How long before someone wins the lottery twice?

[Mr.A.Gibson]

Homework Statement

The chance of winning the Uk lottery is 49C6 = 1 in 13,983,816
So the chance of winning the lottery twice with two tickets is 1 in 195,547,109,856

But what is the chance of anyone person winning the lottery twice. Approx 30million tickets are sold each week, average 3 tickets per person. We'll assume that people carry on playing the same way after they've won.

Also, how long will the lottery have to run for before it is more likely that not that someone will have won it twice.

Homework Equations

The Attempt at a Solution

Similar method to the number of people in a room for the same birthday I suspect.

30million tickets per week, 30M/13,98M gives approx 2 winners per week

chance of uniqueness
$$\frac{9,999,998}{10,000,000}$$

So the chance of uniqueness after n weeks
$$2\left({\frac{4,999,999!}{(4,999,999-n)!50,000,000^n}}\right)$$
So to work out how long the lottery need to run before its more likely than not that someone will have one twice is:
$$1-2\left({\frac{4,999,999!}{(4,999,999-n)!5,000,000^n}}\right) >0.5$$

Is this method correct and then how could I solve this equation?

 

[nate808]

I would approach this problem by using probability theory and statistics. Firstly, we can calculate the probability of a single person winning the lottery twice by using the formula 1/13,983,816 * 1/13,983,816 = 1/195,547,109,856.

Next, we can calculate the probability of a person winning the lottery twice out of a group of people. Assuming that there are 30 million tickets sold each week and each person buys an average of 3 tickets, we can estimate that there are 10 million people playing the lottery each week (30 million tickets / 3 tickets per person = 10 million people). The probability of a single person winning the lottery twice out of a group of 10 million people is (1/195,547,109,856)^2 * 10 million = 5.12 x 10^-16.

To calculate the probability of at least one person winning the lottery twice, we can use the binomial distribution formula: P(X≥1) = 1 - (1-p)^n, where p is the probability of a single person winning twice (5.12 x 10^-16) and n is the number of people playing (10 million). This gives us a probability of approximately 5.12 x 10^-6, or 0.000512%.

To answer the question of how long the lottery needs to run before it is more likely than not that someone will have won it twice, we can use the same binomial distribution formula. We want to find the number of weeks (n) where the probability of at least one person winning twice is greater than 0.5: 1 - (1-5.12 x 10^-16)^n > 0.5. Solving for n, we get approximately 2.15 x 10^15 weeks, or 4.14 x 10^13 years.

However, it's important to note that this is just an estimation and does not take into account other factors such as the possibility of multiple winners in a single week or changes in the number of people playing the lottery. It's also worth mentioning that the probability of a person winning the lottery twice may decrease if they continue to play after winning, as they may not continue to buy the same number of tickets each week.