat-is-the-expected-percentage-of-targets-hit-when-n-arrows-are-shot-to-n-targe%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
$begingroup$
Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$
After $n$ tries, the probability that a target in not hit is $(1-p)^n$.
The probability that a target is hit is therefore
$$P_h = 1 - (1-p)^n = 1 - left(1 - frac{1}{n}right)^n$$
This correspond to the pourcentage of hit targets.
As $n to infty$, this probability converge to $1-e^{-1}$, about 63.2%.
The probabilty that one target is hit by exactly one arrow is equal to
$$np(1-p)^{n-1} = left(1-frac{1}{n}right)^{n-1}$$
If we consider the hit targets only, the ratio is equal to
$$frac{left(1-frac{1}{n}right)^{n-1}}{1 - left(1 - frac{1}{n}right)^n} $$
As $n to infty$, this probability converges to
$$frac{e^{-1}}{1-e^{-1}} = frac{1}{e-1} $$
About 58%
answered Dec 6 '18 at 20:42
DamienDamien
58214
$endgroup$
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
add a comment |
$begingroup$
Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$
After $n$ tries, the probability that a target in not hit is $(1-p)^n$.
The probability that a target is hit is therefore
$$P_h = 1 - (1-p)^n = 1 - left(1 - frac{1}{n}right)^n$$
This correspond to the pourcentage of hit targets.
As $n to infty$, this probability converge to $1-e^{-1}$, about 63.2%.
The probabilty that one target is hit by exactly one arrow is equal to
$$np(1-p)^{n-1} = left(1-frac{1}{n}right)^{n-1}$$
If we consider the hit targets only, the ratio is equal to
$$frac{left(1-frac{1}{n}right)^{n-1}}{1 - left(1 - frac{1}{n}right)^n} $$
As $n to infty$, this probability converges to
$$frac{e^{-1}}{1-e^{-1}} = frac{1}{e-1} $$
About 58%
answered Dec 6 '18 at 20:42
DamienDamien
58214
$endgroup$
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
add a comment |
$begingroup$
Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$
After $n$ tries, the probability that a target in not hit is $(1-p)^n$.
The probability that a target is hit is therefore
$$P_h = 1 - (1-p)^n = 1 - left(1 - frac{1}{n}right)^n$$
This correspond to the pourcentage of hit targets.
As $n to infty$, this probability converge to $1-e^{-1}$, about 63.2%.
The probabilty that one target is hit by exactly one arrow is equal to
$$np(1-p)^{n-1} = left(1-frac{1}{n}right)^{n-1}$$
If we consider the hit targets only, the ratio is equal to
$$frac{left(1-frac{1}{n}right)^{n-1}}{1 - left(1 - frac{1}{n}right)^n} $$
As $n to infty$, this probability converges to
$$frac{e^{-1}}{1-e^{-1}} = frac{1}{e-1} $$
About 58%
answered Dec 6 '18 at 20:42
DamienDamien
58214
$endgroup$
Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$
After $n$ tries, the probability that a target in not hit is $(1-p)^n$.
The probability that a target is hit is therefore
$$P_h = 1 - (1-p)^n = 1 - left(1 - frac{1}{n}right)^n$$
This correspond to the pourcentage of hit targets.
As $n to infty$, this probability converge to $1-e^{-1}$, about 63.2%.
The probabilty that one target is hit by exactly one arrow is equal to
$$np(1-p)^{n-1} = left(1-frac{1}{n}right)^{n-1}$$
If we consider the hit targets only, the ratio is equal to
$$frac{left(1-frac{1}{n}right)^{n-1}}{1 - left(1 - frac{1}{n}right)^n} $$
As $n to infty$, this probability converges to
$$frac{e^{-1}}{1-e^{-1}} = frac{1}{e-1} $$
About 58%
answered Dec 6 '18 at 20:42
DamienDamien
58214
edited Dec 6 '18 at 20:49
answered Dec 6 '18 at 20:42
DamienDamien
58214
answered Dec 6 '18 at 20:42
DamienDamien
58214
answered Dec 6 '18 at 20:42
DamienDamien
58214
58214
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
add a comment |
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
$begingroup$
I hope there is no mistake ! No need to cite me, happy to have helped you
$endgroup$
– Damien
Dec 7 '18 at 10:36
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%2f3028943%2fwhat-is-the-expected-percentage-of-targets-hit-when-n-arrows-are-shot-to-n-targe%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
