Statistics - not my strong point

[Lincolnshirebodger]

I don't think that the odds work like that either.

50 lottery tickets simply means that you have 50 chances each of which has a chance of 1 in 14 million. Having more than 1 ticket does not improve your chances since each ticket has the same odds.

regards

Alan

Yes it does, if all the tickets are in the same draw.

1 ticket gives you 1 chance in 14 million.
2 tickets give you 2 chances in 14 million.
14 million tickets give you 14 million chances in 14 million (ie a dead cert) .

PROVIDED all the tickets are different and in the same draw.

Think about it. If you were correct i would have exactly the same chance of winning with one ticket as someone with a complete set, which is clearly wrong, since its a 100% likely (ie probability of 1.0 ) that the person with a full set will have a winning ticket. The probability of winning with one ticket is 7.14x10e-8

If all the tickets were in different draws then its 14 million each time each ticket.

Cumulative probability is most clearly demonstrated in Pascals Triangle http://en.wikipedia.org/wiki/Pascal_triangle and in the Cumulative Probability Function http://en.wikipedia.org/wiki/Cumulative_distribution_function

The best way to learn to understadn cumulative probability is to learn to play 5 card Stud Poker (at which i excel ) :lol:

This might explain it better http://www.probabilitytheory.info/topics/pascal_combinations_permutations.htm

 

[RogerS]

And, although intuitively we might run a mile from such a selection, my understanding is that one day, it's perfectly conceivable that the winning numbers for the lottery will be

1 2 3 4 5 6

Still decided to have a blast at the 45000:1 shot.

 

[Karl]

And, although intuitively we might run a mile from such a selection, my understanding is that one day, it's perfectly conceivable that the winning numbers for the lottery will be

1 2 3 4 5 6

Still decided to have a blast at the 45000:1 shot.

Those numbers are in fact just as likely as any other combination to come in.

Take this saturdays numbers for eg - 20, 21, 23, 24, 27 & 28 (I think).

Cheers

Karl

 

[Lincolnshirebodger]

And, although intuitively we might run a mile from such a selection, my understanding is that one day, it's perfectly conceivable that the winning numbers for the lottery will be

1 2 3 4 5 6

Still decided to have a blast at the 45000:1 shot.

Absolutely. Its no less or more likely to win than any other combination.

HOWEVER

If you read the last link that guy has analysed what numbers win, and certain combination of numbers are picked more frequently by the general population than other combination, so if you pick one of those you end up sharing with a lot of people.

You stand more chance of winning the Pools (2 million to 1) or a scratch card ( 50,000 - 200,000 to 1 )

45,000 to one is pretty good odds relatively speaking.

Go learn Poker FREE on Pokerstars . However, the number of possible hands in Poker is alarming:

Number of opponents - possible no of hands:
1 1,225
2 690,900
3 238,360,500
4 56,372,258,250
5 ≈9.7073 × 1012 (more than 9 trillion)
6 ≈1.2620 × 1015 (more than 1 quadrillion)
7 ≈1.2674 × 1017 (more than 126 quadrillion)
8 ≈9.9804 × 1018 (almost 10 quintillion)
9 ≈6.2211 × 1020 (more than 622 quintillion)

 

[Anonymous]

Provided you have 50 different sets of 6 numbers, your chance of winning the jackpot is indeed 50 in 14 million, or 280,000 to 1.
Steve.

This is incorrect.

Roger, you will have 50 chances of winning the lottery, each of which is still (just under) 14 million to one.
You do not increase your chance of wining to 280,000 to 1

This is a common mistake for people unfamiliar with, or new to stats and probability - it is also a way that criminals rip people off


Chances in the lottery:
6 balls (6/49) x (5/48 ) x (4/47)x(3/46) x (2/45) x (1/44) = 1 in 13983816
5 balls (5/49) x (4/48 ) x (3/47) x (2/46) x (1/45) = 1 in 1906884
4 balls (4/49) x (3/48 ) x (2/47) x (1/46) = 1 in 211876
3 balls (3/49) x (2/48 ) x (1/47) = 1 in 18424

In excel, use the Combin function

combin(x,y)

For 49 balls and 6, 5, 4, 3 chances, enter the following into some cells
=combin(49, 6)
=combin(49, 5)
=combin(49, 4)
=combin(49, 3)

and press enter after entering each into a cell

 

[Lincolnshirebodger]

My links disagree.....................

If you have all 14 million tickets, each ticket has 1 in 14 million chance of winning ,but cumulatively you have 14 million in 14 million of winning, giving you a cumulative probability of 1.0

Dont confuse combinatorials with cumulative probabilities, its not the same thing.

Also, your maths is not correctly applied. You have to factorise the numbers.

The chance of winning the lottery, using 1 combination of 6 numbers out of 49 is:

(49!+48!+47!+46!.......2!+1) - (43!+42!+41!+....2!+1) to 1

 

[Rich]

I have never been very good at numbers, but, in my opinion this is an infinite quandary.
If the prize was guaranteed to be won each week I suppose one could set up a model to work out possibilities, as it is, the more people that enter, then the bigger chance of a duplication of numbers and the top prize having to be split, although most unlikely, 14 million people may select the same numbers and NOT win, or all may win and get 1/14millionth,
chance is a very difficult area to lay on a theoretical proposition, how many are likely to be struck by lightning once, let alone twice.
It's endless.

Rich.

 

[bjm]

The only way to reduce the odds is to buy nine tickets and cover every number - that way you are guaranteed to get the first number so the odds go down to whatever the odds are for five balls (?) That just leaves you down £9!

Having 50 tickets (or more) doesn't increase your chances as none of them are guaranteed to have any of the winning numbers, or a subset of them. Individually they are all at the same odds. Cumulatively, I cannot see how the odds improve because the numbers on each ticket are not transferrable and it's the correct single combination that wins, not the cumulative collection. It's like having nine tickets covering all the numbers - that doesn't necesarily amount to a win unless the draw is 1,2,3,4,5,6 or 7,8,9,10,11,12 etc?. You could also cover the numbers with non-sequential combinations and the odds are the same?


Brian

 

[Lincolnshirebodger]

The only way to reduce the odds is to buy nine tickets and cover every number - that way you are guaranteed to get the first number so the odds go down to whatever the odds are for five balls (?) That just leaves you down £9!

Having 50 tickets (or more) doesn't increase your chances as none of them are guaranteed to have any of the winning numbers, or a subset of them. Individually they are all at the same odds. Cumulatively, I cannot see how the odds improve because the numbers on each ticket are not transferrable and it's the correct single combination that wins, not the cumulative collection. It's like having nine tickets covering all the numbers - that doesn't necesarily amount to a win unless the draw is 1,2,3,4,5,6 or 7,8,9,10,11,12 etc?. You could also cover the numbers with non-sequential combinations and the odds are the same?

Because they are all in the same draw. Your logic works if each ticket is in a separate draw. Ill reiterate the example the doubters keep ignoring:

1 ticket gives you 1 chance in 14 million.
2 tickets give you 2 chances in 14 million.
14 million tickets give you 14 million chances in 14 million (ie a dead cert) - if you have all 14 million tickets, each ticket has 1 in 14 million chance of winning ,but cumulatively you have 14 million in 14 million of winning, giving you a cumulative probability of 1.0

Be realistic - if you hold every possible combination on tickets, then you WILL hold the winning ticket. Therefore the probability of winning rises to 1.0. its not possible to NOT win.

One ticket has a probablity of 7.14x10e-8 ( ie 1/14000000) Each and every ticket you buy in that draw adds to that probability, until you own all the possible tickets in which case the cumulative effect is the sum of all the probabilities, which in this case is 1.0 ( - ie 7.14x10e-8 x 14000000 / 14000000 ) - its 100% certain you will win - after all you hold a ticket for every and any possible winning combination.

The probability of ONE ticket in ONE draw is 14 million. If you hold more tickets, your chances rise. If you hold all tickets your chances rise to being a dead cert you will win.

I already posted links to other site that support my point of view.
Brian[/quote]

 

[filsgreen]

Deleted

 

[Smudger]

Tonyw-- agree with you.

There are very genuine helpfull people who post on the other parts of this forum, hats off to them, who you never see post on the section, they are here for woodwork.

Then you have the few who send most of the time here talking complete dribble, thats the ones i refer to, the ones that need to get out more..

I think that would be 'drivel'. If they did. Which they sometimes do, as do we all...

 

[TonyW]

.....I'm still amazed by the politeness with which others on this Forum treat such obnoxious posts.
Ray.

I understand your point, and yes guilty of politeness.
Another view is that Trolls thrive by receiving emotional responses to their often outrageous posts therefore if you feed them with kindness they may either go away or change
On the other hand perhaps Do Not Feed the Trolls is a more appropriate response :?

Cheers
Tony

 

[StevieB]

Wow, this has taken off big time!

I still stand by my post Tony, although am happy for anyone to convince me I am wrong!

Imagine rolling a dice. You have a 1 in 6 chance of rolling a 6. If I buy a ticket in the lottery I have a 1 in 14 million chance of winning the lottery. Now what is the chance if I buy an extra ticket? Its the same as picking an extra number the dice can roll. (NOT giving the dice an extra roll). Thus the chance of getting a 5 or a 6 is now 2 in 6 or 1 in 3. Buying a ticket in the lottery is the same principal - you are increasing the chance of winning in the SAME draw. Thus the odds when having two tickets is 2 in 14 million because 2 of the 14 million possible combinations of numbers will win you the jackpot.

Now, I am happy to concede that the correct way to express the odds with 50 tickets is 50 in 14 million rather than 1 in 280,000 and if that is what you meant in your post I apologise, but to go back to the dice analogy, if I want to roll a 5 or a 6, in an infinite series or rolls with an unloaded dice a 5 or a 6 will come up 1 in 3 rolls of the dice. Thus expressed as either 2 in 6 or 1 in 3 is semantics I think :?

Or I could be horribly wrong...:roll:

Steve.

 

[woodbloke]

Drummer

Come on, surprise us all and stick something worth reading into the finished projects forum.

Cheers

Mark

Mark - clearly, this would be interesting to see, but what sort of project?...maybe a little trolly of some sort knocked up from rough pallet wood to keep the knuckles clear of the ground...
Btw, Mark, expecting to see your name highlighted in the next issue of F&C :wink: - Rob

 

[TonyW]

I have read the replies about calculating the odds and must say that most have gone way over my head - I was hoping that a foolproof method of winning would be revealed

My simple take on this:
The UK national lottery is a pure game of chance, there is only one way of increasing your chances of winning - enter more times. So buying 50 tickets rather than 1 gives you 50 chances to win - for each ticket the odds remain the same

According to the official statistics, the odds of winning the UK National Lottery jackpot prize are 1 in 13,983,816. The chances of winning any prize at all is around 1 in 54. So maybe I should look at playing a game where the odds of winning jackpot are higher - perhaps the Irish lottery which I believe are 1 in 8 mill. Or the football pools?

Another statistic is that two thirds of ALL winners of the big prizes are not individual ticket buyers, but instead, are members of the many lottery syndicates. Seems to confirm buy more improve your chances.

Quote about the lottery "A tax for people who are bad at math". Well thats for me then - what is the jackpot this week

 

[TrimTheKing]

Btw, Mark, expecting to see your name highlighted in the next issue of F&C :wink: - Rob

I must admit to having a quick flick through this months just to see if MH had got that far through the project. :wink:

Mark

 

[woodbloke]

Btw, Mark, expecting to see your name highlighted in the next issue of F&C :wink: - Rob

I must admit to having a quick flick through this months just to see if MH had got that far through the project. :wink:

Mark

Mark - as far as I can remember, MH is looking at the screw threads for his bench project in the next issue - Rob

 

[Anonymous]

My links disagree.....................

If you have all 14 million tickets, each ticket has 1 in 14 million chance of winning ,but cumulatively you have 14 million in 14 million of winning, giving you a cumulative probability of 1.0

Dont confuse combinatorials with cumulative probabilities, its not the same thing.

Also, your maths is not correctly applied. You have to factorise the numbers.

The chance of winning the lottery, using 1 combination of 6 numbers out of 49 is:

(49!+48!+47!+46!.......2!+1) - (43!+42!+41!+....2!+1) to 1

Disagree all you want. The numbers I posted are correct. I carried out each of the calculations on a calculator and in excel using the inbuilt combin function (as a check) before posting!!!!

My Links disagree

Don't believe all you read on the web.

I didn't use the web as I was taught stats 20+ years ago whilst studying for an engineering degree and have used it many times since .

The fact is that you have 50 chances, each of which is (just less than) 1 in 14 million to one. This is a statistical fact, not my opinion :roll: you do NOT have a 1 in 280000 chance of winning by doing this.

With the first ticket, you are using one of 13983816 chances to win.
With the second ticket, you are using 1 of 13983815 chances to win
With the 3rd ticket, you are using 1 of 13983814 chances to win
etc. etc.

Therefore, the chances of wining with 50 tickets are still just above "not a hope in hell"



If you toss a (balanced and fair) coin 20 times and it comes up heads each time, the probability of it coming up heads next time is still 1 in 2 or 0.5 regardless of the past history. The coin has no more chance of comig up heads agian than it did on the first toss.

 

[Karl]

If you toss a (balanced and fair) coin 20 times and it comes up heads each time, the probability of it coming up heads next time is still 1 in 2 or 0.5 regardless of the past history. The coin has no more chance of comig up heads agian than it did on the first toss.

Which is the same reason why numbers 1 through 6 are just as likely to come up on the lottery as any other combination.

Tony's "calculations" regarding the 50 ticket scenario are correct. Each ticket has the same (miniscule) chance of being a winner. So your chances of winning remain at 50 in 14,000,000. Not 14,000,000 divided by 50.

Cheers

Karl

 

[TonyW]

The penny has finally dropped :idea:

After taking note of how to do the calculations and running them through my spreadsheet I have discovered that I have the same chance of winning the lottery whether I play or not

Cheers
Tony