How to calculate lotto odds?

Question

Mathematics

There is a game similar to Keno , where 10 out of 40 numbers are selected.
The player can choose his set of up to 10 numbers and, depending on the number of selected and guessed numbers, his bet is multiplied by a different coefficient. For example, the player has chosen 3 numbers. Then his chances are:

• guess 0 numbers: 41.093%
  
• guess 1 number: 44.028%
  
• guess 2 numbers: 13.664%
  
• guess 3 numbers: 1.215%
  

For each of the outcomes, we calculate the coefficient according to the formula 100 / вероятность. It turns out:

• if guessed 0 numbers: 0
  
• if you guessed 1 number: 2.271
  
• if guessed 2 numbers: 7.319
  
• if guessed 3 numbers: 82.305
  

But with such odds, the "casino" is a big loser. For example, 100 players bet $1 each. Of these, according to probabilities, 1 number will be won by 44,028 and taken away, at a rate of 2.271 2.271 * 44.028 = $99.9- i.e. the entire prize fund. And they guess the 2nd and 3rd numbers - there, too, each option spends $100. Total payouts are 3 times higher than the bets made.
How to calculate the odds so that the "casino" remains in the black, and the bets reflect the real probability of winning?
Also tell me, please, literature about this kind of games and their mathematics: both probabilistic and financial.

Answer 1

10 out of 40 are randomly chosen by the machine, 3 out of 40 are chosen by the player. It is necessary, for example, to find the probability of falling out only 2.
The total number of cases C(10,40) * C(3,40)- we divide the number of favorable options by it. This is a choice of 10 winning C(10,40); out of 10 you need to choose 2 hits C(2,10); from the remaining 30, you need to choose 1 missed C(1,30) and multiply them all. Total

Q(40,10,3,2) = C(10,40) * C(2,10) * C(1,30) / (C(10,40) * C(3,40))

Surely something can be cut down nicely.
C(N, M) – number of combinations
C(N, M) = M! / ( N! * (M-N)! )