Probability Mass Function For Winning the Lottery

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SUMMARY

The discussion centers on the probability mass function for winning a lottery, specifically when participating in two different lotteries with probabilities p1 and p2 for winning at least one million euros. The number of attempts, M, until winning is modeled as a geometric distribution, with the parameter defined as p = p1 + p2 - p1 * p2. This formulation accounts for the probability of winning in either lottery while considering the independence of events. The conclusion confirms that the geometric distribution is appropriate for modeling the scenario of stopping after the first win.

PREREQUISITES
  • Understanding of geometric distribution and its properties
  • Familiarity with probability theory, specifically independent events
  • Knowledge of binomial distribution and its applications
  • Basic algebra for manipulating probability equations
NEXT STEPS
  • Study the properties of geometric distributions in-depth
  • Explore the concept of independent events in probability theory
  • Learn about the applications of binomial distribution in real-world scenarios
  • Investigate advanced probability concepts such as Markov chains and their relation to lottery models
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Students in statistics or probability courses, mathematicians interested in lottery probabilities, and anyone analyzing risk and reward in gambling scenarios.

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Homework Statement



You decide to play monthly in two different lotteries, and you stop playing
as soon as you win a prize in one (or both) lotteries of at least one million
euros. Suppose that every time you participate in these lotteries, the probability
to win one million (or more) euros is p1 for one of the lotteries and p2
for the other. Let M be the number of times you participate in these lotteries
until winning at least one prize. What kind of distribution does M have, and
what is its parameter?

Homework Equations



Binomial Distribution: bin(n, p)
px(k) = p(X = K) = (n choose k) * p^k * (1 -p)^(N-k) for k = 0, 1, ..., n

Gemetric distribution: Geo(p):
Px(k) = P(X = k) = (1-p)^(k-1)p for k = 1, 2, ...



The Attempt at a Solution



I'm pretty sure this is a geometric distribution.

However, I'm not quite sure what the p is. I think it's
p1 * (1 - p2) + p2 * (1 - p1) + p1 * p2 =
p1 + p2 - p1p2

so would the parameter be (p1 + p2 - p1p2)?
 
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I think you are right. Your probability at stopping at each stage is the probability of winning lottery 1 OR lottery 2. As you say, that's p=p1+p2-p1*p2. So to reach the kth stage you have have not stopped k-1 times and stopped once. (1-p)^(k-1)*p.
 

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