Probability of drawing numbers based on past drawing counts? Just a mathematical thought I had, and I couldnt find a relevant answer. 
Lets assume that 3 numbers are drawn from the following list: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The drawing is without replacement (if we draw 1, we have 2, 3, 4, 5, 6, 7, 8, 9, 10 left, and so on). Now, if we know that there were 10 previous drawings with the following statistics:


*

*2, 4, 7

*1, 4, 8

*3, 8, 10

*1, 5, 6

*2, 7, 10

*5, 7, 9

*6, 7, 10

*1, 4, 7

*4, 8, 9

*8, 9, 10
So now we know that the following numbers were drown x times as follows:


*

*1 = 3 times total

*2 = 2 times total

*3 = 1 times total

*4 = 4 times total

*5 = 2 times total

*6 = 2 times total

*7 = 5 times total

*8 = 4 times total

*9 = 3 times total

*10 = 4 times total


How can one calculate the probability of drawing, X, Y, Z on the 11th drawing? 
Is this probability purely based on the count of previous number drawings? Or also on the number combinations of each individual drawings? 

You don't have nearly enough data to estimate dependencies.  
The best you can do is estimate weighted probabilities.   Such as:
$$\Pr(\{1, 7,10\}) \simeq \dfrac{3}{30}\dfrac{5}{27}\dfrac{4}{22}+\dfrac{3}{30}\dfrac{4}{27}\dfrac{5}{23}+\dfrac{4}{30}\dfrac{3}{26}\dfrac{5}{23}+\dfrac{4}{30}\dfrac{5}{26}\dfrac{3}{21}+\dfrac{5}{30}\dfrac{3}{25}\dfrac{4}{22}+\dfrac{5}{30}\dfrac{4}{25}\dfrac{3}{21}$$
