I've been staring at this one for a few hours and not sure where to start (caveat: my statistics is EXTREMELY rusty), but Google didn't offer up much to help.
The scenario is an age-old lottery of 49 choose 6, except for the fact your ticket has a variable number of balls. If you choose 6 balls on your ticket, then the answer is covered ad nauseam (https://en.wikipedia.org/wiki/Lottery_mathematics#Information_theoretic_results). However, the catch is members may opt, for a lower fee, to enter only 5 numbers on the ticket. They are excluded from the possibility of matching 6 balls, obviously, however they can still make all sub-prizes of matching 1 through 5 balls, albeit the odds are worse.
How would one go about calculating the odds for the secondary prizes when members play 5, 4, 3 etc balls on a draw that is always 6. Conversely, how could the odds be calculated when the allowed number of picks on a ticket are MORE than the number of draws (6). Obviously in this case, there is not a single "jackpot" and odds overall improve.
Thanks very much in advance!
I believe I was able to solve this, using a modification of the hypergeometric distro. Feel free to check work, but this seems to pass a sanity check:
N = ball pool depth
H = house ticket size
Y = your ticket size (less, equal, or greater than H)
C = correct hits
$\binom{H}{C} {N - H \choose Y - C}$
Where the denominator is the part I'm slightly fuzzy on, thinking whereby the maximum possibilities is based on the house's draw, which would be ${N \choose H}$.