Lottery Probability

[Iron4ever]

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

[Boerseun]

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

The chances of picking 1-2-3-4-5-6 is exactly the same as picking any other set of numbers. And the chances suck. Consider:

 

Your chances of getting the right combination, is 49x48x47x46x45x44. And that equals one out of 10,068,347,520. One out of more than ten billion.

 

Those odds kinda suck.

 

No wonder they say the lottery is the tax people pay who can't do math!:confused:

[CraigD]

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

Yes.

 

You occasionally find articles, books, and software that claims that some special combinations of numbers in this kind of random drawing are mathematically more likely to occur. These claims, no matter how impressive the analysis behind them, are bogus. “1 2 3 4 5 6” is as good a pick-6 guess as any other (and a darned sight easier to remember!)

[Iron4ever]

Your chances of getting the right combination, is 49x48x47x46x45x44. And that equals one out of 10,068,347,520. One out of more than ten billion.

 

I thought the odds were 49 x 48 x 47 x 46 x 45 x 44 / 6 x 5 x 4 x 3 x 2 = 13,983,81 ??????

[Boerseun]

I thought the odds were 49 x 48 x 47 x 46 x 45 x 44 / 6 x 5 x 4 x 3 x 2 = 13,983,81 ??????

Why would that be?

[Iron4ever]

Why would that be?

 

There are 49 possible choices for the first ball

For each of these 49 choices of the first ball,

there are 48 possible choices for the second ball

(because one ball has already been taken out)

meaning there are 49 x 48 ways of choosing the first two balls.

However, because the order doesn't matter, this has to be divided by 2

(because the second ball can come before or after the first ball)

 

So the number of different possible choices for two balls can be written as 49 x 48 /2

 

There are then 47 possible choices for the third ball,

but since it doesn't matter which position the third ball goes in we divide by 3, giving 49 x 48 x 47 / 3 x 2

 

This sequence continues for all six balls which are chosen hence my calculation above.

[Boerseun]

Interesting point, but I can't see the order actually having anything to do with the calculation?

 

Fact is - there's 49 balls to begin with. Only one of them will fall on the first selection. Now this can be any one of your six numbers. Let's say your numbers are 1-2-3-4-5-6, and the first ball to come out is six. The fact that it was indeed one of your numbers, doesn't change the fact that there's now 48 balls left. So now you're down to 49x48. And so on. I can honestly not see why the order of the selection should have anything to do with your odds? I mean - 6-5-4-3-2-1 is valid for your selection, but the odds are exactly the same as for 4-3-5-2-6-1, which is also valid.

 

Maybe I'm just slow on the uptake, but I don't get it.

[C1ay]

There are 49 possible choices for the first ball

For each of these 49 choices of the first ball,

there are 48 possible choices for the second ball

(because one ball has already been taken out)

meaning there are 49 x 48 ways of choosing the first two balls.

However, because the order doesn't matter, this has to be divided by 2

(because the second ball can come before or after the first ball)

 

So the number of different possible choices for two balls can be written as 49 x 48 /2

 

There are then 47 possible choices for the third ball,

but since it doesn't matter which position the third ball goes in we divide by 3, giving 49 x 48 x 47 / 3 x 2

 

This sequence continues for all six balls which are chosen hence my calculation above.

 

Correct, so the odds that any six numbers in any order will come up are 13,983,816:1 but, you asked, "What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out?" which implies that you are also specifying the order of the draw which would be 10,068,347,520:1.

[C1ay]

Interesting point, but I can't see the order actually having anything to do with the calculation?

As far as the lottery is concerned the combinations 1,2,3,4,5,6;1,3,2,4,5,6;1,2,4,3,5,6, etc for all 720 combinations of 1,2,3,4,5,6 are all the same draw.

[Bo]

I would think:

there are 49*48*47*46*45*44 possible sixnumber combinations (basicly this is 49!/(49-6)! );

however with this calculation 123456 is a different combination from 654321, so you have to devide by the number of ways in which you could arange 6 given numbers; this indeed is 6*5*4*3*2*1.

or: P=N! /(x!(N-x)!)

(where N is the total number of things you can take, and x is the number of things you actually take)

these are the so called binomial coefficients.

 

Bo

[Turtle]

The probability of a multistage experiment is the the product of all the seperate experiments. The lottery drawing you describe has 6 such intermediate experiments. Now think this one over; is the probability of drawing six consecutive numbers the same as drawing 6 numbers that are factors of 3. e.g. 3,6,9,12,15,18? I say hardly.

___The very fundamental axiom of probability is false. i.e. 'assume equal probability', Otherwise called 'fairness', there is no such thing.

___It is possible to handicap these lotteries by analyzing the past draws. For example in the history of the Washington Lottery when it used 6/49, no drawing ever produced more than 3 of the numbers drawn in the previous game and by playing combinations with 0, 1, 2, or 3 such numbers hundreds of thousands of combinations get eliminated thus increasing your odds for winning.

[Boerseun]

It is possible to handicap these lotteries by analyzing the past draws. For example in the history of the Washington Lottery when it used 6/49, no drawing ever produced more than 3 of the numbers drawn in the previous game and by playing combinations with 0, 1, 2, or 3 such numbers hundreds of thousands of combinations get eliminated thus increasing your odds for winning.

I don't think so.

 

When there's a draw and three numbers correspond to the previous draw, that in itself is already remarkable. But the chances of being the exact same set of numbers as the previous draw should be the same as being any other set of chosen numbers - i.e. very small.

 

I cannot see history having any effect on the game at all.

 

You can say that to play the same set of numbers that got selected in the previous draw is silly, but the chances are one to ten billion for it, same as any other set.

[Turtle]

I don't think so.

 

Of course you don't think so; never the less, it is the case. Years ago I wrote a proof of it which most mathematicians refused to even look at & the one who did spent 3 days with it & on returning it he said he found no mistakes but that it 'just can't be right'. If you understand something about Chaos Theory, you must understand it shows that 'assuming equiprobability' is a false assumption. The 2 guys that started much of chaos down at Stanford used it to win at Roulette, but being mathematicians and not gamblers they sold their method to gambers & moved on. I hear they work for Wall Street now. Rerad about it in James Gleicks book on Chaos.

[Boerseun]

I'm still not too sure.

 

The same should apply to roulette, then. And, I've played a little in my time. After the ball has landed on a certain number, the next number to fall has a 1:38 chance. And the same number falling twice does indeed happen roughly once every 40 to 50 spins, much to everyone's dismay.

 

I simply can't see history having an effect, but I'm open to suggestions!

[Turtle]

___A quick Google of "Stanford Chaos Theory guys" gave this:

http://www.cs.brown.edu/research/ai/dynamics/tutorial/Documents/CrackingWallStreet.html

___It's a long read, but it's a seriously complex subject. Even though Gleick's book isn't the most current, it's a good introduction. The gambling business I mentioned is in the linked article about 1/3 of the way in. :confused:

[cwes99_03]

History of the numbers held before would of course be quite important if the creation of the parts of the "random" drawing were made by anything. If the numbers are purely mathematical, then yes fairness exists. If the numbers are based upon improperly shaped balls and a vacuum, then yes history is imporant. Though, according to the theory above, the opposite result of what is expected is true. Therefore, I propose that the system actually is made up of several machines (as it is in Illinois). However, this would not preclude the same number from occuring more than once, unless the same balls from the first machine that were not drawn are parsed into the second machine. Then it becomes a statistical analysis of which ball is most likely to be picked up by each machine, and a running statistical analysis of which number would be most likely in the second machine if the first machine picks up number x.

 

Quite the confoundary techniques they've created. But so is the way the gambling commision varies things.

[Turtle]

History of the numbers held before would of course be quite important if the creation of the parts of the "random" drawing were made by anything. If the numbers are purely mathematical, then yes fairness exists.

___There is only psuedo-random number generation that is "purely mathematical". Chaos Theory is far more significant than a 'technique'. I found nothing confounding in the article; what confounds me is how slowly people come to understand this subject. :confused: