MHB Generating Function for Gambler's Probs of Broke at Time n

[oyth94]

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
m_x(s) = r_x(e^s).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the r_x(t) = (t x theta + 1 - theta)ⁿ
so m_x(s) = r_x(e^s) = (e^stheta + 1 - theta)ⁿ
am i on the right track? i think i am not.. please help?

 

[chisigma]

Re: generating function

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
m_x(s) = r_x(e^s).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the r_x(t) = (t x theta + 1 - theta)ⁿ
so m_x(s) = r_x(e^s) = (e^stheta + 1 - theta)ⁿ
am i on the right track? i think i am not.. please help?

The first step is the computation of $p_{n}$, i.e. the probability that he/she first becomes broke at the n-th step. It is not too difficult to realize that $p_{n}=0$ for n even and for n odd is... $\displaystyle p_{2 n+ 1} = (1 - p)\ h_{n}\ [p\ (1-p)]^{n}\ (1)$ ... where $h_{n}$ obeys to the recursive relation... $\displaystyle h_{n+1} = h_{n} + n,\ h_{1}=1\ (2)$... so that is...

$\displaystyle p_{n}= (1-p)\ \frac{n^{2} - n + 2}{2}\ [p\ (1-p)]^{n}\ (3)$

The (3) can now be used to valuate the generating function... Kind regards $\chi$ $\sigma$

 

[Opalg]

Re: generating function

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
m_x(s) = r_x(e^s).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the r_x(t) = (t x theta + 1 - theta)ⁿ
so m_x(s) = r_x(e^s) = (e^stheta + 1 - theta)ⁿ
am i on the right track? i think i am not.. please help?

This is a version of the gambler's ruin problem. As chisigma points out, the probability $f_n$ of becoming broke at the $n$th step is $0$ if $n$ is even. In the case where it is odd, $f_{2n+1} = C_np^n(1-p)^{n+1}$, where $C_n$ is the $n$th Catalan number. Using the first of those two links, you can check that the generating function for these probabilities can be expressed in the form $$\frac{1-\sqrt{1-4p(1-p)x^2}}{2px}.$$

 

[oyth94]

Is there any other way to solve it besides using Catalan number and recursive relation and gamblers ruin?? Like using expected value or different distributions of some sort?