If you buy one lotto ticket, what is the theoretical probability of matching at least one of your numbers? In this lotto you pick 6 numbers between 1 and 49. How do I go about finding this?
[Math] Probability of matching at least one number in lotto
probability
Related Solutions
We would intuit that the probability of drawing a tight sequence of numbers is less than than a wide spread.
This is indeed true, but that is simply because there are more possible wide spreads than possible tight sequences.
What you have is a particular tight sequence, and a particular wide spread. In a fair draw, no particular spread of numbers is more likely to result than any other.
Lotto balls do not magically repel closer numbers more than distant numbers.
Select any array of $6$ different, valid lotto numbers by whatever method you choose, as long as it is completely independent of knowing what numbers will be drawn.
The Lotto draw then selects $6$ balls from $45$ (here; may vary by locality). Compare the first of your numbers with the results. Whatever number it is, there is a $6/45$ probability that it is on one of the balls drawn. When given that, there is a $5/44$ conditional probability that the second number is on one of the remaining numbers drawn. And so on, et cetera.
Thus there is a $\frac{6!\;39!}{45!}$ probability that all six of your numbers will be drawn, however you chose them; including by using Biff's method, or by using George's method.
$$P = \frac 1{8145060}$$
So Biff's method and George's method are both equally unlikely to win.
Edit: I will note that very few Lottos use number higher than 60, and thus George may be certain to lose (should 67 be an invalid number).
So this question deals with a binomial distribution, as we want to know how many tickets must be purchased for $Pr(X \ge 1) > 0.9$, where $X$ is the number of wins. So as you already noted $Pr(win) = 1/3$, and since this is binomial distribution then $X \sim Bin(1/3,n)$ where $n$ is what we are looking for, number of tickets purchased.
Observe: $Pr(X \ge 1) = 1 - Pr( X = 0) = 1 - (2/3)^n $ So we want $ 1- (2/3)^n > 0.9 $, which will give you $ n > \frac {ln(0.1)}{ln(2/3)}$, so $ n \ge 6 $.
This is a binomial distribution as we are counting two possible outcomes, and the winning or losing of one ticket does not affect the other tickets. And we may organize the tickets in any way we choose to.
Best Answer
The chance to match at least one number is the complement of matching none of the numbers.
There are $\binom {49-6}{6}$ possibilities to match none out of a total of $\binom {49}{6}$ possibilities.
Therefore the probability to match at least one is: $$1 - \frac{\binom {49-6}{6}}{\binom {49}{6}}$$