I'm making this table.
You have classic cards from A to 10. 4 of every number. So 40 cards.
It's a hand, so you have one less card on every draw.
I want to calculate the repeats. (only in the numbers)
* numbers are rounded
I'm basically ignorant on maths, even this basic, so I didn't understand how I to continue now.
I was calculating the probabilities to at least one repeat, but now I want to calculate the probabilities to obtain a double, a triple and a quadruple.
I have to take in account that now when you have a triple, you didn't have a double and the same for the quadruple and triple.
How can I calculate the rest?
Suppose you have a deck of 4 Ones, 4 Twos, ... ,4 Tens.
You can write the individual situations as tupels. The tupel will denote how many Ones, Twos, .... , Tens you have after drawn K cards. E.g. $(0,0,2,0,0,0,0,0,0,1)$ denotes that you have drawn 3 cards in total, your draw consisting of 2 Threes and 1 Ten.
Now the probability for drawing such a tupel is given by the multivariate hypergeometric distribution. When you have drawn $K$ cards, consisting of $k_1$ Ones, $k_2$ Twos, ..., $k_{10}$ Tens, you have
$$ p(k_1,k_2, ... , k_{10}) = \frac{{4 \choose k_1} {4 \choose k_2}\cdots {4 \choose k_{10}} }{{40 \choose K}} $$
So for the above example, $$ p(0,0,2,0,0,0,0,0,0,1) = \frac{{4 \choose 2} {4 \choose 1}}{{40 \choose 3}} = 0.0024 $$
Now for your table, you can sum up the probabilities of all tuples which lead to the desired result.
E.g. 2 draws, 1 double (you have that already):
$$ p(2,0,0,0,0,0,0,0,0,0) = \frac{{4 \choose 2}}{{40 \choose 2}} = 0.0077 $$
You can have the 2 at ten positions, so you get
$P($ 2 draws, 1 double$) = 10 \; p(2,0,0,0,0,0,0,0,0,0) = 0.077$
Next: 3 draws, 1 double:
$$ p(2,1,0,0,0,0,0,0,0,0) = 0.0024 $$ (see above)
You can have the 2 at ten positions, and for each of these positions you can have the 1 at nine positions, so you get
$P($ 3 draws, 1 double$) = 10 \cdot 9 \cdot 0.0024 = 0.2160$
Next: 3 draws, 1 triple:
$$ p(3,0,0,0,0,0,0,0,0,0) = \frac{{4 \choose 3} }{{40 \choose 3}} = 0.000405 $$
You can have the 3 at ten positions, so you get
$P($ 3 draws, 1 triple$) = 10 \cdot 0.000405 = 0.00405$
This concludes the 3 draws.
You will have to sum up increasingly many different situations as you go to 4 draws and higher. It may be a bit tedious but it works.