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February 21st, 2010, 03:13 PM   #1
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5 Blue, 10 Red, 15 green balls. Odds I will draw 5 blue 1st?

Suppose I have 30 balls: 5 are blue, 10 are red, and 15 are green. I will draw balls until I have reached 5 of the same color, so there is no specific number of balls I am drawing (I am NOT just drawing 5, unless it is the case that my first five are all the same color). How likely is it that blue will be my winner (I draw 5 blue balls before I draw 5 reds or 5 greens). How likely is it that red will be my winner?

I tried figuring out the total number of ways I can arrange 5 B's, 10 R's and 15 G's. But then I realized that if I used the Pigeon Hole Principle I can see that after I've drawn 13 balls someone will have won, so I don't need to look at all the combinations. Then, I used 5 C 13 to see if I had my 5 of the winning color, how many ways they could be disbursed among the 13 selections. But it will only be the winning color if the other 8 only consist of 4 of each of the other two colors. I feel like I'm on the right track, but I don't know how to combine all of this together. Could someone help me figure out how to do this?

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February 21st, 2010, 03:24 PM   #2
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Re: 5 Blue, 10 Red, 15 green balls. Odds I will draw 5 blue 1st?

Let n loop from 5 to 13.

For each n, find the number of ways to choose exactly 5 blue balls in the first n draws, including one as the nth draw. Then subtract from this the number of ways to choose n balls with exactly 5 blue balls in the first n draws, including one as the nth draw, and also including at least 5 of some other color.

Add all these together.
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