In an urn, I can put 20 balls having a value from 1 to 7.
So I got Ni, with i $\in$ [1,2,3,4,5,6,7], where Ni is the number of balls with a value i in the urn.
And $\sum_{i=1}^7 Ni = 20$
The game is played like this :
At the first round I pick 3 balls randomly.
Every round I Need to play the round's number in value of balls. (At the second round, I need to play one balls with a value of two, or two balls with a value of 1)
The the balls used on a turn disapears
Every round I pick one balls randomly
If I dont do so there is a malus. The malus is counted like that :
malus = $(N_r - V_r ) * M_r$
$N_r$ = round's number
$V_r$ = Value played at this round
Mr = Malus coefficient of this round (there is one for all the round) , $M_r$ $\ge$ 1
My goal, is to calculate this malus given the urn.
My method for this calcul go like this :
First round, I got two possibilities :
I can play 1 ball of value 1 : 1 - $\binom{20-N1}{3}$
I cant play : $\binom{20-N1}{3}$
Second round, I got 8 possibilities :
Turn 1, I played :
I can play 1 ball of value 2
I can play 2 ball of value 1
I can play 1 ball of value 1
I cant play
And same 4 possibilities if I dont played..
So every turn the calcul become harder..
If there is only 7 turn, it resul in :
2 * 4 * 7 * 12 * 19 * 30 * 45 = 17236800 possibilities to calculate
Is there a better way to do it ?
I tried to use the Law of total probability to find a series that I could sum, but I didnt suceeded in it.
( I also did a simulation before that, but I want the real result and not the result $\pm \epsilon$
Tell me if you need more information.
Thanks for any and all help.