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

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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







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  • $begingroup$
    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




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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






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oldest

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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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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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


















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