From what I understand of the question the chance will always be as the events are independent.
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.
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.
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 . Which means finding the smallest integer value of n that solves:
.
This can be done by trial and error using a simple scientific calculator. Alternatively an an exact algebraic solution to 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.