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In my country we have a special lottery

There are 45 numbers and you can choose as many as you like in order to get right the 5 winning numbers.

For example you can choose 10 out of the 45 numbers asking to get right the 5 winning numbers - this means that the choosen numbers will use if n=10 (n*n-1*n-2*n-3*n-4)/5*4*3*2*1 = 252 columns

My questions are

1) if i finally get the 5 winning numbers right so this mean I will get 1 first category win (5 numbers) but how many wins will I have of the second category (4 numbers), how many of the third category (3 numbers), how many of the forth category (2 numbers) and how many of the fifth category (1 number).

2) if i finally get the 4 out of 5 winning numbers right so this means that I will get at least 1 second category win (4 numbers), how many second category wins will i have, how many wins will i have of the third category (3 numbers), how many of the forth category (2 numbers) and how many of the fifth category (1 number).

3) if i finally get the 3 out of 5 winning numbers right so this means that I will get at least 1 third category win (3 numbers), how many third category wins will i have, how many wins will i have of the third category (3 numbers), how many of the forth category (2 numbers) and how many of the fifth category (1 number).

and so on

Anyone knows the math type that will help me answer my questions?

Teo Chris
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  • Not sure I am following. If I have guessed all $5$ winning numbers correctly, then I suppose I have five "second category" wins, no? If the $5$ numbers are ${a,b,c,d,e}$ then I claim second category victory for ${a,b,c,d},{a,b,c,e},{a,b,d,e},{a,c,d,e},{b,c,d,e}$. Or is this not how the game works? Not sure the rules are clear. If I can truly "choose as many as I like" why don't I just choose all $45$? – lulu Jan 04 '16 at 13:05
  • Lulu thanks for your answer, if I choose all 45 numbers then I will have to pay for 1221759 columns and each column costs 0.50 euros - If i had 611000 euros then i would not care to play this lottery. I have heard this (n!/n!)*(n-k)! is it related? – Teo Chris Jan 04 '16 at 13:23
  • There is a table that shows what happens if someone chooses up to 20 out of the 45 numbers asking for the 5 winning numbers http://www.opap.gr/en/web/guest/360 does it help? I need to fill this table for all 45 numbers – Teo Chris Jan 04 '16 at 13:29
  • The question (as posted) did not mention that you have to pay for each column. It seems this option in this lottery is essentially allowing you to buy $\frac{10!}{(10-k)!k!}$ "tickets", each containing a single 5-number guess, one ticket for each possible set of $5$ numbers drawn from the $k$ numbers you choose. In other words it is a shortcut for playing many different combinations of numbers without having to write all those guesses one by one on a series of betting slips. You also don't mention the "Joker"; do you care about it? – David K Jan 04 '16 at 13:52
  • By the way, depending on how the pari-mutuel awards are determined, someone with $611000$ euros might want to play the lottery sometimes. It has happened before: http://www.nytimes.com/1992/02/25/us/group-invests-5-million-to-hedge-bets-in-lottery.html?pagewanted=all (at least the plan was to buy tickets for all possible combinations, but the Virginia lottery required you to place individual bets on each combination of numbers--no "full development" option on their betting slips--and the syndicate ran out of time before they managed to place all the bets). – David K Jan 04 '16 at 14:01
  • David at the moment i do not care for the joker - guessing all 5 numbers right will give enough profit! - So can you please help me to calculate the number of smaller wins when all 5 numbers are right - when only 4 numbers are right - when only 3 numbers are right - when only 2 numbers are right - when only 1 number is right. For example lets say i choose 25 out of the 45 numbers and i am asking for the 5 winning numbers! – Teo Chris Jan 04 '16 at 14:07

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