# Probability of winning at least one bet when the bets have different probabilities?

Let's say you have a lottery, in which you select 4 numbers between 1 and 36. The person running the lottery draws 4 numbers as well. If you have any numbers that match (1, 2, 3, or 4) you win a prize. Of course, each of these has a different probability of winning––it's much more likely that you'll have 1 number that matches instead of all 4.

We're interested in calculating the probability of winning at least one of these. How do we do this?

We could calculate the probabilities of winning any combination of these and add them up, but my intuition tells me that there's a different relationship between these. Since matching 2+ numbers means that you'd have to match at least 1 number, the probability of winning at least one of these lotteries simply be the probability of matching only one number?

Edit: Similarly, would the probability of winning all of these be the probability of matching all 4 numbers?

• Often the easiest way to proceed with "at least one" questions is to consider the complementary event (zero). i.e. what's the probability of matching none of the numbers? (Now imagine that it's a ball drawing problem -- you have 36 balls, four of which are black ... the ones drawn by the person running the lottery -- and your four numbers choose four balls at random from the 36. What's the chance none of those are black? ... or you can think of a card game problem -- take A-9 -- 36 cards in all -- from a deck of cards, and shuffle. What's the chance the top four cards contains no A's?). – Glen_b Jan 12 '17 at 5:06

Often the easiest way to proceed with "at least one" questions is to consider the complementary event (zero). i.e. what's the probability of matching none of the numbers?

Now you might find it easier to imagine that it's a ball drawing problem -- you have 36 balls, four of which are black (representing the ones drawn by the person running the lottery), with the rest being something other than black (white, say) - and your four numbers choose four balls at random from the 36 (treating the numbers as randomly allocated to the physical balls). What's the chance none of those drawn balls are black?

(or you can think of a card game problem -- take A-9 -- 36 cards in all -- from a deck of cards, and shuffle. What's the chance the top four cards contains no A's?)

Then if it's still not quite obvious, consider the chance that first drawn ball is not black, then the chance that the second is not black given that the first was not black, and so on for all four balls, to you get the chance none of the balls are black. (Then remember that you want the chance of this not happening.)