A problem over morning coffee

[jay_shark]

OK here is a fun little exercise to introduce to probability students .

A bag contains A red marbles and B black marbles . Marbles are withdrawn at random until only balls of the same color remain . What is the probability that only red marbles are left ?

[AaronBrown]

This only uses some probability if you do it the hard way. It's a brainteaser, with a bit of insight you can see the answer immediately, otherwise you could slog through some combinatorics.

It is a clever problem, I have used it in job interviews.

[DiamondDog]

I'll have a go.

For there to be only red marbles remaining, the first B marbles picked all have to be black.

P(first one picked is black) = B/(A+B)

assuming that happens, P(second one picked is black) = (B-1)/((A+B-1)

etc

Probability that first B marbles picked are all black
= B/(A+B) x (B-1)/(A+B-1) x (B-2)/(A+B-2) X ….. x 1/(A+1)


= B!/[(A+B)!/(A!)]

= A!B!/[(A+B)!]

Is that right?
Is there a quicker way?

[AaronBrown]

That is the probability that the first B are all black, but that is not required for the problem. We only require that B of the first B+n be black, for any n < A. You have solved for n = 0.

You can do this problem without any combinatorics at all, in a simple one-line argument.

[Dale Dough]

EDIT this is too easy, I don't have to wait for someone to tell me I was right. So I won't screw it up for the rest.

[DiamondDog]

[ QUOTE ]
That is the probability that the first B are all black, but that is not required for the problem. We only require that B of the first B+n be black, for any n < A. You have solved for n = 0.

[/ QUOTE ]
You're right. I can see that now.

[ QUOTE ]
You can do this problem without any combinatorics at all, in a simple one-line argument.

[/ QUOTE ]
I'm going to have to take your word for that. [img]/images/graemlins/crazy.gif[/img]

[Silent A]

Think of it this way ...

Let's say you keep picking until no marbles are left.

What's the probability that the last one picked will be red?

Then think about how this relates to the OP.

[neverforgetlol]

Ah I see, clever problem.

[rjp]

I must be missing the obvious. What's the answer?

[jay_shark]

The answer is a/(a+b) .


It is equivalent to the probability that the last ball is red which is obviously a/(a+b) .

Think about it for a while . Once only red marbles remain , you can continue discarding red marbles until the last one is left since this won't change the probability . In other words , all that matters is the last ball .