Lottery Probability: Find 7 Numbers Out of 20

AI Thread Summary
The discussion focuses on calculating the probability of matching 7 numbers out of 20 drawn from a pool of 80 in a lottery game. Various formulas are proposed, but the correct approach is identified as using the combination formula P = C(n,k)C(80-n,20-k)/C(80,20), where n is the total numbers chosen and k is the numbers matched. The participants confirm that the probabilities calculated using different methods yield the same result, specifically P = C(73,13)/C(80,20) for matching all chosen numbers. The game in question is identified as Keno, where players choose n numbers from 80, and the discussion clarifies the conditions for winning based on matching k numbers. The conversation concludes with a consensus on the correct probability calculation method.
jk22
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I watch lastly a lottery game : there are 80 numbers, from them 20 are selected, and you can choose for example to find 7 numbers out of the 20.

I'm looking for the probability to find 7 numbers, is it simply C(80,7)/C(80,20) or is it 20!/13!/C(80,20) or C(80,13)/C(80,20) ?
 
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I can tell that none of those are right. Consider the extreme case where the game show picked 80 (instead of 20) out of 80 . Then C(80,80)=1, so all those formulas would give numbers much larger than 1. They must all be wrong. At the moment, I'm not sure what the correct answer is. What about C(20,7) / C(80,7)? That is the ratio of all your possible picks from the special 20 over all your possible picks from the original 80.
 
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I thought a bit about the following : I fix 7 numbers over 80, there remains 13 numbers to choose freely among 73, these are the favorable cases, whereas all the cases are 20 numbers chosen between 80, hence : C(73,13)/C(80,20) ?
 
jk22 said:
I thought a bit about the following : I fix 7 numbers over 80, there remains 13 numbers to choose freely among 73, these are the favorable cases, whereas all the cases are 20 numbers chosen between 80, hence : C(73,13)/C(80,20) ?

My equation above, C(20,7)/C(80,7) gives exactly the same answer as yours (C(73,13)/C(80,20) = 2.4402556E-5). I'm not sure I follow your reasoning, but since we came at it from two different logic directions, I bet they are both right. I tried to see how they could be identical, but it got too messy for me.
 
Your expressions are the same. The game is Keno. More generally than your case, you choose n numbers from 80 where n is less than or equal to 20. 20 numbers are then drawn and you win if you match k of them where k is less than or equal to n. There are C(80,20) total combinations of 20 drawn from 80. You are holding n numbers. There are C(n,k) combinations of k from n. The rest of the numbers drawn do not match yours, that is 20-k drawn from the remaining 80-n. There are C(80-n,20-k) ways to do that. So the probability of matching k of your n numbers is C(n,k)C(80-n,20-k)/C(80,20). In the special case where you match all of your n numbers, k=n and P=C(80-n,20-n)/C(80,20). If n=7 then P=C(73,13)/C(80,20).
 
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