How to win the lottery? Don't play it.
Not playing it wins it?
Bit radical that me thinks.
Right, ive had a few drinks, i wont deny! However and stay with me on this now, ive got it cracked !!!
There are 49 numbers, take off the six numbers from the previous weekend, they wont come out again......they just wont ok?
Count up how many odds or even numbers there are.
whichever has the most go with the opposite...better still if its all odd numbers go with evens...or vice versa.
Simples........i think.:thinking about::thinking about::thinking about:
You're describing, in mathematical terms, a rectangular distribution.
In layman's terms, after a long period of time, you would expect all of the balls to have been picked the same amount of time.
So if numbers appear one week, by a distributional term, they are less likely to be picked the following week.
You actually increase your chances of winning by using this method and keeping a track of previously drawn balls.
Statistically, you don't improve your chances by anything 'significant'. But they are increased none the less.
Run the sequence enough times and you will find that it's possible the same set could come out more than twice consecutively. Each draw is completely independent to the previous one and each set of variables has an EXACT EQUAL CHANCE.
The series of 49 balls is only an exageration of a coin spin (a perfectly flat and even coin spin as we know many coins hold a slight bias over the long term) where even the longest patterns and sequences will be replicated.
It's similar to that thing where you have the option of 3 things hidden and in one is a big prize with 2 duffs in the others. Say you pick item 1 but it isn't opened, 1 of the other hidden prizes is removed. Naturally you think only 2 things left so it's a 50/50 chance, but in fact it's not, it's 33.3% of getting the big prize if you stick with your original option, but 66.6% if you change to the other item. Can't remember what it's called, anybody enlighten me?
It's similar to that thing where you have the option of 3 things hidden and in one is a big prize with 2 duffs in the others. Say you pick item 1 but it isn't opened, 1 of the other hidden prizes is removed. Naturally you think only 2 things left so it's a 50/50 chance, but in fact it's not, it's 33.3% of getting the big prize if you stick with your original option, but 66.6% if you change to the other item. Can't remember what it's called, anybody enlighten me?
What you are saying is not fact.
For independent events (which is what each draw is) you multiply probabilities.
For example...
What is the probability of getting a head when flipping a coin? Ans: 0.5
What is the probability of getting a head when you next flip it? Ans: 0.5
However, what is the probability that you get two heads in a row? Ans: 0.5 x 0.5 = 0.25
Because there are four outcomes:
HH
HT
TH
TT
Each has a probability of 0.5 x 0.5 = 0.25
So when you add up the four outcomes (0.25 x 4) you get 1.
If you flip a coin 9 times and get a head each time, what is the probability that of getting a head when you flip the coin for the 10th time?
For the event on its own it is still 0.5
But if the question you are asking is 'what is the probability of getting 10 heads in a row' - then the answer is 0.5^10.
The individual probability for the single event doesn't change. But when you look at a sequence of events then there are many different permutations and combinations.
For mutually exclusive (independent events) this is true.
So you would not expect the same lottery numbers to come out 10 weeks in a row (an exaggerated example).
The probability of those exact numbers occurring for each event is the same.
But the probability of those same numbers being chosen every week for 10 weeks is not the same as them being picked one week and then not happening for the other 9 weeks.
You then look to distributions to work out how many different ways the numbers can be picked.
In the case of a coin being flipped you can use a binomial distribution - which is nice and simple to follow as there are only ever two outcomes.
In the case of the lottery, you would look to a rectangular distribution, as over time you would expect the balls to be picked (within reason) a similar amount of time.
So are you saying that if the lottery draw was made (assuming approx 14m cobinations) that if we ran it 14m x 14m times, that you wouldn't get two consecutive draws of the same number anywhere within that sequence? Surely this goes against every statistical probability.
I'm not that strange!If I had a flawless plan to win the lottery I'd definitely share it with dozens of strangers on a football forum.
^^^ What you've just described is known as "The Gambler's Fallacy"
Try using it to play the even chance odds at a roulette table and see how far you get?
Past events can have no bearing on future outcomes.
Is that the thing where if you lose you just keep doubling up, so a £5 bet becomes £10 then £20 then £40, then £80? I know it's the way a lot play and potentially as long as you have enough money, you should always be coming out with a small profit. There will be the odd occasion where it comes out 10 times in a row but this is very rare.
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