A certain lottery, called 6-49, draws six different numbers at random from $1$ to $49$, and then draws a seventh, called the bonus number. The player provides a list of six numbers. If the first six numbers drawn match the six numbers on the players list in any order, the player wins the grand prize. If five of the six drawn numbers match numbers on the players list, and the bonus number matches the remaining number, the player wins second prize. The third prize goes to those players whose list contains any five of the six lottery numbers, but not the bonus number.
What is the probability of getting the grand prize? the second prize? and the third prize?
I am confused about the fact whether the order in which the cards are drawn matters or not ... if order matters then I think the sample space has size $7! {49\choose 7}$ and if order doesn't matter then the the sample space has size ${49\choose 7}$. Depending on order, no. of favorable outcomes also varies: If order matters, then I think the no. of favorable choices for winning grand prize is $6!\times 43$ and if order doesn't matter then that would be $43$.
Please help. Thanks in advance
This game is called "lolería primitiva" in Spain and it is widely used (not played) here in Spain for educational purposes, as a source for combinatory and probability problems. It is also widely played for ... well, playing, but this has nothing to do with maths.
To the point: the probability to win the first prize is $$\frac 1{\binom{49}6}$$ (not $7$), since the player doesn't have to guess the order in which the numbers are drawn. Among the ways to choose the numbers that the player has, only one wins the first prize.
The probability for the second prize is $$\frac 6{\binom{49}6}$$ since now extaclty six of those ways win that prize (the bouns number and five out ofr the six winning numbers).
The probability for the third prize is $$\frac{42\cdot6}{\binom{49}6}$$ Now choose five of the winning numbers (six ways) and one of the $42$ remaining numbers (remember that the bonus number can't be counted here).