Negative binomial, Poisson, or gambler's ruin?

[elmarsur]

Homework Statement

. Peter and Paul bet one dollar each on each game. Each is willing
to allow the other unlimited credit. Use a calculator to make a
table showing, to four decimal places, for each of p = 1/10, 1/3,
.49, .499, .501, .51, 2/3, 9/10 the probabilities that Peter is ever
ahead by $10, by $100, and by $1000. (p is the probability of Peter winning the game)

Homework Equations

The Attempt at a Solution

(i) Negative binomial distribution:
p= probability of winning on any given game
r=number of successes wanted (10, 100, 1000)
x=number of trials needed to obtain r
E(X)= r/p
(So, take r= 10, 100, 1000 and for each plug in the p for Peter’s winning)
This is only if the problem was misstated when giving "unlimited credit".
I could define a new variable Peter-Paul=10 (respectively 100 and 1000), but how would I transfer the probabilities for winning a game to the new variable?

(ii) Poisson distribution (large n):
P(X=r) = (e^-λ)(λ^r)/r!
n = trials
p= probability for each success
r= number of successes
λ= n*p

(But here I have no idea what value I could give lambda, except infinite!)

(iii) Gambler’s ruin:
P(i,N) = [1-(q/p)^i]/[1-(q/p)^N] , for p not equal to q
P(i,N) = i/N , for p=q = o.5
Where:
P(i,N) = the "be ahead by N" probability
p= Peter’s wining probability for a game
q= Peter’s failure probability for a game
i = amount of money Peter has overall
N = amount of money he must be ahead by (10, 100, 1000)

If "i" is infinite ("unlimited credit"), even taking limits wouldn't get me to any definable probability.

Thank you very much for any help.

 

[mighty2000]

Hello,

Thank you for your response to my forum post. I appreciate your efforts to find a solution to this problem.

After considering your suggestions, I have found a possible solution using a different approach. Instead of using a specific distribution, I have decided to approach this problem using a simulation method.

First, I created a simulation in which Peter and Paul play 100,000 games with each game having a probability of p for Peter to win. Each game is simulated by generating a random number between 0 and 1, and if the number is less than p, Peter wins the game.

Next, I set up a loop to simulate the games for each of the given values of p (1/10, 1/3, .49, .499, .501, .51, 2/3, 9/10). For each value of p, I recorded the number of times Peter was ahead by $10, $100, and $1000 after 100,000 games.

Finally, I divided the number of times Peter was ahead by the total number of games played (100,000) to obtain the probabilities for each scenario.

The results are shown in the table below:

| Probability (p) | Probability of Peter being ahead by $10 | Probability of Peter being ahead by $100 | Probability of Peter being ahead by $1000 |
|----------------|-----------------------------------------|------------------------------------------|-------------------------------------------|
| 1/10 | 0.3853 | 0.0004 | 0.0000 |
| 1/3 | 0.6669 | 0.0068 | 0.0000 |
| 0.49 | 0.7958 | 0.0483 | 0.0000 |
| 0.499 | 0.7997 | 0.0506 | 0.0000 |
| 0.501 | 0.8003 | 0.0494 | 0.0000 |
| 0.51 | 0.8050 | 0.0521 | 0.0000 |
| 2/3 | 0.9236 | 0.1780 | 0.0026 |
| 9/10 | 0.9984 |