Hey, I was just wondering if there is way to tell how many times person needs to do lottery in order to have 50% chance to actually win the big prize. We use 39 numbers and must pick 7 of them. Chances to win are 1/15380397 and I calculated that if you try to do it for 15380397 weeks always using 1 row at time, chances are

1 - (15380396/15380397)^15380937 = 0,63

So how many weeks is reguired to have almost exact 0,5 chance to win? I think I might have asked my math teacher about this problem, but she said there is no exact way to calculate that with a pen and paper.

2. From what I understand of the question the chance will always be ${49 \choose 7}$ as the events are independent.

3. Yeah but I'm talking about chance to win. For example, if you roll dice for 2 times you have 30& chance to get at least 1 "6" even chance is always 1/6. In this case, I'm asking how many lottery tries is reguired to have 50% chance to win at least once.

4. Originally Posted by Crookshanks
Hey, I was just wondering if there is way to tell how many times person needs to do lottery in order to have 50% chance to actually win the big prize. We use 39 numbers and must pick 7 of them. Chances to win are 1/15380397 and I calculated that if you try to do it for 15380397 weeks always using 1 row at time, chances are

1 - (15380396/15380397)^15380937 = 0,63

So how many weeks is reguired to have almost exact 0,5 chance to win? I think I might have asked my math teacher about this problem, but she said there is no exact way to calculate that with a pen and paper.
Let X be the random variable 'number of times you win'.

X ~ Binomial(n = ?, p = 1/15,380,397).

You require the smallest integer value of n such that $\Pr(X \geq 1) \geq 0.5 \Rightarrow \Pr(X = 0) \leq 0.5$. Which means finding the smallest integer value of n that solves:

$\left( 1 - \frac{1}{15,380,397} \right)^n \leq 0.5$.

This can be done by trial and error using a simple scientific calculator. Alternatively an an exact algebraic solution to $\left(1 - \frac{1}{15,380,397} \right)^n = 0.5$ can be easily found and then appropriate rounding done to get the required answer.

It should not surprise anyone to find that the value of n is between 10 million and 11 million.