Probability of a result of a game.
$begingroup$
Problem statement:
Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.
It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.
My approach:
First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.
Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Hence, the final probability:
$$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).
probability combinatorics game-theory
$endgroup$
add a comment |
$begingroup$
Problem statement:
Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.
It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.
My approach:
First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.
Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Hence, the final probability:
$$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).
probability combinatorics game-theory
$endgroup$
add a comment |
$begingroup$
Problem statement:
Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.
It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.
My approach:
First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.
Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Hence, the final probability:
$$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).
probability combinatorics game-theory
$endgroup$
Problem statement:
Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.
It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.
My approach:
First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.
Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Hence, the final probability:
$$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).
probability combinatorics game-theory
probability combinatorics game-theory
edited Dec 6 '18 at 22:28
Elman
asked Dec 6 '18 at 22:14
ElmanElman
584
584
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).
$endgroup$
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
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%2f3029125%2fprobability-of-a-result-of-a-game%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 seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).
$endgroup$
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
add a comment |
$begingroup$
It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).
$endgroup$
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
add a comment |
$begingroup$
It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).
$endgroup$
It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).
edited Dec 7 '18 at 23:26
answered Dec 6 '18 at 22:43
RobRRobR
113
113
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
add a comment |
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
$begingroup$
Good catch. Do you see how to fix the calculation?
$endgroup$
– saulspatz
Dec 6 '18 at 22:57
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%2f3029125%2fprobability-of-a-result-of-a-game%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