Re: Recursive Probability?
I used fishkebent's first approach and got 0.102134.
The table below gives the possible number of marbles to end up with (assuming you stop if you get all 9 non-special and don't keep drawing forever), the probability of getting that number and the probability of having at least two reds if you draw that number.
2 0.888889 0.083333
3 0.097222 0.226190
4 0.011905 0.404762
5 0.001653 0.595238
6 0.000265 0.773810
7 0.000050 0.916667
8 0.000011 1.000000
9 0.000006 1.000000
The computation is easy. If you hold n balls in your hand, for n=0 to 9, the chance of drawing the special ball is 1/(10-n). For n=0 and n>8 we can ignore all the special draws. For n in between you either draw the special ball or some other ball and end. If you draw the special, I don't care how many times you do it in a row, you always get one more ball, then another chance to get the special or another ball.
Given the number of balls you end up with, the chance having at least two red balls is a simple combinatoric exercise.
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