Lottery Probability: Calculating Win Chances in 5 Years

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Discussion Overview

The discussion revolves around calculating the probability of winning a lottery over a span of five years, given a specific winning chance per ticket. Participants explore different scenarios, assumptions, and mathematical approaches related to lottery probabilities, including independent draws and the implications of buying multiple tickets.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Exploratory

Main Points Raised

  • Some participants assert that the probability of not winning with three tickets in a year is (19/20)^3, leading to a cumulative probability of (19/20)^15 over five years.
  • Others argue that the calculation is only an approximation and that the independence of draws may not hold if the total number of tickets is small compared to the number of tickets purchased.
  • A participant suggests that if one holds onto tickets without submitting them, it raises questions about the likelihood of those numbers winning, though this does not change the fundamental probabilities of future draws.
  • Some participants clarify that the probability of winning remains 1/20 for each draw, regardless of past outcomes, emphasizing that past losses do not influence future chances.
  • There is a discussion about the difference between the probability of winning at least once over multiple years versus the probability of winning in a specific year.
  • One participant mentions that if all tickets were purchased, the probability of winning would be 1, contrasting with the independent failure calculation which approaches zero but never reaches it.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the (19/20)^N calculation, with some supporting it as a good approximation while others highlight its limitations. The discussion remains unresolved regarding the implications of ticket independence and the effects of purchasing strategies on winning probabilities.

Contextual Notes

Participants note that the calculations depend on assumptions about ticket independence and the total number of tickets available. There are also discussions about the nuances of conditional probabilities that are not fully resolved.

speg
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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 let's 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?
 
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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).
 
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.
 
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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!
 
speg said:
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!

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 let's 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 :biggrin:

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... :smile:
 
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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!

Sorry about the toaster!
 
techwonder said:
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!

Sorry about the toaster!

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.
 
techwonder said:
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.

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.
 
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
techwonder said:
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.

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
speg said:
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 let's 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?

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
Mathematicians never buy lottery tickets?

Gokul43201 said:
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.

Mathemataicians never buy lottery tickets? No, you are NOT CORRECT in insisting that it is always "not worth it" to buy lottery tickets. With progressive jack pots it can occur that the total prize is so large that the expectation exceeds the cost of the ticket. THIS HAS HAPPENED. In Ohio, for example, a group from Australia bought all the possible choices in a pick 6 game. However, you never know just how many other winners might come along that drawing, and then there is the huge question of taxes, but, as I say, "Buying the pot," has been done.
 

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