# Math Help - Lottery Probability

1. ## Lottery Probability

Well firstly I would just like to say, that I don't only want to work out the probability, I also want to make some exceptions. We will use a typical lottery, 6 numbers are drawn from a range of 49 and if the 6 numbers on a ticket match the numbers drawn, the ticket holder is a jackpot winner.

I understand that the numbers 1,2,3,4,5,6 have the same probablility as any other six numbers. BUT! I want to exclude all consecutive numbers so I would like to calculate the probability of winning the lottery, but I don't want to take into account any consecutive numbers. For example:

1, 13, 23, 45, 46, 7

I know that this sounds stupid, but I just want to exclude them.

This is how far I've got... (Skip red if you know)

Starting with a bag of 49 differently-numbered lottery balls, there is clearly a 1 in 49 chance of predicting the number of the first ball selected from the bag. When the second number is picked there are now only 48 balls so there is a 1 in 48 chance and so on.

49 x 48 x 47 x 46 x 45 x 44 = 10,068,347,520

Next the order in which the numbers are selected doesn't matter, so...

1 x 2 x 3 x 4 x 5 x 6 = 720

10,068,347,520 / 720 = 13,983,816

So there is a 1 in 13,983,816 chance of winning?

Ok now this is the bit I am struggling with! How do I now ignore any set that has 2 or more consecutive numbers? Any help would be great, it's just something I am interested. I guess it's the fact that my mind is telling me that consecutive numbers isn't going to happen frequently, and It'll make my chances look mor elikely if I ignore them, and will make me think I'm going to win. 5% of my winnings to the first person to help

2. Let’s discuss the commonly understood lottery game.
If we draw out six numbered balls from forty-nine, order makes no difference.
Only content matters. The combination {2,4,27,33,41,42} is the same as {41,2,33,4,27,42}. If the first is the winning combination then so is the second.
${{49} \choose 6}=13983816$ is the number of possible winning combinations.

However, if you note that example contains a pair of consecutive numbers, 41 & 42. That is what you do not want. Well sometime ago some players, in I think Louisiana, noticed that quite often winning combinations did have consecutive pairs. They want to file a legal complaint. But as it turns out it not that unreasonable for it to happen.
We can think of that a string of six ones and 43 zeros as one ticket. Any arrangement of that string could represent a winning combination. So your question comes down to how many rearrangements of that bit string have no two consecutive ones. The answer is ${{44} \choose 6}=7059052$ the zeros create 44 places to put the ones. So for this particular set of numbers there almost a 50/50 chance a winning ticket will have consecutive entries.

Now of course, if your own lottery game turns on the order in which the balls come out of the hopper, then a totally different analysis must be done. But in the usual lottery order does not matter.

3. The order does not matter, what I want to do it remove all cases of 2 or more consecutive numbers in a set and remove that and then get a probibility.

4. Originally Posted by tumbleweed
The order does not matter, what I want to do it remove all cases of 2 or more consecutive numbers in a set and remove that and then get a probibility.
I suppose that you did not read my response very carefully.
If there is a pool of N numbers from which k are selected to determine a winning ticket, then the probability that two consecutive integers are not chosen is
$\frac{{_{N - k + 1} C_k }}{{_N C_k }} = \frac{{\left( {\begin{array}{c}
{N - k + 1} \\
k \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
N \\
k \\
\end{array}} \right)}}$

I have generated a table to show the probabilities for several different values of N & k.
I have highlighted your particular case of N=49 and k=6.