A mathematical paradox

Today was my last day of lectures for the quarter. My how time flies! And today's was a doozy: a 3 hour barn-burner Physiology wrap-up on the Respiratory system that turned my grey matter into a puddle of primordial goo. I know it had something to do with ventilation, perfusion, PAO2, PaO2, the A-a gradient, PACO2, PICO2, PaCO2, and laws by fellows such as Boyle, Hentry, Dalton, LaPlace, Fick, and Poisseuille but after that it gets kind of hazy ... though I do have all weekend to figure it out. So far, I've taken 2 finals so far -- Clinical Medicine and Epidemeology. They mercifully split our finals up so they wouldn't all be on finals week. That being said, I have a final every day of the week (Mon-Fri) next week and my meta-analysis paper the following Monday so it's going to be a long haul regardless.

Thought I'd shift gears a bit and talk about Math today (is this the nerdiest blog you've ever read or what?) This is something I picked up from Dee -- she is taking a graduate level statistics course, and so the bulk of the material in her class is beyond the grasp of my pre-Calc intellect. After all, I'm just a simple caveman. However, there is one thing I do know! And that is the strange case of Simpson's Paradox, which I will now present to you in a hopefully coherent and understandable manner.

Imagine 2 colored urns, one black and one white. Now imagine there being a mixture of red and green balls in each urn. Finally, imagine a game where you must select either the black or white urn, and then draw a ball at random from that urn. If you happen to pick a red ball, you win $1,000,000! Sounds simple enough so far, right?

Okay, in the 1st round, the black urn contains 5 Red balls and 6 Green balls. The white urn contains 3 Red balls, and 4 Green balls. Which urn do you pick to draw from, the black or the white?

Let's look at the probabilities of picking a red ball:
black urn - probability of red ball = 5/11 = .455
white urn - probability of red ball = 3/7 = .429

So, as you can see, you would likely pick the black urn, as the percentage of getting a red ball is higher.

Now let's consider 2 more urns. In this case, the black urn contains 6 Red balls and 3 Green balls. The white urn contains 9 Red balls and 5 Green balls. Again, which urn do you pick to draw from?

Here are the probabilities this time of picking a red ball:
black urn - probability of red ball = 6/9 = .667
white urn - probability of red ball = 9/14 = .643

Once again, it should be apparent that the black urn would be the one to pick, as it has the better odds of yielding a red ball.

Now, we will consider a third choice. Imagine if we combined the balls from the 2 black urns used in the previous examples. We also combine the balls from the 2 white urns. Now which urn would you pick to draw from? Since in each example, the black urn had the higher probabability of a red ball pick, you would no doubt pick the black urn once again.

(Right about now is when I lean in and whisper -- this is where the story gets w-e-i-r-d!)

Well, lets go ahead and check the math on this, just to make sure we chose right. Looking at the combined urns, in the black urn there are now 11 Red balls and 9 Green balls. In the white urn, there are now 12 Red balls and 9 Green ones.
black urn - probability of red ball = 11/20 = .55
white urn - probability of red ball = 12/21 = .571

What you talkin' bout Willis? I'm sure those of you who have made it this far are left scratching your heads a little. This one fooled me too. However, there is a rational and reasonable explanation for this, which can be found at this link: The Secret of Simpson's Paradox

Considering the quantum leap in racial equality made 2 nights ago in the U.S. presidential election, I will leave you with a fittling mosaic. (for full effect, look at each block close up and then back away from the screen -- not that you probably needed me to mention that, but that's why my friends call me Capt. Obvious)