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#1 2006-10-08 05:38:29

MarkusD
Member
Registered: 2006-10-08
Posts: 28

Conditional Probability and Lottery Question

Hey everyone, first post here and I have a few questions about a lottery:

If you have a lottery and you have to pick 5 numbers between 1-37 (with each number only able to be drawn once) whats the probability of getting all 5 numbers the same? Whats the probability of getting all 5 numbers in the same order as the drawing. And why are the odds of winning in the former case  1 in 435,897?


I tried using conditional probability but Im not sure how to use beyond two events.

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#2 2006-10-08 06:24:42

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Conditional Probability and Lottery Question

The conditional probably isn't too hard for this, because you need all the balls to win. It would only start getting hard if you could win with only 4 balls matched or something.

The first ball that is drawn out is one of a possible 37, and there are 5 possible numbers it could be for it to be on your list. Assuming the 1st ball is matched, the second ball will be one out of 36, and there are 4 possibilities for it being on your list. This continues until you get to the 5th ball, and in that case there are 33 possible numbers it could be and only one possible way for it to be on your list.

So the overall probability is (5*4*3*2*1)/(37*36*35*34*33) = 120 / 52307640 = 1/435897.

The second question is simpler, because each ball that is drawn, you need to match exactly, so the probability is just 1/(37*36*35*34*33) = 1/52307640.


Why did the vector cross the road?
It wanted to be normal.

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#3 2006-10-08 07:20:32

MarkusD
Member
Registered: 2006-10-08
Posts: 28

Re: Conditional Probability and Lottery Question

Thanks for the reply!

I see that youre using combinations or at least I got the same answers as you did using

n!/r!(n-r)!

for the first case, and:

n!/(n-r)!

for the second.


I was trying to use conditonal probability earlier like P(B|A) = P(A&B)/P(A). I was trying to answer the question: what is the probability of getting the 5th number right provided that the first 4 were correct? Is that equivalent to the question "what is the probability of getting all 5 numbers correctly (assuming order doesnt matter, and what is the answer if order does matter)?"

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#4 2006-10-08 08:33:33

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Conditional Probability and Lottery Question

Not exactly.

"What is the probability of getting the 5th number right provided that the first 4 were correct?" means that the first 4 are already correct and so you don't need to work out probabilities for them, you just need to work out the chance that the 5th will be correct as well.

In "What is the probability of getting all 5 numbers correctly?", you need to work out the chances of the first 4 balls as well, and multiply them together.

The second answer will be significantly lower than the first.


Why did the vector cross the road?
It wanted to be normal.

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#5 2006-10-08 09:06:20

MarkusD
Member
Registered: 2006-10-08
Posts: 28

Re: Conditional Probability and Lottery Question

mathsyperson wrote:

Not exactly.

"What is the probability of getting the 5th number right provided that the first 4 were correct?" means that the first 4 are already correct and so you don't need to work out probabilities for them, you just need to work out the chance that the 5th will be correct as well.

hmmm, what if the word "provided" was changed to "condtioned on"? Would that make a difference?

I guess my confusion is that we know before making
the last draw that the first four draws were correct, which is contrasted from the other example of getting all 5 numbers correct because this probability is calculated before the result of the drawing is known.

In my head Im thinking P(5th draw being correct|1st, 2nd, 3rd, and 4th number was correct) and I dont know what the bayesian formula would be. Or why I wouldnt need it.

Last edited by MarkusD (2006-10-08 09:08:13)

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