Lottery Odds

Freakbro · 9 Jan 2012 at 13:21

Oh dear, the fail is strong in this thread

Haircut · 9 Jan 2012 at 13:21

I think what a lot of people are failing to understand is the difference between doubling your chance of winning and halving your chance of not winning.

Participant · 9 Jan 2012 at 13:22

People are getting confused between 1 in 7 and 1 in 7 million.

Jono8 · 9 Jan 2012 at 13:22

Freakbro said:
Oh dear, the fail is strong in this thread

indeed

Irish_Tom · 9 Jan 2012 at 13:27

Haircut said:
I think what a lot of people are failing to understand is the difference between doubling your chance of winning and halving your chance of not winning.

So a 13,999,998 in 14,000,000 chance of losing = 1 in 7 million chance of winning?

Haircut · 9 Jan 2012 at 13:27

Panzerbjorn said:
So a 13,999,998 in 14,000,000 chance of losing = 1 in 7 million chance of winning?

Yes

crisbduck · 9 Jan 2012 at 13:31

If I buy a ticket for number 3 in a dice rolling game I have a 1/6 chance of winning
If I buy a ticket for numbers 3 and 5 in a dice rolling game I have a 2/6 chance of winning, or 1/3
Whys the lottery any different?

Cosimo · 9 Jan 2012 at 13:32

Freakbro said:
Oh dear, the fail is strong in this thread

Just like the fail in the hope winning the lottery.

Irish_Tom · 9 Jan 2012 at 13:32

Haircut said:
Yes

Then I do understand it!

Kenai · 9 Jan 2012 at 13:34

krooton said:
If you have 2 tickets, you have a 13,999,998 in 14,000,000 chance of losing.

...or a 6,999,999 in 7,000,000 chance of losing?

How you write the odds is completely irrelevant.

13,999,998 in 14,000,000 is the same thing as 6,999,999 in 7,000,000, the odds are the same, just expressed differently.

If you were to express the odds as decimals or percentages rather than fractionally, they'd come out exactly the same.

Abraham · 9 Jan 2012 at 13:35

for the people disagreeing, just take a smaller example.
the Abramillions lottery has 20 numbers, you choose one.
if you buy one ticket, you have a 1 in 20 chance of winning.
if you buy 4 tickets you have a 4 in 20 chance of winning, this is the same as a 1 in 5 chance of winning.
if you buy 10 tickets you have a 10 in 20 chance of winning, this is the same as 1 in 2.

Do you disagree with it on that scale?

Bloomfield · 9 Jan 2012 at 13:40

amigafan2003 said:
Buying two tickets does not increase your odds to 1 in 7 million. If you bought two tickets (and assuming they are for different numbers) the odds would be 2 in 14 million.

You cannot apply fraction logic to this scenario.

G-MAN2004 said:
How on earth does buying an extra ticket increase your chances to 1 in 7m? It'll be 2 in 14m.

krooton said:
This.

If you reverse the numbers it's easier to see.

If you have 1 ticket, you have a 13,999,999 in 14,000,000 chance of losing.
If you have 2 tickets, you have a 13,999,998 in 14,000,000 chance of losing.

Doubling the number of tickets does not halve the odds.

Each additional unique ticket increases your percentile chance of winning by 1/14,000,000 percent.

.

Shayper · 9 Jan 2012 at 13:42

Panzerbjorn said:
So a 13,999,998 in 14,000,000 chance of losing = 1 in 7 million chance of winning?

Yes. It's binomial statistics. I could explain it, but I'm on my phone and can't be arsed.

MooMoo444 · 9 Jan 2012 at 13:42

Now I see why they say the maths is on the decline in this country.

Vonhelmet · 9 Jan 2012 at 13:48

I can't work out which way it works better, if either, but I imagine any difference is academic.

Think about it in terms of, say the £10 prize for 3 numbers. If you have 12 identical lines, then if they have that £10 prize, you'll get £120. However, you could equally say that 12 different lines are better as that way you are more likely to hit at least one 3 number combo and get £10 more often.

I don't think it will make any practical difference either way, to be honest, but I can't be bothered to do the maths to prove it.

krooton · 9 Jan 2012 at 13:48

Abraham said:
for the people disagreeing, just take a smaller example.
the Abramillions lottery has 20 numbers, you choose one.
if you buy one ticket, you have a 1 in 20 chance of winning.
if you buy 4 tickets you have a 4 in 20 chance of winning, this is the same as a 1 in 5 chance of winning.
if you buy 10 tickets you have a 10 in 20 chance of winning, this is the same as 1 in 2.

Do you disagree with it on that scale?

I've never studied statistics, which might be why I am perceiving this differently.

Having 1/2 odds means that you have 1 of the 2 possible outcomes.
So if 2 unique draws were made, your number would definitely come up.

If you have 10 out of 20 possible draws, it does not equal 1/2, as 2 unique draws could still result in you winning nothing, if 2 of the numbers you don't have are drawn. If would take 11 unique draws to guarantee you a win

So in order to guarantee a win from odds of 4 in 14 million, 13,999,997 unique numbers need to be drawn.

If you change to the lowest common denominator, it becomes 2 in 7million, meaning that a guaranteed win would need 6,999,999 unique numbers to be drawn, which doesn't match the above.

Vonhelmet · 9 Jan 2012 at 13:49

Abraham said:
for the people disagreeing, just take a smaller example.
the Abramillions lottery has 20 numbers, you choose one.
if you buy one ticket, you have a 1 in 20 chance of winning.
if you buy 4 tickets you have a 4 in 20 chance of winning, this is the same as a 1 in 5 chance of winning.
if you buy 10 tickets you have a 10 in 20 chance of winning, this is the same as 1 in 2.

Do you disagree with it on that scale?

This is true, but the question also depends on what happens if, say, you picked 10 of the same number and that one came up. Do you win 10 times? Or just win once, essentially, because of how the prizes are divided?

Shayper · 9 Jan 2012 at 13:52

krooton said:
I've never studied statistics, which might be why I am perceiving this differently.

Having 1/2 odds means that you have 1 of the 2 possible outcomes.
So if 2 unique draws were made, your number would definitely come up.

If you have 10 out of 20 possible draws, it does not equal 1/2, as 2 unique draws could still result in you winning nothing, if 2 of the numbers you don't have are drawn. If would take 11 unique draws to guarantee you a win

So in order to guarantee a win from odds of 4 in 14 million, 13,999,997 unique numbers need to be drawn.

If you change to the lowest common denominator, it becomes 2 in 7million, meaning that a guaranteed win would need 6,999,999 unique numbers to be drawn, which doesn't match the above.

Dude, 10/20 most definitely does equal 1/2. I do statistics at uni.

geuben · 9 Jan 2012 at 13:52

krooton said:
I've never studied statistics, which might be why I am perceiving this differently.

Having 1/2 odds means that you have 1 of the 2 possible outcomes.
So if 2 unique draws were made, your number would definitely come up.

If you have 10 out of 20 possible draws, it does not equal 1/2, as 2 unique draws could still result in you winning nothing, if 2 of the numbers you don't have are drawn. If would take 11 unique draws to guarantee you a win

So in order to guarantee a win from odds of 4 in 14 million, 13,999,997 unique numbers need to be drawn.

If you change to the lowest common denominator, it becomes 2 in 7million, meaning that a guaranteed win would need 6,999,999 unique numbers to be drawn, which doesn't match the above.

No. Having 1/2 odds means you have HALF of the possible outcomes. There might be 2 outcomes, there might be 2000000000 outcomes (in which case you would need 1000000000 tickets).

Burnsy2023 · 9 Jan 2012 at 13:52

Vonhelmet said:
This is true, but the question also depends on what happens if, say, you picked 10 of the same number and that one came up. Do you win 10 times? Or just win once, essentially, because of how the prizes are divided?

I think most people are assuming that each ticket has a unique set.