Probability Problem (non-poker)

[DrSavage]

You are wrong.
The probability of drawing 5 blue balls from 9 blue and 1 red ball is (9/10)*(8/9)*(7/8)*(6/7)*(5/6) = 5/10=0.5 exactly.
your (0.9)^5 applies to the case when balls would be put back into the box after each removal, which was not the case here.

[thylacine]

Don't ask

What is the smallest number of balls and how many of them blue should he have placed in the box to get that result?

I've just been talking to a number theorist friend and the interesting question is this:

What is the LARGEST number of balls and how many of them blue should he have placed in the box to get that result?

Is there a largest? Is there any solution other than the obvious one? Same question with any number 3 or more balls. (With the 2 ball version there are infinitely many solutions.)

[slider77]

Here's a simple way without a computer.

Since it asks for the minimum number of balls, the number of blue balls (B) would have to be just 1 less than the number of total balls (N). It can't be like 5000 total and 4852 are blue - even though something like that could work.

Prob(5 blue balls in a row) =
B/N * B-1/N-1 * B-2/N-2 * B-3/N-3 * B-4/N-4 = .50

Substitute B=N-1...do the cancelling....and you get:

(N-5)/N = .5

N=10
B=9

[Copernicus]

The original post asked for a solution without iteration. Guessing (or "assuming") IS iteration when you dont guess right the first time!

[thylacine]

They are using a method referred to by mathematicians by the technical term "cracking a walnut with a sledgehammer".

[Copernicus]

those are the applied mathematicians of course.