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?

 

[blue_raver22]

Yes, you are on the right track. The probability of becoming broke at time n can be represented by a binomial distribution. However, the generating function for this scenario can be calculated using the formula:

G(s) = (1-p + ps)^n

Where p is the probability of winning, (1-p) is the probability of losing, and n is the number of trials (in this case, the number of times the gambler plays the game).

Using this formula, we can find the generating function for the probabilities of becoming broke at time n:

G(s) = (1-p + ps)^n = (1-p + p)^n = (1)^n = 1

Therefore, the generating function for the probabilities of becoming broke at time n is 1. This means that the probabilities for becoming broke at different times are all equal to 1. This may seem counterintuitive, but it is because the probabilities for winning and losing are equal (p=1-p), so the chances of becoming broke at any given time is the same.

I hope this helps clarify the concept of a generating function for this scenario. It is important to note that generating functions are used to represent the probabilities of different outcomes in a more compact and efficient way. They are a useful tool in probability and statistics, but they may not always be intuitive.