Originally Posted by CardDead711
I hate math too, unless its poker math, so...
50/50


That's a good answer, and arguably it's correct. As we study math and introductory combinatorics and optimization, we learn about discrete events. Once you've flipped one card, you use that information to recalculate your odds of predicting the result of another flip. We know that there are 4 aces of 52 cards, so if you get one ace in the hole, then there are 3 aces left of 50 cards. So the odds of the first flop card being an ace are 3 in 50. If it's not an ace, the odds of the second flop card are slightly improved to 3 in 49, and if it's not an ace then 3 in 48 for the third flop card. This is poker math, how to quickly figure out the approximate odds of getting your pair of aces on the flop.
As I recall, it was Bertrand Russell who in the early 20th century refined the relationships between philosophy/mathematics/logic. He didn't invent it all, but Russell was remarkable both for being not insane and for being a gifted speaker, teacher, and writer. Russell used my example of 3 pairs of cards (he used coins I think) to demonstrate that often the events which we think are discrete, or mutually exclusive, are not so.
According to Russell, you're hunting for the pair of aces so if you flip an ace you've flipped one of the three. For two of the three aces on the board, the other card of its pair is an ace, and one of the aces has a king with it. Russell would say that the flipping of the cards isn't mutually exclusive in the information you've gleaned. Rather than simply eliminating the pair of Kings by flipping an ace and thus making it a 50/50 that you picked the pair of aces, you're at 66.6% of flipping a second ace once you've flipped the first.
Thinking about it in this way leads different approaches of solving a problem, to Heisenberg's uncertainty principle and Shroedinger's cat, essentially saying that the unflipped cards literally have no value until they are flipped and thus the probability function (which is a wave) collapses and a single result gels. In quantum physics, it's impossible to determine the precise location of an electron and trying to do so affects the path of it; all you can do is create probabilities of where the electron might be along its path depending on your knowledge of its properties.
So back to OP's question. You have 52 cards laid out and you want to pick a spade on your first pick, so one in four. Does selecting four of the 52 cards into a smaller pile and then flipping one of those four change the odds? Does eliminating cards one at a time from the 52 until there's only one left, does that change the odds of the last remaining card being a spade? Does it matter if you look at each of the 51 cards you eliminate as you do so? Most people would say "of course not!" and they are probably right. But most people would also have said that flipping an Ace leads to the odds of a second ace flip being 50%.
So it's no wonder that OP is having a hard time sorting this out, and even with the help of a PhD in statistics the answers aren't clear. (However, I suspect that the PhD is an exaggeration because someone that smart could answer the questions easily I'm sure). I hope that OP comes back so we can discuss it.