Recursive Probability?

[_TKO_]

Let's say the situation is this:
There is a bag containing 10 marbles. There are 3 red marbles, 3 green marbles, 3 blue marbles, and 1 "special" marble. You get to pick two marbles out of the bag. If you draw the special marble, then two things happen:
1. You must put the special marble back in the bag.
2. You get to pick two more marbles, regardless of how many picks you have already had (ie: If you draw the special marble on your first pick, you get to pick two more marbles, and if you draw the special marble on your second pick, you still only get to pick two more marbles).

What is the probability that you will end up with two red marbles, and how do you calculate this?

[fiskebent]

I'd calculate the odds that you end up 2 to 10 marbles. Then for each category calculate the odds that 2 are red.

The odds of missing the special on your first draw is 80% (9/10)*(8/9). The odds of those two being red are 3/9*2/8 or 8.333%

The odds of getting the special on the first draw and missing it on the second is 0.2 * (8/9) * (7/8) or 15.555%
When you draw 3 non-special marbles, the odds of two of them being red are 21.4%

So the odds of ending up with 2 or 3 marbles and getting 2 reds are
0.8 * 0.083 + 0.155 * 0.214

You should be able to extend the method to include the rest of the possibilities.

[Copernicus]

If thats the total problem then enumeration cant be that hard. Note that its irrelevant what the other colors are, they could be 6 blacks and 1 special.

If you need to generalize it to 3n+1 non-reds then BruceZ will come up with something Im sure!

[Moose747]

I isolated 14 possible unique game states and, using backwards induction, solved the probability of winning the game at each state. Each state is enumerated by the number of red marbles followed the number of non-red, non-special marbles. The initial game state is (3,6). The game state (2,3) occurs after when two red marbles and three blue or green marbles remain, along with the special marble.
Working backwards, I find the following probabilities of winning at each of the 14 possible game states:
(2,0)=1
(2,1)=1
(3,0)=1
(2,2)=0.9
(3,1)=0.69
(2,3)=0.738
(3,2)=0.436
(2,4)=0.607
(3,3)=0.292
(2,5)=0.516
(3,4)=0.212
(2,6)=0.452
(3,5)=0.164
(3,6)=0.119

[Copernicus]

I dont follow (3,1), and is this a complete enumeration? Dont you need to know the number of balls remaining to be picked?

Eg (3,1) = 1 if there are 2 remaining to be picked because you can never pick two non-reds/non-special in a row, but (3,1) = .2 if there is only one pick left.

[Moose747]

(3,1) is the game state where there are 3 red marbles, the special marble and one blue or green marble remaining. There's a 30% chance you pick a red and the non-red, which ends the game before the player has acquired 2 red marbles. The extra 1% is a mistake; I'll post corrections in a minute.

[Moose747]

Think I got it now.

(2,0)=1
(2,1)=1
(3,0)=1
(2,2)=0.9
(3,1)=0.7
(2,3)=0.74
(3,2)=0.46
(2,4)=0.611
(3,3)=0.314
(2,5)=0.520
(3,4)=0.228
(2,6)=0.455
(3,5)=0.175
(3,6)=0.120

[fiskebent]

I get 11.72%

[AaronBrown]

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.

[fiskebent]

[ QUOTE ]
2 0.888889 0.083333

[/ QUOTE ]
Isn't the chance that you miss the special on the first draw 80%?

9/10 * 8/9 = 0.8