Mixing
Finding out the probability of the first person to win
Two people are playing a dice, they take turns to roll a dice and if one person rolls a "1", the game ends and the person who rolls a "1" wins. If he does not roll a "1", then the game continues until one person roll a "1", what is the probability of the first person to win?
I have no idea how to calculate this.
probability
asked Nov 21 '18 at 10:35
Learning Mathematics
726
add a comment |
Two people are playing a dice, they take turns to roll a dice and if one person rolls a "1", the game ends and the person who rolls a "1" wins. If he does not roll a "1", then the game continues until one person roll a "1", what is the probability of the first person to win?
I have no idea how to calculate this.
probability
asked Nov 21 '18 at 10:35
Learning Mathematics
726
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48
add a comment |
Two people are playing a dice, they take turns to roll a dice and if one person rolls a "1", the game ends and the person who rolls a "1" wins. If he does not roll a "1", then the game continues until one person roll a "1", what is the probability of the first person to win?
I have no idea how to calculate this.
probability
asked Nov 21 '18 at 10:35
Learning Mathematics
726
Two people are playing a dice, they take turns to roll a dice and if one person rolls a "1", the game ends and the person who rolls a "1" wins. If he does not roll a "1", then the game continues until one person roll a "1", what is the probability of the first person to win?
I have no idea how to calculate this.
probability
probability
asked Nov 21 '18 at 10:35
Learning Mathematics
726
asked Nov 21 '18 at 10:35
Learning Mathematics
726
asked Nov 21 '18 at 10:35
Learning Mathematics
726
asked Nov 21 '18 at 10:35
Learning Mathematics
726
asked Nov 21 '18 at 10:35
Learning Mathematics
726
726
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48
add a comment |
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48
add a comment |
4 Answers
4
active
oldest
votes
Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in ${1,2,3,4,5,6}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ _ ,_1$ or $_,_,_,_1$ and so on where $_$ are numbers in the set ${2,3,4,5,6}$. So now, that happens with probability $frac{1}{6}+(frac{5}{6})^2frac{1}{6}+...= frac{1}{6}sum_{k=0}^infty{(frac{5}{6}})^{2k}=frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:54
mathnoob
1,797422
add a comment |
It is not really necessary to apply geometric distribution together with infinite sums.
If $p$ denotes the probability of the first player to win then we have the equality:$$p=frac16+frac56(1-p)=1-frac56p$$leading to $$p=frac6{11}$$
Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.
Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(Wmid D=1)+P(Dneq1)P(Wmid Dneq1)=frac16cdot1+frac56cdot(1-p)$$
answered Nov 21 '18 at 11:25
drhab
98.2k544129
add a comment |
Using $p = 1/6$, we have
$P($the first player wins$)$ = $sum_{i=0}^{infty} P($the first 1 is in throw $ 2i+1) = sum_{i=0}^{infty} (1-p)^{2i}p = frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:57
Stockfish
52726
add a comment |
Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of ${2,3,4,5,6}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $frac{25}{36}$, and the first term as $frac{1}{6}$. From here, you can just compute the infinite sum to be $frac{6}{11}$.
answered Nov 21 '18 at 10:57
Boshu
705315
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%2f3007545%2ffinding-out-the-probability-of-the-first-person-to-win%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in ${1,2,3,4,5,6}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ _ ,_1$ or $_,_,_,_1$ and so on where $_$ are numbers in the set ${2,3,4,5,6}$. So now, that happens with probability $frac{1}{6}+(frac{5}{6})^2frac{1}{6}+...= frac{1}{6}sum_{k=0}^infty{(frac{5}{6}})^{2k}=frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:54
mathnoob
1,797422
add a comment |
Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in ${1,2,3,4,5,6}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ _ ,_1$ or $_,_,_,_1$ and so on where $_$ are numbers in the set ${2,3,4,5,6}$. So now, that happens with probability $frac{1}{6}+(frac{5}{6})^2frac{1}{6}+...= frac{1}{6}sum_{k=0}^infty{(frac{5}{6}})^{2k}=frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:54
mathnoob
1,797422
add a comment |
Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in ${1,2,3,4,5,6}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ _ ,_1$ or $_,_,_,_1$ and so on where $_$ are numbers in the set ${2,3,4,5,6}$. So now, that happens with probability $frac{1}{6}+(frac{5}{6})^2frac{1}{6}+...= frac{1}{6}sum_{k=0}^infty{(frac{5}{6}})^{2k}=frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:54
mathnoob
1,797422
Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in ${1,2,3,4,5,6}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ _ ,_1$ or $_,_,_,_1$ and so on where $_$ are numbers in the set ${2,3,4,5,6}$. So now, that happens with probability $frac{1}{6}+(frac{5}{6})^2frac{1}{6}+...= frac{1}{6}sum_{k=0}^infty{(frac{5}{6}})^{2k}=frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:54
mathnoob
1,797422
edited Nov 21 '18 at 11:00
answered Nov 21 '18 at 10:54
mathnoob
1,797422
answered Nov 21 '18 at 10:54
mathnoob
1,797422
answered Nov 21 '18 at 10:54
mathnoob
1,797422
1,797422
add a comment |
add a comment |
It is not really necessary to apply geometric distribution together with infinite sums.
If $p$ denotes the probability of the first player to win then we have the equality:$$p=frac16+frac56(1-p)=1-frac56p$$leading to $$p=frac6{11}$$
Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.
Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(Wmid D=1)+P(Dneq1)P(Wmid Dneq1)=frac16cdot1+frac56cdot(1-p)$$
answered Nov 21 '18 at 11:25
drhab
98.2k544129
add a comment |
It is not really necessary to apply geometric distribution together with infinite sums.
If $p$ denotes the probability of the first player to win then we have the equality:$$p=frac16+frac56(1-p)=1-frac56p$$leading to $$p=frac6{11}$$
Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.
Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(Wmid D=1)+P(Dneq1)P(Wmid Dneq1)=frac16cdot1+frac56cdot(1-p)$$
answered Nov 21 '18 at 11:25
drhab
98.2k544129
add a comment |
It is not really necessary to apply geometric distribution together with infinite sums.
If $p$ denotes the probability of the first player to win then we have the equality:$$p=frac16+frac56(1-p)=1-frac56p$$leading to $$p=frac6{11}$$
Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.
Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(Wmid D=1)+P(Dneq1)P(Wmid Dneq1)=frac16cdot1+frac56cdot(1-p)$$
answered Nov 21 '18 at 11:25
drhab
98.2k544129
It is not really necessary to apply geometric distribution together with infinite sums.
If $p$ denotes the probability of the first player to win then we have the equality:$$p=frac16+frac56(1-p)=1-frac56p$$leading to $$p=frac6{11}$$
Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.
Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(Wmid D=1)+P(Dneq1)P(Wmid Dneq1)=frac16cdot1+frac56cdot(1-p)$$
answered Nov 21 '18 at 11:25
drhab
98.2k544129
edited Nov 21 '18 at 11:43
answered Nov 21 '18 at 11:25
drhab
98.2k544129
answered Nov 21 '18 at 11:25
drhab
98.2k544129
answered Nov 21 '18 at 11:25
drhab
98.2k544129
98.2k544129
add a comment |
add a comment |
Using $p = 1/6$, we have
$P($the first player wins$)$ = $sum_{i=0}^{infty} P($the first 1 is in throw $ 2i+1) = sum_{i=0}^{infty} (1-p)^{2i}p = frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:57
Stockfish
52726
add a comment |
Using $p = 1/6$, we have
$P($the first player wins$)$ = $sum_{i=0}^{infty} P($the first 1 is in throw $ 2i+1) = sum_{i=0}^{infty} (1-p)^{2i}p = frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:57
Stockfish
52726
add a comment |
Using $p = 1/6$, we have
$P($the first player wins$)$ = $sum_{i=0}^{infty} P($the first 1 is in throw $ 2i+1) = sum_{i=0}^{infty} (1-p)^{2i}p = frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:57
Stockfish
52726
Using $p = 1/6$, we have
$P($the first player wins$)$ = $sum_{i=0}^{infty} P($the first 1 is in throw $ 2i+1) = sum_{i=0}^{infty} (1-p)^{2i}p = frac{1}{6}frac{36}{11}=frac{6}{11}$.
answered Nov 21 '18 at 10:57
Stockfish
52726
answered Nov 21 '18 at 10:57
Stockfish
52726
answered Nov 21 '18 at 10:57
Stockfish
52726
answered Nov 21 '18 at 10:57
Stockfish
52726
52726
add a comment |
add a comment |
Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of ${2,3,4,5,6}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $frac{25}{36}$, and the first term as $frac{1}{6}$. From here, you can just compute the infinite sum to be $frac{6}{11}$.
answered Nov 21 '18 at 10:57
Boshu
705315
add a comment |
Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of ${2,3,4,5,6}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $frac{25}{36}$, and the first term as $frac{1}{6}$. From here, you can just compute the infinite sum to be $frac{6}{11}$.
answered Nov 21 '18 at 10:57
Boshu
705315
add a comment |
Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of ${2,3,4,5,6}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $frac{25}{36}$, and the first term as $frac{1}{6}$. From here, you can just compute the infinite sum to be $frac{6}{11}$.
answered Nov 21 '18 at 10:57
Boshu
705315
Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of ${2,3,4,5,6}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $frac{25}{36}$, and the first term as $frac{1}{6}$. From here, you can just compute the infinite sum to be $frac{6}{11}$.
answered Nov 21 '18 at 10:57
Boshu
705315
answered Nov 21 '18 at 10:57
Boshu
705315
answered Nov 21 '18 at 10:57
Boshu
705315
answered Nov 21 '18 at 10:57
Boshu
705315
705315
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%2f3007545%2ffinding-out-the-probability-of-the-first-person-to-win%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
Consider the sequence of $X_1, dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins.
– Stockfish
Nov 21 '18 at 10:48