Thread: chance of winning the lottery

I was discussing the odds of winning the lottery with some people earlier. The chances of winning are roughly 14 million to 1. I said if i bought two tickets my chances would now be 7 million to 1. If i bought 4 tickets my chances would now be 3.5 million to 1 etc. Everyone seems to think this is wrong. They tried to explain to me why but it didn't make any sense. Apparently it's something to do with the numbers being so large the odds end up changing. I just can't see the logic here. If i'm wrong can someone explain why?

P.S - The tickets have different groups of numbers.

It depends on what lottery you're talking about

If the lottery is such that 14 million tickets are printed each with a different number on it, and at the end the winning ticket is selected at random, then the odds of any ticket being picked are a million to one, and if you have 2 tickets the odds that one of these two tickets will be picked is 14 million to 2, or 7 million to one.

However, since you brought up the "14 million to 1" number, I assume you're talking about the 6:49 lottery, where 6 numbers are selected at "random" from a pool of 49 different numbers, and thus the odds of any combination of being the winning one is 13 983 816, which is the total amount of combinations of 6 numbers you can make with 49 numbers. (49 C 6 = 13 983 816)

Now in this situation, you could say that every DIFFERENT combination of 6 numbers selected from 49 has exactly the same odds of winning. So for your odds of winning to increase with every ticket, you have to put a different number on each one. Indeed, if I have 13 983 816 tickets each with a different combination on it, then I have ALL combinations so I'm guaranteed to win.

The only possible objection I could imagine is if you have a different definition of "winning". So far we have considered winning having the 6 numbers match, but with some lotteries you can win smaller prizes if you get 2, 3, 4 or 5 numbers right. In this case, the odds of "winning" as a broad term has completely different odds.

Thanks for the reply. Yes this question is related to the national lottery. You said at the start of your post the odds of any ticket being picked is a million to one. I assume you meant to type 14 million to one?

Yes i'm just interested in an outright win for this question, so smaller prizes are irrelevant. So just to confirm you agree with me that if i had 2 tickets with different number groups to each other then i'd have a 7 million to 1 chance of winning? If i had 4 tickets i'd have a 3.5 million to 1 chance of winning. If i had 10 tickets my chance of winning would be 1.4 million to 1 and so on.

Originally Posted by phatus

You said at the start of your post the odds of any ticket being picked is a million to one. I assume you meant to type 14 million to one?

My bad, you're right, I meant 14 million to one.

Yes, I agree with what you were saying in the original post. An easy way to show this by using basic probability theory is the following:

If A and B are two different events, and you know the probability that A and B occur, respectively noted P(A) and P(B), then the probability that (A or B) occurs is given by:
P(A or B) = P(A)+P(B)-P(A and B)
Where P(A and B) is the probability that both A and B occur.

Now when we apply this to our situation, we could consider the set
C={C1, C2, ... ,C14 000 000} of all possible combinations for the lottery, and each has an equal probability of 1 over 14 000 000 to be selected.

Then if you have 2 tickets with different combinations on them, say Cn and Cm, well you win if Cn or Cm is selected, so your odds of winning are:
P(Cn or Cm)=P(Cn) + P(Cm) + P(Cn and Cm)

We immediately see that P(Cn and Cm) is zero, because only one combination has been selected as the winning one, so the probability that any two combinations win at the same time is zero. And you can add up a third and a fourth ticked and so on, dividing the odds against you by two every time.