What is the Probability of Losing in a Lottery with 20,000,000 Tickets?

In summary, the conversation discusses the probability of losing in a lottery with 20,000,000 tickets, which is approximately 86%. The calculations for this probability are based on the number of possible numbers/choices and the rules of the lottery. The conversation also mentions the use of ln(1+x) ~ x for small x in approximating probabilities.
  • #1
O Great One
98
0
If you have a lottery (Megamillions) and you sell 20,000,000 tickets, the probability of them all losing is given by:
(135,145,919/135,145,920)^20,000,000 = 0.862448363

A close approximation is given by:

e^-(20,000,000/135,145,920) = 0.8624413

I just learned this from a book. That's pretty cool!
 
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  • #2
I suggest that you go back and read that book again!

Did it say anything about where it got the figures
"135,145,919" and "135,145,920". For that matter, did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?
 
  • #3
Originally posted by HallsofIvy
For that matter, did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?
I bet people would still play... ;)

- Warren
 
  • #4
This is because ln(1+x) ~ x for small x.
=> ln[(k/(k+1))^n] = n ln [k/(k+1)] = n ln [1 - 1/(k+1)]
~ -n/(k+1)
=> (k/(k+1))^n ~ e^[-n/(k+1)].
See?

However I'm puzzled because it doesn't say how many tickets there are in total. Don't we have to know this to figure out the 135,145,919?
 
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  • #5
did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?
Ever hear of PowerBall?
 
  • #6
arcnets
This is because ln(1+x) ~ x for small x
At face value this doesn't seem like a good approximation for some similar applications, like 1/(1-ln(1+x)) where x=1.
 
  • #7
Loren Booda, by 'small x' I meant |x| << 1. So it's not valid for x=1, of course.
 
  • #8
I assume there are 135,145,920 possible numbers/choices, with one chosen as the winner.

The rules say there is a power-number 1-52, and 5 distinct numbers 1-52 which can be in any order. So that's 52 * 52! / (5!*47!) , which is right.
 
  • #9
mea culpa, arcnets
 

What is the Poisson distribution?

The Poisson distribution is a discrete probability distribution that is used to model the number of events that occur in a fixed interval of time or space. It is often used to predict the likelihood of rare events, such as accidents or natural disasters.

What are the key characteristics of the Poisson distribution?

The Poisson distribution is characterized by two parameters: λ, which represents the average number of events that occur in a given interval, and t, which represents the length of the interval. It is also a memoryless distribution, meaning that the probability of an event occurring in a given interval is not affected by previous events.

How is the Poisson distribution related to other distributions?

The Poisson distribution is closely related to the binomial distribution, as it can be thought of as a limiting case of the binomial distribution when the number of trials (n) is large and the probability of success (p) is small. It is also related to the exponential distribution, which models the time between events in a Poisson process.

What are some real-world applications of the Poisson distribution?

The Poisson distribution is commonly used in a variety of fields, including insurance, finance, and biology. It can be used to predict the number of accidents, the number of customers arriving at a store, or the number of mutations in a DNA sequence, among other things.

How is the Poisson distribution calculated?

The probability mass function for the Poisson distribution is given by P(X = k) = (λ^k * e^(-λ)) / k!, where k is the number of events and λ is the average number of events in the interval. This formula can be used to calculate the probability of a specific number of events occurring in a given interval.

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