Gamblers ruin difference equation says he'll never reach goal if $p=frac 1 2$












2












$begingroup$


I was reading about the gamblers ruin problem here, and they approach it with a difference equation.



The issue is that when I plug $p=frac 1 2$, I get that the probability that the gambler will ever reach his goal becomes zero which is clearly not true.



Here is how I'm getting that:



The gambler starts with $a$ dollars and aims to get to $c$ dollars. He tosses a coin that has a probability $p$ of heads until he reaches either $c$ or $0$. The difference equation is given by:



$$s_c(a) = p s_c(a+1)+ (1-p)s_c(a-1)$$



Then, assuming $s_c(a) = z^n$ we get the characteristic polynomial:



$$pz^2-z+(1-p) = 0$$



We get the two solutions: $z = 1$, $z=frac 1 p -1$.



So the general solution becomes:



$$s_c(a) = c_1 + c_2left(frac 1 p - 1right)^a$$



Then we plug in the boundary conditions, one of which is $s_c(0)=0$.



Now here is where things get weird. If we plug $p =frac 1 2$, the equation becomes:



$$s_c(a) = c_1$$



And since we know $s_c(0)=0$, this means $c_1 = 0$. So this implies that for $p=frac 1 2$, $s_c(a)=0$. But this is obviously not true. What am I missing here?






probability sequences-and-series markov-chains







share|cite|improve this question











$endgroup$












  • $begingroup$
    No matter, it's a constant and the constant is 0.
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:50
















2












$begingroup$


I was reading about the gamblers ruin problem here, and they approach it with a difference equation.



The issue is that when I plug $p=frac 1 2$, I get that the probability that the gambler will ever reach his goal becomes zero which is clearly not true.



Here is how I'm getting that:



The gambler starts with $a$ dollars and aims to get to $c$ dollars. He tosses a coin that has a probability $p$ of heads until he reaches either $c$ or $0$. The difference equation is given by:



$$s_c(a) = p s_c(a+1)+ (1-p)s_c(a-1)$$



Then, assuming $s_c(a) = z^n$ we get the characteristic polynomial:



$$pz^2-z+(1-p) = 0$$



We get the two solutions: $z = 1$, $z=frac 1 p -1$.



So the general solution becomes:



$$s_c(a) = c_1 + c_2left(frac 1 p - 1right)^a$$



Then we plug in the boundary conditions, one of which is $s_c(0)=0$.



Now here is where things get weird. If we plug $p =frac 1 2$, the equation becomes:



$$s_c(a) = c_1$$



And since we know $s_c(0)=0$, this means $c_1 = 0$. So this implies that for $p=frac 1 2$, $s_c(a)=0$. But this is obviously not true. What am I missing here?






probability sequences-and-series markov-chains







share|cite|improve this question











$endgroup$












  • $begingroup$
    No matter, it's a constant and the constant is 0.
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:50














2












2








2


1



$begingroup$


I was reading about the gamblers ruin problem here, and they approach it with a difference equation.



The issue is that when I plug $p=frac 1 2$, I get that the probability that the gambler will ever reach his goal becomes zero which is clearly not true.



Here is how I'm getting that:



The gambler starts with $a$ dollars and aims to get to $c$ dollars. He tosses a coin that has a probability $p$ of heads until he reaches either $c$ or $0$. The difference equation is given by:



$$s_c(a) = p s_c(a+1)+ (1-p)s_c(a-1)$$



Then, assuming $s_c(a) = z^n$ we get the characteristic polynomial:



$$pz^2-z+(1-p) = 0$$



We get the two solutions: $z = 1$, $z=frac 1 p -1$.



So the general solution becomes:



$$s_c(a) = c_1 + c_2left(frac 1 p - 1right)^a$$



Then we plug in the boundary conditions, one of which is $s_c(0)=0$.



Now here is where things get weird. If we plug $p =frac 1 2$, the equation becomes:



$$s_c(a) = c_1$$



And since we know $s_c(0)=0$, this means $c_1 = 0$. So this implies that for $p=frac 1 2$, $s_c(a)=0$. But this is obviously not true. What am I missing here?






probability sequences-and-series markov-chains







share|cite|improve this question











$endgroup$




I was reading about the gamblers ruin problem here, and they approach it with a difference equation.



The issue is that when I plug $p=frac 1 2$, I get that the probability that the gambler will ever reach his goal becomes zero which is clearly not true.



Here is how I'm getting that:



The gambler starts with $a$ dollars and aims to get to $c$ dollars. He tosses a coin that has a probability $p$ of heads until he reaches either $c$ or $0$. The difference equation is given by:



$$s_c(a) = p s_c(a+1)+ (1-p)s_c(a-1)$$



Then, assuming $s_c(a) = z^n$ we get the characteristic polynomial:



$$pz^2-z+(1-p) = 0$$



We get the two solutions: $z = 1$, $z=frac 1 p -1$.



So the general solution becomes:



$$s_c(a) = c_1 + c_2left(frac 1 p - 1right)^a$$



Then we plug in the boundary conditions, one of which is $s_c(0)=0$.



Now here is where things get weird. If we plug $p =frac 1 2$, the equation becomes:



$$s_c(a) = c_1$$



And since we know $s_c(0)=0$, this means $c_1 = 0$. So this implies that for $p=frac 1 2$, $s_c(a)=0$. But this is obviously not true. What am I missing here?






probability sequences-and-series markov-chains




probability sequences-and-series markov-chains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 6:45







Rohit Pandey

















asked Dec 16 '18 at 5:35









Rohit PandeyRohit Pandey

1,4421023




1,4421023












  • $begingroup$
    No matter, it's a constant and the constant is 0.
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:50


















  • $begingroup$
    No matter, it's a constant and the constant is 0.
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:50
















$begingroup$
No matter, it's a constant and the constant is 0.
$endgroup$
– Rohit Pandey
Dec 16 '18 at 5:50




$begingroup$
No matter, it's a constant and the constant is 0.
$endgroup$
– Rohit Pandey
Dec 16 '18 at 5:50










1 Answer
1






active

oldest

votes


















2












$begingroup$

What you are missing is that if $p=frac{1}{2}$, then the two roots of the characteristic polynomial coincide. The general solution becomes $$s_c(n)=c_11^n + c_2n1^n=c_1+nc_2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. What is the reasoning for this though?
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:58






  • 2




    $begingroup$
    This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
    $endgroup$
    – vadim123
    Dec 16 '18 at 6:01











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042273%2fgamblers-ruin-difference-equation-says-hell-never-reach-goal-if-p-frac-1-2%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









2












$begingroup$

What you are missing is that if $p=frac{1}{2}$, then the two roots of the characteristic polynomial coincide. The general solution becomes $$s_c(n)=c_11^n + c_2n1^n=c_1+nc_2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. What is the reasoning for this though?
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:58






  • 2




    $begingroup$
    This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
    $endgroup$
    – vadim123
    Dec 16 '18 at 6:01
















2












$begingroup$

What you are missing is that if $p=frac{1}{2}$, then the two roots of the characteristic polynomial coincide. The general solution becomes $$s_c(n)=c_11^n + c_2n1^n=c_1+nc_2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That makes sense. What is the reasoning for this though?
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:58






  • 2




    $begingroup$
    This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
    $endgroup$
    – vadim123
    Dec 16 '18 at 6:01














2












2








2





$begingroup$

What you are missing is that if $p=frac{1}{2}$, then the two roots of the characteristic polynomial coincide. The general solution becomes $$s_c(n)=c_11^n + c_2n1^n=c_1+nc_2$$






share|cite|improve this answer









$endgroup$



What you are missing is that if $p=frac{1}{2}$, then the two roots of the characteristic polynomial coincide. The general solution becomes $$s_c(n)=c_11^n + c_2n1^n=c_1+nc_2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 5:55









vadim123vadim123

76.3k897191




76.3k897191












  • $begingroup$
    That makes sense. What is the reasoning for this though?
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:58






  • 2




    $begingroup$
    This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
    $endgroup$
    – vadim123
    Dec 16 '18 at 6:01


















  • $begingroup$
    That makes sense. What is the reasoning for this though?
    $endgroup$
    – Rohit Pandey
    Dec 16 '18 at 5:58






  • 2




    $begingroup$
    This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
    $endgroup$
    – vadim123
    Dec 16 '18 at 6:01
















$begingroup$
That makes sense. What is the reasoning for this though?
$endgroup$
– Rohit Pandey
Dec 16 '18 at 5:58




$begingroup$
That makes sense. What is the reasoning for this though?
$endgroup$
– Rohit Pandey
Dec 16 '18 at 5:58




2




2




$begingroup$
This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
$endgroup$
– vadim123
Dec 16 '18 at 6:01




$begingroup$
This is the standard technique for repeated roots. See, e.g., brilliant.org/wiki/linear-recurrence-relations
$endgroup$
– vadim123
Dec 16 '18 at 6:01


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042273%2fgamblers-ruin-difference-equation-says-hell-never-reach-goal-if-p-frac-1-2%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more