# Thread: I have a probability question

1. ## I have a probability question

Lets assume that Jim is playing a 6/49 lottery and that he purchases 91 ticket with 8 encore plays.

The 6/49 works that you choose 6 numbers 1-49 and 7 numbers are drawn. A 3 number match is considered a win.

With encore (keeping it simple) a player has 2, 1 in 10 chances of winning something per play.

What are the odds that nothing will be won from this lot of tickets?

2. Please note: probability is not my strong suit, so you should probably double-check this solution. However, I think it is correct...

Originally Posted by D. Bronson
Lets assume that Jim is playing a 6/49 lottery and that he purchases 91 ticket with 8 encore plays.

The 6/49 works that you choose 6 numbers 1-49 and 7 numbers are drawn. A 3 number match is considered a win.
The winning possibilities are picking 6, 5, 4 or 3 correct numbers. The total probability of winning, then, is the sum of the probability of getting 6, 5, 4 and 3 correct numbers. It's easy to find the probability for getting all six:

$P(X=6)=\frac{43!7!}{49!}$

To find the probability of five successes, we must first determine the total possible combinations:

${6\choose5}=\frac{6!}{5!}=6$

The probability of five successes is the sum of the probabilities of each individual combination of five successes:

$P(X=5)=6\frac{43!7!42}{49!2}$

In fact, we can generalize this:

$P(X\geq3)=\sum_{x=3}^6{6\choose x}\frac{43!42!7!}{49!(36+x)!(7-x)!}=\sum_{x=3}^6\frac{43!42!7!6!}{49!(36+x)!(7-x)!(6-x)!x!}$

Conversely, the probability of not winning is simply:

$1-\sum_{x=3}^6\frac{43!42!7!6!}{49!(36+x)!(7-x)!(6-x)!x!}$

and the probability of not winning 91 times in a row is:

$[1-\sum_{x=3}^6\frac{43!42!7!6!}{49!(36+x)!(7-x)!(6-x)!x!}]^{91}$

With encore (keeping it simple) a player has 2, 1 in 10 chances of winning something per play.
The probability of losing each encore ticket:

$(\frac{9}{10})^2=\frac{81}{100}$

The probability of losing eight of these is:

$(\frac{81}{100})^8$

What are the odds that nothing will be won from this lot of tickets?
Just multiply the probabilities:

$[1-\sum_{x=3}^6\frac{43!42!7!6!}{49!(36+x)!(7-x)!(6-x)!x!}]^{91}(\frac{81}{100})^8$

I'll let you compute that.

3. Thats a really nice formula... wish I could solve it, but...

Here's the deal, I am not versed in advanced mathematics and the reason that I am posing such a question is just this: Jim. I know him, he purchased these tickets for our lottery pool as he does every week. This week in particular I was injured and not at work.... maybe you can see where I am going with this. Two other of my co-workers are in different yet similar situation; we paid for the Saturday draw but were not available to contribute for the following Wednesday draw. Low and behold the Wednesday draw nets a rather substancial return.

Now, I figure that if money or free tickets were won on that saturday draw, then surely I have a vested interest in a portion of the winnings. I ask Jim. He tells me that they got caught up in the moment and he will make this right ... I say "well, the only way that I can see that we don't have interest in some of this money is if nothing was won on the Saturday.".... The conversation ends. Several days later I call Jim to see what is going on.... "Jim tells me that the pool went dead and nothing was won". Funny that he didn't mention that in the first conversation. Funny that though we always keep records of the purchased tickets posted for everyone to have a look at, that that weeks have disappeared. Funny that Jim can't look me straight in the eye and tell me the pool went dead. Funny that the office was "restaffed" and the three of us were all let go within three weeks of returning to work.

Basically am trying to figure out what the odds are that Jim is lying to us.

Can anyone solve those odds for me?

4. I don't have time to calculate that out, but I can tell you that the second term is about .19, which would be an upper bound. The term on the left (to the 91st power) I just don't have the patience to compute.

5. Originally Posted by hatsoff
I don't have time to calculate that out, but I can tell you that the second term is about .19, which would be an upper bound. The term on the left (to the 91st power) I just don't have the patience to compute.

Its a tough one is it? Well I appreciate your help thus far. Thank you.

6. Originally Posted by D. Bronson
Its a tough one is it? Well I appreciate your help thus far. Thank you.
Actually, it didn't take as long as I thought. The probability of nothing at all winning is approximately .004289366051, or, 0.4289366051%. In other words, there is less than one half of one percent chance that nothing won.

7. Originally Posted by hatsoff
Actually, it didn't take as long as I thought. The probability of nothing at all winning is approximately .004289366051, or, 0.4289366051%. In other words, there is less than one half of one percent chance that nothing won.
So basically 1 in 233... for the lay person

I appreciate the effort very much. And thank you for helping without judgement.

8. Originally Posted by D. Bronson
And thank you for helping without judgement.
I deal in math. What people do with it is their own business.