An UNLUCKY Lottery Win!? - Can anyone work out the odd's..

[fataddick]

Statistically you are more likely to die on your way to buy a national lottery ticket than you are of winning the main prize. An actuary told me that so it must be true.

That's why I buy mine online from home.

 

[Triggaaar]

A winning line (as in the real numbers that are drawn) would leave you in this position if it had (a) ANY three from Ticket A, and (b) ANY three from Ticket B. There are 20 ways of picking 3 numbers out of 6, so 20 possible ways of satisfying (a) and 20 possible ways of satisfying (b).

Ignoring the losing numbers for a second (which we can't do, but we'll come to that) there are as you say, 20 ways of picking 3 out of 6 to satisfy (a) (in my workings, combination 6,3). There aren't however 20 ways of satisfying (b), because you've already used 3 of the winning numbers for (a). There is (since we're still ignoring losing numbers) only 1 way to satisfy (b), and that's to get all 3 numbers left (combination 3,3).

You are dividing those into the total number of combinations (13,983,816) and this is why the losing numbers are important. Out of those 14 million combinations, there's a lot more than 20 that satisfy (a), there's 20 ways of getting 3 correct numbers times 12,341 ways of getting 3 incorrect ones (combination 43,3) = 246,820 out of 14m = 1 in 56.66 of getting 3 numbers right.

To highlight the problem with your workings, your workings would say that to get 2 numbers out of 6 in each ticket it would be (a) ANY two from Ticket A, and (b) ANY two from Ticket B. There are 15 ways of picking 2 numbers out of 6, so 15 possible ways of satisfying (a) and 15 possible ways of satisfying (b).
You'd then say there are 225 (15x15) different lottery draws that would leave you in this position, so 1 in 62,150 - ie, your workings would say it's harder to get 2 numbers in each ticket than to get 3.

 

[fataddick]

As attempts to snare Jeremy Forrest go, this thread must be the best. #sussexpolice #budgetcuts

 

[pastafarian]

Ignoring the losing numbers for a second (which we can't do, but we'll come to that) there are as you say, 20 ways of picking 3 out of 6 to satisfy (a) (in my workings, combination 6,3). There aren't however 20 ways of satisfying (b), because you've already used 3 of the winning numbers for (a). There is (since we're still ignoring losing numbers) only 1 way to satisfy (b), and that's to get all 3 numbers left (combination 3,3).

You are dividing those into the total number of combinations (13,983,816) and this is why the losing numbers are important. Out of those 14 million combinations, there's a lot more than 20 that satisfy (a), there's 20 ways of getting 3 correct numbers times 12,341 ways of getting 3 incorrect ones (combination 43,3) = 246,820 out of 14m = 1 in 56.66 of getting 3 numbers right.

To highlight the problem with your workings, your workings would say that to get 2 numbers out of 6 in each ticket it would be (a) ANY two from Ticket A, and (b) ANY two from Ticket B. There are 15 ways of picking 2 numbers out of 6, so 15 possible ways of satisfying (a) and 15 possible ways of satisfying (b).
You'd then say there are 225 (15x15) different lottery draws that would leave you in this position, so 1 in 62,150 - ie, your workings would say it's harder to get 2 numbers in each ticket than to get 3.

my testicles are much bigger than the average bear and i use their mean size(not average size) to push me toward higher or lower numbers dependent of course on wind and yaw at the time of betting.

with that in mind and your statistical analysis should i put my nuts on ticket A or ticket B

 

[The Camel]

The real question isn't what are the odds of 6 numbers appearing on two consecutive tickets. As you suggest, it's about 14 million to 1 against. The interesting question is:- what are the odds of you knowing the person that it happened to?

About the same chance I know a lottery winner I suppose?

Maybe slightly less, because most people only buy one ticket.

 

[Questions]

If we work on the probability or odds of 14 million to one, that would equate - and of course I know these are just figures - that you should win it, if you do it once a week, once every 269,610 years.

 

[DTES]

Ignoring the losing numbers for a second (which we can't do, but we'll come to that) there are as you say, 20 ways of picking 3 out of 6 to satisfy (a) (in my workings, combination 6,3). There aren't however 20 ways of satisfying (b), because you've already used 3 of the winning numbers for (a). There is (since we're still ignoring losing numbers) only 1 way to satisfy (b), and that's to get all 3 numbers left (combination 3,3).

Edited to make this bit clearer:

I don't mean 3 out of the 6 winning numbers will satisfy (b); I mean any 3 of the 6 numbers written on Ticket B. We know there are no duplicates, so any 3 of those 6 will do.

The 400 combinations that I'm referring to are: Pick any 3 out of the 6 written on Ticket A (20 combinations), and then pick any 3 out of the 6 written on Ticket B (20 combinations). Any of the 20 from A could be matched with any of the 20 from B, giving 400 "winning lines" that do the job,

You are dividing those into the total number of combinations (13,983,816) and this is why the losing numbers are important. Out of those 14 million combinations, there's a lot more than 20 that satisfy (a)

Correct. There are 13,983,816/20 = roughly 700,000. But that's not the point - the point was how many satisfy (a) AND (b).

To highlight the problem with your workings, your workings would say that to get 2 numbers out of 6 in each ticket it would be (a) ANY two from Ticket A, and (b) ANY two from Ticket B. There are 15 ways of picking 2 numbers out of 6, so 15 possible ways of satisfying (a) and 15 possible ways of satisfying (b).
You'd then say there are 225 (15x15) different lottery draws that would leave you in this position, so 1 in 62,150 - ie, your workings would say it's harder to get 2 numbers in each ticket than to get 3.

No. Once you've picked 2 from 6 on each ticket, you're only restricting 4 of the 6 numbers in total, giving you more 'freedom' with the remaining 2 numbers drawn.

There are 225 different lottery draws of the four matching numbers that leave you in that position, plus any two other numbers (that aren't on either ticket = 49 - 4 already drawn - 8 other numbers across the two tickets = 37), so 225 x (37C2) = 666, so 225 x 666 = 149,850 different lottery draws, so 1 in 93.3.

This extra level doesn't come in on the 3 + 3 case that the OP posted, because once you've drawn 6 numbers there aren't any more to be drawn.

 

[Triggaaar]

You are dividing those into the total number of combinations (13,983,816) and this is why the losing numbers are important. Out of those 14 million combinations, there's a lot more than 20 that satisfy (a)

Correct. There are 13,983,816/20 = roughly 700,000. But that's not the point - the point was how many satisfy (a) AND (b).

I know what your point is, but I’m trying to explain that your workings are wrong. You’re saying that there are roughly 700,000 (699191) ways of getting 3 numbers in the first ticket, so the odds of a person buying 1 ticket and getting 3 numbers is 1 in 20, purely because there are 20 ways of having 3 numbers from a choice of 6. That’s obviously wrong. The odds of getting 3 numbers on 1 ticket is 56.66. Anyway, I doubt you’re going to like that, so let’s leave it for a minute and concentrate on your overall method:

Let’s say our lottery has a total of 10 numbers, and 6 are drawn, and you need all 6 to win. The odds of winning with one ticket are 1/combin(10,6) = 1/210 I expect you’d agree with that. So, if you bought 2 tickets for our new 10 number lottery what are the odds of getting 3 in the first, and 3 in the second? I’ll cut and paste your method of working this out:
‘A winning line (as in the real numbers that are drawn) would leave you in this position if it had (a) ANY three from Ticket A, and (b) ANY three from Ticket B. There are 20 ways of picking 3 numbers out of 6, so 20 possible ways of satisfying (a) and 20 possible ways of satisfying (b). There are then 400 (20 x 20) different lottery draws that would leave you in this position if you buy 2 tickets.’
(That is all exactly the same in our new lottery as the old lottery, as your method up to this point makes no reference to how many numbers there are in the lotter draw, either 49 or 10)

‘There are 210 possible lottery draws, of which 400 do the job - if you only buy 2 tickets.’

That proves that your method doesn’t work.

 

[Aadam]

Some people need to go back to school. Buying two tickets just means you have a 2 in 13,983,816 chance of winning, not 1 in 6,991,908. The odds don't half when you buy more tickets, they increase your chances as you have two chances in c.14m to win the jackpot rather than one chance. Statistically speaking, you're still not going to win.

 

[Triggaaar]

Some people need to go back to school. Buying two tickets just means you have a 2 in 13,983,816 chance of winning, not 1 in 6,991,908.

Surely the odds of you not winning with the first ticket are (a) 13,983,815/13,983,816 (= 0.99999992848876) and the odds of you not winning with the second ticket (assuming the ticket is random, and could even be the same as the first) is the same, so the odds of you winning neither time are (a) squared = 0.999999859775 (call that x).
So the odds of not losing both times a 1-x = 0.000000143 or 1 in 6991908

 

[DTES]

I know what your point is, but I’m trying to explain that your workings are wrong. You’re saying that there are roughly 700,000 (699191) ways of getting 3 numbers in the first ticket, so the odds of a person buying 1 ticket and getting 3 numbers is 1 in 20, purely because there are 20 ways of having 3 numbers from a choice of 6. That’s obviously wrong. The odds of getting 3 numbers on 1 ticket is 56.66. Anyway, I doubt you’re going to like that

Well I'm not going to like it because you've completely misunderstood my point. I am saying nothing about the chances of winning with 3 numbers on just one line. All I am doing is listing the number of possible "winning" combinations, where in this case "winning" is defined as having exactly 3 on Ticket A and 3 on Ticket B.

We can (begin to) list them out if you like. If the numbers on Ticket A are a b c d e f and the number on Ticket B are A B C D E F then the following lottery draws will result in a (3+3) "win":

a b c A B C
a b c A B D
a b c A B E
a b c A B F
a b c A C D
etc etc

There are 20 ways to pick 3 from a b c d e f (I am assuming we agree on 6C3 being 20). There are 20 ways to pick 3 from A B C D E F. We know that none of a, b, c, d, e, f, A, B, C, D, E or F are the same as the OP stated there are no duplicates. Therefore the completed list above would contain 400 items.

If the actual lottery draw is one of them, the 3+3 "win" happens. If it isn't, the "win" doesn't happen. All 13,983,816 lines have an exactly equal chance of being drawn, therefore there is a 400/13,983,816 chance of hitting one.

Let’s say our lottery has a total of 10 numbers, and 6 are drawn, and you need all 6 to win. The odds of winning with one ticket are 1/combin(10,6) = 1/210 I expect you’d agree with that. So, if you bought 2 tickets for our new 10 number lottery what are the odds of getting 3 in the first, and 3 in the second? I’ll cut and paste your method of working this out:
‘A winning line (as in the real numbers that are drawn) would leave you in this position if it had (a) ANY three from Ticket A, and (b) ANY three from Ticket B. There are 20 ways of picking 3 numbers out of 6, so 20 possible ways of satisfying (a) and 20 possible ways of satisfying (b). There are then 400 (20 x 20) different lottery draws that would leave you in this position if you buy 2 tickets.’
(That is all exactly the same in our new lottery as the old lottery, as your method up to this point makes no reference to how many numbers there are in the lotter draw, either 49 or 10)

‘There are 210 possible lottery draws, of which 400 do the job - if you only buy 2 tickets.’

That proves that your method doesn’t work.

No, it doesn't. My working - and the OPs post - assumed that there were no duplicate numbers across the two tickets. You can't apply workings that involve 12 different numbers to a lottery involving 10 numbers as there's no way to pick them; it's nonsensical.

If you want to follow the workings through, you wouldn't be able to list the 400 combinations as there would be duplicates in the list a b c d e f A B C D E F and the actual draw would not contain duplicates, ruling out a high number of the 400 combinations.

 

[DTES]

Some people need to go back to school. Buying two tickets just means you have a 2 in 13,983,816 chance of winning, not 1 in 6,991,908. The odds don't half when you buy more tickets, they increase your chances as you have two chances in c.14m to win the jackpot rather than one chance. Statistically speaking, you're still not going to win.

If both tickets are for the same draw, and both tickets are different (i.e. you're not just buying lucky dips where technically you could end up with the same line twice) then yes the odds do half. If you bought all 13,983,816 tickets in the same week, you'd be guaranteed to win.

Your logic works for buying each ticket for a different draw.

 

[Triggaaar]

Well I'm not going to like it because you've completely misunderstood my point. I am saying nothing about the chances of winning with 3 numbers on just one line.

Although that wasn't the point of your first post, what exactly did you mean by this ‘Correct. There are 13,983,816/20 = roughly 700,000.’

All I am doing is listing the number of possible "winning" combinations, where in this case "winning" is defined as having exactly 3 on Ticket A and 3 on Ticket B.

Yes, I fully understand that, but that’s not how to work out the odds of getting those two tickets.

We can (begin to) list them out if you like…
There are 20 ways to pick 3 from a b c d e f (I am assuming we agree on 6C3 being 20).

There’s no need to list them, we fully agree that combination 6,3 is 20. But out of the 14 million alternative draws, multiplying 20 by 20 is irrelevant. It’s so far wrong I’m not sure where to start. I have already shown you why it’s wrong, and how to do it properly.

No, it doesn't. My working - and the OPs post - assumed that there were no duplicate numbers across the two tickets. You can't apply workings that involve 12 different numbers to a lottery involving 10 numbers as there's no way to pick them; it's nonsensical.

No no no. Use your method to work out the odds, exactly as the OP has asked, but assume there are not 49 numbers in the lottery, but only 10. You’ll hit problems, fast.

 

[Triggaaar]

If both tickets are for the same draw, and both tickets are different (i.e. you're not just buying lucky dips where technically you could end up with the same line twice) then yes the odds do half. If you bought all 13,983,816 tickets in the same week, you'd be guaranteed to win.

Your logic works for buying each ticket for a different draw.

Actually the odds of winning one draw with 2 differing tickets are almost identical to the odds of winning one of 2 draws with 1 ticket for each.

 

[Aadam]

If both tickets are for the same draw, and both tickets are different (i.e. you're not just buying lucky dips where technically you could end up with the same line twice) then yes the odds do half. If you bought all 13,983,816 tickets in the same week, you'd be guaranteed to win.

Your logic works for buying each ticket for a different draw.

What? It doesn't half the odds. One ticket is 1 in 13,983,816 to win. Two tickets is 2 in 13,983,816 to win. Thirteen-million, nine-hundred and eighty-three thousand, eight-hundren and sixteen tickets is 13,983,816 to 13,983,816 or odds on 1/1 to win. Guaranteed. The odds of winning DO NOT half if you buy two tickets, you just have more chances to win. It is ALWAYS the same odds to win the jackpot with each individual ticket. Please don't argue against this because you're looking silly.

 

[DTES]

There's no need to list them, we fully agree that combination 6,3 is 20. But out of the 14 million alternative draws, multiplying 20 by 20 is irrelevant. It’s so far wrong I'm not sure where to start. I have already shown you why it's wrong, and how to do it properly.

Sorry, adding a bit more just to make clear what I'm saying and not saying. 20 x 20 does nothing in terms of odds directly, the only purpose for multiplying 20x20 is to work out how many items are on the list. That is it. Nothing more.

Step 1: Do you agree that if I pick 3 from 6 on Ticket A, and 3 from 6 on Ticket B I am left with a list of 400 possible ways to right these choices down (all of the 20 from A can be matched with all of the 20 from B), regardless of what I'm going to do with that list?

Step 2: If you agree that the list has 400 lines in it (even if you don't agree with why I've created it), go on then, explain:
a) How a line within that list could possibly NOT be a "winner", and
b) How a line that is not on the list could possibly BE a "winner"?

It's impossible. If it is on the list of 400, it "wins". If it is not on the list of 400, it "loses". It is that simple.

No no no. Use your method to work out the odds, exactly as the OP has asked, but assume there are not 49 numbers in the lottery, but only 10. You’ll hit problems, fast.

I will hit problems because you CANNOT pick 12 different numbers from a list of 10 numbers! How is that difficult to understand?!

 

[DTES]

Two tickets is 2 in 13,983,816 to win. Thirteen-million, nine-hundred and eighty-three thousand, eight-hundren and sixteen tickets is 13,983,816 to 13,983,816 or odds on 1/1 to win. Guaranteed.

This is exactly what I said.

The odds of winning DO NOT half if you buy two tickets, you just have more chances to win.

The odds of each ticket winning is 1/13,983,816. The odds of one of the two tickets winning is 2/13,983,816. But then that's what you're saying. So... what?

It is ALWAYS the same odds to win the jackpot with each individual ticket. Please don't argue against this because you're looking silly.

I'm not arguing with that. What are you talking about?

 

[DTES]

Actually the odds of winning one draw with 2 differing tickets are almost identical to the odds of winning one of 2 draws with 1 ticket for each.

I completely agree; they are very, very close to identical. But not quite - to see the difference:
a) Buying all 13,983,816 tickets on any one week would guarantee you a win
b) Buying one ticket for 13,983,816 draws would not guarantee you a win.

 

[The Optimist]

Some people need to go back to school. Buying two tickets just means you have a 2 in 13,983,816 chance of winning, not 1 in 6,991,908. The odds don't half when you buy more tickets, they increase your chances as you have two chances in c.14m to win the jackpot rather than one chance. Statistically speaking, you're still not going to win.

2 in 13,983,816 and 1 in 6,991,908 are the same odds.

 

[The Optimist]

I completely agree; they are very, very close to identical. But not quite - to see the difference:
a) Buying all 13,983,816 tickets on any one week would guarantee you a win
b) Buying one ticket for 13,983,816 draws would not guarantee you a win.

And buying all 13,983,816 tickets in any one week means it's impossible to win 2 or more times, while buying one ticket for 13,983,816 draws would give you a very good chance of winning more than once.