Statistics - not my strong point

[StevieB]

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.

I think you are confusing the issue here Tony - tossing a coin multiple times is like having multiple lottery draws, not having multiple tickets in the same draw. You are equating drawing the balls multiple times, I am saying you have 50 of the 14 million possible combinations of a single draw covered by having 50 tickets. Therefore in an infinite number of draws, you would have a winning ticket once for every 280,000 draws of 6 balls.

Your analogy of tossing a coin is like saying what is the chance of any single ticket winning in successive draws - for each draw a single ticket has the probability of 1 in 14 million irrespective of how many draws are made. Roger is asking what is the probability of any ticket winning, not a single ticket winning.

Read my dice analogy above - its like picking two numbers the dice can roll on, not rolling the dice twice and estimating the probability of getting a 6 either time (which is clearly 1 in 6 for each throw).

Steve.[/b]

 

[StevieB]

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.

No - Roger asked what is the chance of any ticket winning, not what is the chance of a single ticket winning. you are confusing a single ticket with any ticket.

Steve.

 

[Karl]

No - Roger asked what is the chance of any ticket winning, not what is the chance of a single ticket winning. you are confusing a single ticket with any ticket.

Steve.

But surely there can only be one winning ticket. :?

If there are one set of "result" numbers, and Roger were to buy 50 different sets of numbers, he could only win the jackpot with one ticket.

In that case, the odds of winning with ticket 1 are 1 in 14m. The odds of winning with ticket 1 or 2 are 1 in 14m plus 1 in 14m. ie 2 in 14m. And so the sums go for 50 tickets. It is quite wrong to equite 50 in 14m to 14m divided by 50 (ie 1 in 280,000).

I stand to be corrected if my sums are wrong.......

Cheers

Karl

 

[RogerS]

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.

No - Roger asked what is the chance of any ticket winning, not what is the chance of a single ticket winning. you are confusing a single ticket with any ticket.

Steve.

Think so...I think....getting lost.

What I wanted to know was were my odds better having a punt on the flats or an equivalent amount of money played on the lottery (either 50 weeks at £1 a time, 1 week with 50 tickets with different numbers or any combination in between). Kind of think it's the flats.

 

[StevieB]

OK, lets take a simple example - I will stick with dice.

I will ask you to pick a number between 1 and 6. I will then roll a dice. The chance of me rolling the number you chose is 1 in 6 - I think we all agree on that. The chance in any iteration of that experiment will always be 1 in 6 because the events are independent.

Now I am going to ask you to pick two numbers between 1 and 6. I will then roll a dice. The chance of me rolling the number you chose is now 2 in 6.

You have had two guesses at where the dice might land. Your chance of being right has gone up. The chance of it landing on the first number is still 1 in 6. The chance of it landing on the second number is 1 in 6. These are addative events for a single throw - ie 1 in 6 plus 1 in 6 = 2 in 6.

The important point here is the phrase 'for a single throw'.

Now use a dice with 14 million sides. If you have one guess at where it will land, you have a 1 in 14 million chance of being right. repeat the experiment and for each roll you will have a 1 in 14 million chance. Now pick two numbers. You have a 2 in 14 million chance of being right in any one throw. Guess more numbers, and for any one throw of the 14 million sided dice you will have a higher chance of guessing correctly which number the dice lands on for a single throw

Buying more lottery tickets for a single draw increases you chance of winning with any one of those tickets in direct correlation with the number of tickets bought.

However, buy the same ticket for 50 consecutive draws, and your chance of winning is now 1 in 14 million, because each draw is independent.

The conclusion is - if you want to specd £50 on lottery tickets, your best chance of winning the jackpot is to buy 50 tickets with different 6 ball combinations for a single draw - do not buy one ticket for 50 weeks and 50 draws.

Does that help?

Steve.

 

[RogerS]

Yes, it does, thanks Steve.

So what you're saying is that I shouldn't be wasting my £2 a week on two lucky dips but should splash out £100 in one go on just the one lottery draw (100 different numbers). Bit galling if you got diddly squat having lashed out £100 in one go Somehow losing £100 spread over 50 weeks doesn't seem quite so bad.

 

[dedee]

Roger,
buy premium bonds instead, the odds are better and you can cash in the tickets anytime.

Andy who has had letters from Ernie 7 times this year.

 

[Lincolnshirebodger]

Yes, it does, thanks Steve.

So what you're saying is that I shouldn't be wasting my £2 a week on two lucky dips but should splash out £100 in one go on just the one lottery draw (100 different numbers). Bit galling if you got diddly squat having lashed out £100 in one go Somehow losing £100 spread over 50 weeks doesn't seem quite so bad.

Well thats true. Theres no accounting for psychology, mistique and superstition.

 

[Lincolnshirebodger]

Roger,
buy premium bonds instead, the odds are better and you can cash in the tickets anytime.

Andy who has had letters from Ernie 7 times this year.

Bar Steward........

Me and my missus have £500 each, shes won small amounts several times, ive never won squat.

 

[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

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.

No, sorry were going t ohave to agree to disagree. You clearly dont understand the principle of Cumulative Probability.

Take the coin. 50% chance of head or tail every time.

Thats not the same as saying 'whats the chance of throwing heads 10 times on the trot. Its two different measures.


The 1st Law of Probability:

The chances of something never happening: = 0
The chances of getting a head on one toss of a coin = 0.5
The chances of a the sun rising tomorrow = 1.0 (dead cert)

Events like coin tossing as you state are independant of one another. That however is not the same as predicting the odds for sequenced events - its an entirely separate situation.

Ive already pointed out (and no one as yet has addressed or rebutted the fact that if you own all 14 million lottery tickets then the chances of you winning is a dead cert, because you have 14 million chances at an odds of 14 million to one )

2nd Law of Probability:
The probability of two independent events both occurring is the product of there separate probabilities

3rd law of Probability:
The probability of anyone one of n mutually exclusive events occurring is the sum of there separate probabilities

These laws are attributed to
PROBABILITY THEORY: THE LOGIC OF SCIENCE
by
E. T. Jaynes
Wayman Crow Professor of Physics
Washington University
St. Louis, MO 63130, U.S.A.

so apply the above laws:

2 head in two throws: (law 2) 0.5 x 0.5 = 0.25

1 head and one tail in two throws = (law 3 ) 0.5 +0.5 = 1.0

4 heads in 4 throws - (law 2 ) 0.5 x 0.5 x 0.5 x 0.5 =0.0625

The chances of one lottery ticket winning = 1/14000000 = 7.14 x 10e-8

The chances of winning the lottery twice in two consecutive draws is law 2

7.14 x 10e-8 x 7.14 x 10e-8 = 5.15 x 10e-15

Conversely, the chances of winning one lottery draw with one of ten tickets

7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 + 7.14 x 10e-8 = 7.14 x 10e-7.

Cumulative Probability works exactly like that. The more different lottery tickets in the same draw you buy, the more chances you have of winning, its common sense, if nothing else.


To be precise and technical:

Join Probability: The probability of the desired outcomes is the sum of the probability of each event resulting in a desired outcome.

Suppose that the set of desired outcomes A has M different events, i.e., A={am|m=1,2,3,...,M}. The join probability indicates that :

Conditional Probability: Suppose that A and B are two sets of outcomes. The probability of B under the condition that A has happened, denoted by P(B|A), can be expressed as:

sorry about the heavy maths, but its the only way to prove the point.

 

[bjm]

Ive already pointed out (and no one as yet has addressed or rebutted the fact that if you own all 14 million lottery tickets then the chances of you winning is a dead cert, because you have 14 million chances at an odds of 14 million to one )

That's because if you cover every possible permutation you're not playing a game of chance!



Brian

 

[Lincolnshirebodger]

Ive already pointed out (and no one as yet has addressed or rebutted the fact that if you own all 14 million lottery tickets then the chances of you winning is a dead cert, because you have 14 million chances at an odds of 14 million to one )

That's because if you cover every possible permutation you're not playing a game of chance!



Brian

EXACTLY!!!!!! By buying many tickets you improve your odds cumulatively until the probability reaches 1.0 !!!

Hooray, someone got it finally

 

[bjm]

Yes the probability of you winning is 1.0 but the individual odds for all those tickets still remains the same - it's just that you have them all so it's no longer a game of chance!

Brian

 

[Lincolnshirebodger]

Yes the probability of you winning is 1.0 but the individual odds for all those tickets still remains the same - it's just that you have them all so it's no longer a game of chance!

Brian

yes, the individual odds are the same, but the cumulative probability has improved your chances of winning to the point its 1.0. Thats exactly the point !!!!! As I keep pointing out, its two different things!!!!

 

[Karl]

The chances of a the sun rising tomorrow = 1.0 (dead cert)

Events like coin tossing as you state are independant of one another. That however is not the same as predicting the odds for sequenced events - its an entirely separate situation.

1 head and one tail in two throws = (law 3 ) 0.5 +0.5 = 1.0

Errr.... I don't think so. It is surely 0.5 x 0.5 = 0.25.

Cheers

Karl

 

[Lincolnshirebodger]

The chances of a the sun rising tomorrow = 1.0 (dead cert)

Events like coin tossing as you state are independant of one another. That however is not the same as predicting the odds for sequenced events - its an entirely separate situation.

1 head and one tail in two throws = (law 3 ) 0.5 +0.5 = 1.0

Errr.... I don't think so. It is surely 0.5 x 0.5 = 0.25.

Cheers

Karl

on average throws, two throws should produce one head and one tail (0.5 +0.5 - in theory). But i agree, you can also say that there 4 outcomes - HH TH HT TT, each one with a probability of 0.25.

 

[Losos]

Thank you for the pm roger,,might post it here so people can have a look..

Were you always such a proctologist's delight, or did you have to train? I'm still amazed by the politeness with which others on this Forum treat such obnoxious posts.

Ray.

My thoughts exactly Ray, a big round of applause for all those who have been moderate in their replies.

 

[TonyW]

Thank you for the pm roger,,might post it here so people can have a look..

Were you always such a proctologist's delight, or did you have to train? I'm still amazed by the politeness with which others on this Forum treat such obnoxious posts.

Ray.

My thoughts exactly Ray, a big round of applause for all those who have been moderate in their replies.

Well, as I seem to be the one with the moderate replies thank you Losos :wink:

 

[StevieB]

Oh dear LB, trying to blind us with mathamatics is not the answer I am afraid.

Ive already pointed out (and no one as yet has addressed or rebutted the fact that if you own all 14 million lottery tickets then the chances of you winning is a dead cert

I have not commented on it because IMHO it is stating the obvious. It doesn't need a rebuttal or an address. Its the equivalent of picking all 6 numbers then rolling a dice in my analogy above.

1 head and one tail in two throws = (law 3 ) 0.5 +0.5 = 1.0

Just a moments thought should tell you this is an entirely fatuous statement. If your argument is true that means if I toss a coin twice it is impossible for it to come up heads both times or tails both times. What you mean is:

Probability of a head on throw 1 is 0.5
Probability of a tail on throw 2 is 0.5
Probability of a head then a tail is 0.5 x 0.5 = 0.25

Probability of a head and a tail in any order from two throws is 0.25 + 0.25 = 0.5

The reason for this last statement is that two of any 4 possible outcomes leave you with a head and a tail.

The cumulative probability you quote is correct, but probably not as clear as a simple explanation unfortunately.

Steve.

 

[Smudger]

I don't think The Bodger is 'blinding you with mathematics' - I think he is pointing out that the maths is pretty complex!

As far as I can see it is correct (though my stats work extended as far as getting Excel to work it out for me...)