Probability distribution of random events

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  • Thread starter drmalawi
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  • #1
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Hi

Imagine we have a lottery, with chance of winning 1 in 1000 (1/1000). I have made computer simulations in order to find confidence levels for winning. At 1000 bought lottery tickets, the confidence of winning is 64.1% and 2000 bought lottery tickets the confidence of winning is 87.1%

By changing the chance of winning to 1/X, I have verified that there is always 64.1% confidence at X bought lottery tickets, and 87.1% confidence at 2X bought tickets.

Clearly, there must be a probability density function that describe these kind of events but I have failed at finding such in all of my textbooks at home. I was thinking about the Poisson distribution, but that just tell us the distribution of the number of wins for a certain number of bought lottery tickets.

I am only interested if a person wins or not, not how many wins they get. So, should I integrate the Poisson distribution from 1 to infinity? And if Poisson is the correct distribution, how to I figure out the "mean" from the chance of winning?
 

Answers and Replies

  • #2
Nugatory
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It's easier to calculate the probability that all the purchased tickets lose.
 
  • #3
Stephen Tashi
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At 1000 bought lottery tickets, the confidence of winning is 64.1% and 2000 bought lottery tickets the confidence of winning is 87.1%
If you mean "the probability of winning", you should use that phrase. In statistics, "confidence" has a different meaning than "probability".

You didn't define what you mean by "winning". Presumably you mean that you have at least one winning ticket.

You didn't define whether 1000 tickets each have different numbers. For example, in an actual lottery, if the goal is to buy at least one winning ticket, it would be best to buy 1000 tickets, each with a different set of numbers written on them, as opposed to buying 1000 tickets, each with a randomly selected set of numbers written on it - because if you buy randomly selected tickets, you might buy the same set of numbers more than once.


Clearly, there must be a probability density function that describe these kind of events but I have failed at finding such in all of my textbooks at home.
The probability space you have described has only two outcomes "I win" and "I don't win", so a probability distribution on that space involves only two probabilities.

If you wish to consider a probability space with a larger number of outcomes, you need to define the outcomes in that space. Perhaps your computer simulation is not actually a simulation of a lottery. It might be a simulation of M independent trials, where the probability of winning on each trial is 1/N. If an outcome is defined as winning on K of the M trials for K = 0,1,2..N then we have a binomial distribution. A binomial distribution can be approximated by a normal distribution or by a poisson distribution.

If the goal is only to compute the probability of winning on at least 1 of the M trials then take @Nugatory 's suggestion.
 
  • #4
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Ok I figured this one out. It was just to take 100% - P(not winning) :)
 
  • #5
WWGD
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I dont understand, didn't you determine the probability was 1/1000? What are you estimating then, the number of wins after n trials?
 

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