What is the expected return of this hypotetical lottery game?
There are 60 numbers and you must pick 6 different numbers.
Then they pick 6 numbers (that will be different from each other) at random and if your 6 numbers is equal to those 6 numbers (but dont need to be at the same order) you win $3,500,000$ dollars.
Each ticket cost $3.5$ dollars to buy.
This is a simplified version of mega sena from brazil
Best Answer
The probability of winning is 1 / total number of possible outcomes
You're looking for the number of ways to pick 6 numbers at random from 60, I am assuming without considering the order (that is, 1 3 5 7 9 51 is the same as 5 7 9 51 1 3)
This number is equal to $\binom{60}{6}$ so your probability of winning is $1/ \binom{60}{6}$
The expected value of the game therefore is
$$E = \frac 1{\binom{60}{6}} \cdot 3500000 - 3.5\approx -3.43$$
So in average every time you play you lose 3.43 dollars. Don't play.