In game, we randomly generate four grids (cards) $3\times 5$ (row × col), with each column containing 3 numbers randomly selected without replacement from 18 possibilities: first columns from the left are randomly selected from the numbers 1-18, the second columns are selected from 19-36, the third column from 37-54, the fourth columns from 55-72, and the fifth columns from 73-90.
So, for the first columns we fill in 3*4=12 different numbers from the 1-19 without replacement, and so on. Please, see image.
Then, we randomly draw, without replacement, from a pool of balls numbered 1 through 90, 40 balls at ones.
Please find winning patterns on the top of image.
Question is how to calculate the probability, for example, of the event on the attached image. You can see, that the second grid won on two patterns: 8 + 3, and also we have single wins on the first and fourth grids.
I’m kindly asking to show me how to calculate these probabilities. In fact, I need to calculate the expected win in that game. So, basically I need to cover up all outcomes.
I looked at the problem once more, and it looks not so complicated.
Fix a winning pattern $P$ consisting of $k$ numbers and a grid $G$. There are ${90\choose 40}$ different pool draws. Among them ${90-k \choose 40-k}$ provide the pattern $P$ on the grid $G$. Thus, assuming symmetry of pool draw probabilities, the probability that $P$ will be realized on $G$ is ${90-k \choose 40-k}/{90\choose 40}$. So the expected win $E(P)$ of the pattern $P$ is $b(P){90-k \choose 40-k}/{90\choose 40}$, where $b(P)$ is the bonus of pattern $P$.
As I understood from the picture, if several patterns are realized then the total bonus equals the sum of bonuses of the realized patterns. Since an expected sum of wins equals to sum of expected wins, the expected total win of a grid equals to a sum of expected wins for each of the winning patters.
At last, since we have four grids, the expected win of the game equals the expected win of a grid multiplied by four.