# Conditional Probability - Lottery

• Oct 27th 2013, 12:09 AM
UncleFester
Conditional Probability - Lottery
Hi all,I'm working on an assignment but I got an answer for one of the questions that makes no sense. The question is about the lottery (6 numbers picked, from 1 to 49):'What is the conditional probability that a ticket wins the jackpot, given that it matches at least 5 of the numbers drawn?'I used: P(E|F)= P(EF)/P(F)Where P(EF) ~ 1/14,000,000 and P(F) ~ 1/50,000. The answer I got was 1/265 but I don't see how that's possible. Surely the answer should be 1/44?Any help would be very appreciated,Thanks
• Oct 27th 2013, 03:52 PM
chiro
Re: Conditional Probability - Lottery
Hey UncleFester.

Just to clarify, do you need to get 6 numbers correct to get the jackpot? (In other words, you have five numbers and want to get the probability of winning the jackpot given that you already have five winning numbers)?
• Oct 29th 2013, 10:21 AM
UncleFester
Re: Conditional Probability - Lottery
Hey Chiro,

Thanks for the reply. Yes 6 numbers is the jackpot. Also I should clarify that it's the probability given that you have at least 5 numbers, so the answer I think should be actually more likely than 1/44 - I just can't see where I went so wrong.
• Oct 29th 2013, 05:11 PM
chiro
Re: Conditional Probability - Lottery
1/44 sounds like the right answer.

Intuitively you already have 5 numbers so you only have 49-5=44 choices left. All choices are equally likely since we assume that the process of picking the ball is random.

• Oct 31st 2013, 10:25 AM
UncleFester
Re: Conditional Probability - Lottery
My course tutor said:

'1/44 is not the right answer. Note that "you match at least five balls" is different from "you match the first five balls drawn": the conditional probability given the latter would indeed be 1/44'

I think the formula I gave in the OP is what is needed, but I must be using it wrong somehow
• Oct 31st 2013, 02:41 PM
chiro
Re: Conditional Probability - Lottery
I'm not following you.

In your lottery are there 6 different numbers? When a number is selected is it taken out of the urn? If so have you already picked five winning and completely distinct numbers from the urn and taken them out for good?

If the above is the case, and all balls have an equally likely chance of being selected then the probability should be 1/44.

If you have different assumptions, then we need to know them.
• Oct 31st 2013, 02:58 PM
SlipEternal
Re: Conditional Probability - Lottery
Let's work with outcomes rather than probabilities. There are $\binom{49}{6}$ possible outcomes for any given ticket.

There is 1 outcome where your chosen numbers match the jackpot. Now, let's count the outcomes where the ticket matches the winning numbers for at least five of the six numbers.

At least 5 winning numbers means exactly five winning numbers and one non-winning number or exactly six winning numbers. So, let's try that. You have six winning numbers. Choose five of them. You have 43 non-winning numbers. Choose one of them. That is the number of outcomes that give you exactly five winning numbers and one non-winning number. There are a total of $\binom{6}{5}\binom{43}{1} = 6\cdot 43 = 258$ outcomes. Finally, there is exactly 1 outcome where you win the jackpot. That is a total of 259 outcomes for "at least five winning numbers".

So, the conditional probability is the number of outcomes where you win the jackpot divided by the number of outcomes where you have at least five winning numbers. That is $\dfrac{1}{259}$.
• Oct 31st 2013, 03:41 PM
chiro
Re: Conditional Probability - Lottery
I think the mistake I made was that I didn't care about the ordering.

For the OP: I assumed the event was P(Final Number|Other Previous numbers) with no kind of ordering where I treat say the set {A,B,C,D,E,F} as that set with no ordering information what-so-ever.

If you are making different assumptions then you will get different answers and I wanted to clarify my assumptions.
• Oct 31st 2013, 04:25 PM
SlipEternal
Re: Conditional Probability - Lottery
Quote:

Originally Posted by chiro
I think the mistake I made was that I didn't care about the ordering.

For the OP: I assumed the event was P(Final Number|Other Previous numbers) with no kind of ordering where I treat say the set {A,B,C,D,E,F} as that set with no ordering information what-so-ever.

If you are making different assumptions then you will get different answers and I wanted to clarify my assumptions.

You were using the assumption that a psychic predicted five of the six numbers that would come up. So, you knew precisely which five to pick. So, you know you have at least five numbers correct already because you know five of the winning numbers. If you don't assume prior knowledge of five of the six winning numbers, then you get my answer. My answer does not assume any ordering of the numbers.
• Oct 31st 2013, 05:49 PM
chiro
Re: Conditional Probability - Lottery
Yes that sounds about right: I think I was too generous in my assumptions.
• Nov 1st 2013, 05:57 AM
UncleFester
Re: Conditional Probability - Lottery
Quote:

Originally Posted by SlipEternal
Let's work with outcomes rather than probabilities. There are $\binom{49}{6}$ possible outcomes for any given ticket.

There is 1 outcome where your chosen numbers match the jackpot. Now, let's count the outcomes where the ticket matches the winning numbers for at least five of the six numbers.

At least 5 winning numbers means exactly five winning numbers and one non-winning number or exactly six winning numbers. So, let's try that. You have six winning numbers. Choose five of them. You have 43 non-winning numbers. Choose one of them. That is the number of outcomes that give you exactly five winning numbers and one non-winning number. There are a total of $\binom{6}{5}\binom{43}{1} = 6\cdot 43 = 258$ outcomes. Finally, there is exactly 1 outcome where you win the jackpot. That is a total of 259 outcomes for "at least five winning numbers".

So, the conditional probability is the number of outcomes where you win the jackpot divided by the number of outcomes where you have at least five winning numbers. That is $\dfrac{1}{259}$.

Ahh, I think that might be right. When I figured the answer must be 1/44 or close, I was thinking the order mattered too. Thanks both for covering both situations :D