Monday, 23 March 2009

An Improved Solution to the Probability Problem

My last blog gave a labored explanation of why given R red balls and W white balls in a bowl, if they are removed one at a time randomly until one color has been totally removed, what is the probability that White will be exhausted first.

Joshua Zucker gave much more elegant expalantion for the reason the probability of white being exhausted first is red/(red + white)

His solution, in my words.. forget everything except the last ball in the bowl... whatever color it is is Not the color that was exhausted. So All we have to do to know the probability that White is exhausted, is find the probability that the last ball drawn if we emptied the bowl would be red. But if we look at all the possible orders of drawing all the balls, 4/7 of the first balls drawn are red, 4/7 of the second balls drawn are red..and the same continues all the way to the end; 4/7 of the final balls drawn will be red, meaning that in 4/7 of the trials, the white balls were exhausted first.

Now I'm thinking about the case of three colors, red, white and blue. I haven't started, so if you have an easy explanation, lay it on me.... otherwise, more later.

-------------- edited afterward...
ask and you shall receive... Joshua does what he does best, thinking and explaining his thoughts on the fly... see the comment and learn ....

1 comment:

Joshua Zucker said...

Maybe this doesn't work for the RWB case, but it seems plausible to me. Thinking out loud here, could be totally wrong...

Let's find the probability that white is exhausted first, in the case with 3 red, 5 white, and 9 blue.

Well, first of all, we know that white is exhausted LAST in 5/17 cases, so we can forget about those.

We also know red is exhausted LAST in 3/17 of the cases. Now, we can forget about the other two red balls in those cases, so blue is exhausted second-to-last in 9/14 of those, and white in 5/14 of those. So, we get 9/14 * 3/17 chance of white exhausted first (and red last).

Similarly, we get a 9/17 * 3/8 chance of white exhausted first and blue last.

So the total is 3/17 * 9/14 + 9/17 * 3/8 = about 26% for white.

Hm, the numerator is going to always be the product of the other two numbers, cool! My answer for white exhausted first is
(RB)/(R+W+B) * (1/(R+W) + 1/(W+B)) I guess.

Now if I had the energy, I'd give it a couple quick checks:
symmetric when R and B are switched, good.
Goes to 1 as W approaches 0, yup!

The three probabilities add up to 1 (too lazy to check).

And maybe a simulation with Fathom or something .... uh oh, that seems to be showing about 31%, not about 26%.

And 14% blue, and 55% red.

Let's try my formula a little more:
For red I get 9/12 * 5/17 + 5/8 * 9/17, hm, 55%, that matches my formula.

For blue I get 3/12 * 5/17 + 5/14 * 3/17 = about 14%.

Wait a minute, that doesn't add up to 1.

For white I should have 3/8 * 9/17 + 9/14 * 3/17, 31%. How did I get 26% before? Oh well. At least now it all seems consistent and I feel like my formula is plausible.