An easier side game of 5 of 39 allows me to pick more than 5 numbers to match, for example, four of the 5 that are drawn. What would the odds be of choosing 7 to match 4 of the 5 winning numbers of 5 of 39 lottery?
How do I calculate the odds of a "39" lottery side-game where 4 numbers are matched to a 5 of 39 draw but selecting 7 (not just 4) as candidates?
I understand the probability of matching 5 of 39 is 1/(39C5), matching 4 of 5 of 39 is (5C4)(39-5 C 5-4)/(39C5), matching 3 of 5 of 39 is (5C3)(34C2)/(39C5), etc.
But there's a side game that allows me to play up to all the 39 numbers (if I really want to go broke) with the option of matching only 2 or 3 or 4 to the 5 that are drawn.
I'm stuck at what happens when I choose more than the number of the draw, for example, If I select 7 numbers between 1 and 39, what are the odds of matching 4 to the winning 5 that are drawn from 39?
Best Answer
The sample space consists of the $$\binom{39}{7}$$ ways of selecting $7$ of the $39$ numbers in question.
The number of ways of selecting exactly four of the five winning numbers is $$\binom{5}{4}\binom{34}{3}$$ since such a selection requires selecting four of the five winning numbers and $7 - 4 = 3$ of the $39 - 5 = 34$ non-winning numbers.
Hence, the probability of matching exactly four of the five lottery numbers when you are allowed to select seven of the $39$ numbers is $$\frac{\dbinom{5}{4}\dbinom{34}{3}}{\dbinom{39}{7}}$$