# Lottery Probability

1. May 24, 2004

### speg

Ok guys, it has been a while since I worked with probability so I might need some things cleaned up.

Basically the scenerio is this: each ticket has a 1/20 chance of winning.
Let's say you buy three tickets a year. ( you get a discount :P )

The chance of NOT winning is (19/20)^3 correct?

And lets say you did this every year, does that mean that in say year five you would have a (19/20)^15 chance of NEVER winning in the five years?

2. May 24, 2004

### master_coda

That sounds about right. You have a 19/20 chance of losing a draw, and assuming that the drawings are independent your probability of losing on all of them is (19/20)^n (where n is the number of drawings).

3. May 24, 2004

### Gokul43201

Staff Emeritus
The first part of this calculation is only approximately correct, but is probably as good an approximation as you'll ever need. Actually for each year the odds of failure are very slightly smaller than (19/20)^3. The second part is fine.

The reason that the above calculation is usually good enough is that the total number of tickets out there is large compared to 3. If not, you do have to calculate conditional probabilities. The outcomes are not strictly independent.

Here's an illustration of a case where your calculation would not work. Suppose you bought ALL the tickets; you have to win - in fact you'll win on multiple tickets. So the probability of failure is 0. But the "independent failure" calculation will give you (19/20)^N. This gets to be really small for large N, but never goes to 0.

anyways, mathematically...it never pays to buy lottery tickets.

Last edited: May 24, 2004
4. May 24, 2004

### speg

Ok so on the fifth year I'll have a 54% shot at winning at least once in those years :) That's not too bad... although when I do eventually win it'll probably just be a crappy prize like a toaster!

5. May 25, 2004

### Olias

You can also throw caution to the wind?

For instance in the posts above everyone has worked out probability functions relating to your quest for a lottery return? Ok lets change the scenario but keep the rules the same?

You write out your numbers on lottery tickets, but do not hand them in! what is the likelyhood of your numbers coming up? would holding on to the tickets for the first 4 yrs alter the outcome for the projected returns over the five years?

How many time have you put down numbers for lottery, and had NONE of your numbers come up, quite a few times, I would say more often you get no numbers at all on a single lottery ticket.

The amazing thing would be if you keep writing single lines (then disgard the tickets), you would actually increase the chance of getting some numbers actualy on a ticket

Of course we have to have to tell ourselves that the first line of numbers will not come up, thereby reduce the total field of numbers a little in our favour..and so on...

Last edited: May 25, 2004
6. May 28, 2004

### techwonder

It is actually only partially true - the (19/20)^N solution.

If you buy a ticket this year then your chance for winning is 1/20 that's true. Then next year you again buy a ticket. Your chances have not increased! you still only have 1/20 chances to win and not 1-(19/20)^2.

The calculation is ONLY true if you buy all the tickets up front. Your chances of winning this particular draw does not increase due to whatever you did in the past!

7. May 28, 2004

### master_coda

You don't actually have to buy all your tickets up front. You just have to recognize that this is the probability of not winning on any of the draws.

Of course, the probability of not winning of any of the draw given that you lost the first X draws will be different. But that's a different question. So there really isn't any conflict.

8. May 28, 2004

### Gokul43201

Staff Emeritus
1/20 is the probability of winning in the 2nd year.
1 - (19/20)^2 is the probability that you win at least once in the 2 years. And no one said your chances got better for the second year.

The objective would be to win at least once in so many years.

You are comparing 2 different quantities (and finding them to be unequal)...but other than that, there really is no conflict here.

9. May 29, 2004

### techwonder

1/20 is the probability of winning in the 2nd year.
1 - (19/20)^2 is the probability that you win at least once in the 2 years. And no one said your chances got better for the second year.

The objective would be to win at least once in so many years.

That is true when you look at your chances "now". Next year, assuming you lost last year, your chances are when buying a ticket are 1/20 once again, because now you have centainty what happened last year.

10. May 29, 2004

### master_coda

Yes, but it isn't really interesting that the answer to the question changes if you change the question. The original question was basically asking "what are my chances of not winning any draws in 5 years" and the answer was (19/20)^15.

If you change the question to "given that I lost the first X draws, what are my chances of winning the next draw" then the answer will be 1/20. But since that answer was included in the original statement of the problem, I don't think it was really necessary for you to point it out.

11. May 29, 2004

### loseyourname

Staff Emeritus
Let's say each ticket has a two-digit number, 0-9. There are 100 possible numbers that can be generated, giving each ticket a 1/100 chance of winning. If you buy three tickets (assuming each has a different number), you would then have a 3/100 chance of winning. By the same token, your chances of not winning are then decreased from 99/100 to 97/100, not 99/100^3.

12. May 30, 2004