# How would you work out the odds of you winning the lottery

• Apr 29th 2010, 06:14 PM
JohnLord123
How would you work out the odds of you winning the lottery
Hi, I drag at math can someone please explain this to me:

How would you work out the odds of you winning the lottery.
• Apr 29th 2010, 06:25 PM
Gusbob
Quote:

Originally Posted by JohnLord123
Hi, I drag at math can someone please explain this to me.

Really depends what kind of lottery we're talking about - most importantly, we need to know how many digits you need to choose.
• Apr 29th 2010, 06:41 PM
JohnLord123
Quote:

Originally Posted by Gusbob
Really depends what kind of lottery we're talking about - most importantly, we need to know how many digits you need to choose.

The possible numbers are 1 to 40 (the chosen number is replaced before each draw). You choose 6 numbers, you must get all of them correct to win.
• Apr 29th 2010, 06:56 PM
Gusbob
Quote:

Originally Posted by JohnLord123
The possible numbers are 1 to 40 (the chosen number is replaced before each draw). You choose 6 numbers, you must get all of them correct to win.

Ok so you have a 1/40 chance of getting the first number correct. You also have a 1/40 chance of getting the second number correct and so on.

So you need to beat the odds of

1/40 x 1/40 x 1/40 x 1/40 x 1/40 x 1/40 = $1/40^6$

To understand why we multiply the odds together, we can use a simplified model. Say we want 6 heads in a row in a coin toss. For the first flip, you got the possibility for getting a head or a tail. So the odds are 1/2 for a head.

Now for 2 flips, you have four possibilities.
-Tail - Tail

Now there is only one possibility out of four for a head-head combination. So the odds are 1/4.

For every additional flip, each of the combination branches off into two more with an added heads or tails, doubling the # of possible combinations whilst there will only be one correct all heads combination.
• Apr 30th 2010, 04:24 AM
JohnLord123
Thank you so much, you did a wonderful job of explaining that. I understand perfectly now.

What if the ball is not returned after it is drawn?

I suppose it would

1/40 1/39 1/38 e.t.c and your chances of getting all six would be eqaul to timesing them all together.

I have another question though, in this case, what would the chances of getting say, two balls correct? How would you figure that out?

Thanks again for the great explanation.
• Apr 30th 2010, 04:26 AM
Gusbob
Quote:

Originally Posted by JohnLord123
Thank you so much, you did a wonderful job of explaining that. I understand perfectly now.

What if the ball is not returned after it is drawn?

I suppose it would

1/40 1/39 1/38 e.t.c and your chances of getting all six would be eqaul to timesing them all together.

That is correct.

Quote:

I have another question though, in this case, what would the chances of getting say, two balls correct? How would you figure that out?
Would this be two consecutive balls or any two balls?
• Apr 30th 2010, 04:32 AM
JohnLord123
Quote:

Originally Posted by Gusbob
That is correct.

Would this be two consecutive balls or any two balls?

Just the odds of getting any two out of six (any two, in any order), i.e.
• Apr 30th 2010, 05:01 AM
mr fantastic
Quote:

Originally Posted by JohnLord123
The possible numbers are 1 to 40 (the chosen number is replaced before each draw). You choose 6 numbers, you must get all of them correct to win.

Quote:

Originally Posted by Gusbob
Ok so you have a 1/40 chance of getting the first number correct. You also have a 1/40 chance of getting the second number correct and so on.

So you need to beat the odds of

1/40 x 1/40 x 1/40 x 1/40 x 1/40 x 1/40 = $1/40^6$

[snip]

This number needs to be multiplied by 6! because the order of the numbers is not important.

Quote:

Originally Posted by JohnLord123
[snip]
What if the ball is not returned after it is drawn?

I suppose it would

1/40 1/39 1/38 e.t.c and your chances of getting all six would be eqaul to timesing them all together.

[snip]

That is not correct since the numbers you choose could be drawn, but not in the order that you chose them in.

The correct answer is $\frac{1}{{40 \choose 6}}$, where ${40 \choose 6}$ is the number of ways of choosing 6 balls from 40 balls.
• Apr 30th 2010, 04:54 PM
Gusbob
Quote:

Originally Posted by mr fantastic
This number needs to be multiplied by 6! because the order of the numbers is not important.

Sorry JohnLord123 and Mr. Fantastic. In the part of the world where I came from, the order does matter :/ I didn't know there was such a thing as a lottery where you can pick the numbers in any order.
• May 1st 2010, 04:09 AM
JohnLord123
Quote:

Originally Posted by mr fantastic
This number needs to be multiplied by 6! because the order of the numbers is not important.

That is not correct since the numbers you choose could be drawn, but not in the order that you chose them in.

The correct answer is $\frac{1}{{40 \choose 6}}$, where ${40 \choose 6}$ is the number of ways of choosing 6 balls from 40 balls.

I'm really confused now. In the first instant (when the balls are returned), you say the chances are 49!, what does this mean and why is it the case (can you show the working).

Same with when the ball is not returned.

Sorry for all the questions, my math is really bad and i'm determinded to understand this.

Thanks for all the help so far.