My actual problem is way more complex than the example I'll give, but I'm not asking for a solution as much as I'm asking if it's plausible I can solve it on my own with only high school algebra.
Say I blindly draw 5 slips of paper from a hat, each marked with a digit from 0 to 9. I magically know that the first will be even, the second will also be even but smaller than 5 (so 0, 2, or 4), the third will be odd, the fourth will be any number, and the fifth will be any number smaller than the fourth.
I need to assign probabilities per slip, for each number it can possibly be. And I need to update these probabilities each time a number on a slip is revealed.
That's the sort of problem I have, and I need to come up with a generalizable formula that I can use to just plug-in the known slips to get the probabilities.
I am hoping the math just involves high school stuff, am I correct or do I need to get a mathematician to do this for me?
I hope I'm correct that this problem is called combinatorics.
The usual formulas for combinatorics are based on symmetries of the problem. All the numbers or all the slips are equivalent or some such. Your rules are complicated enough so that a hand calculation will be too difficult.
There are only $\frac {10!}{5!}=30240$ possible draws of five slips without the restrictions. You can have a program go through all of them and find which draws are acceptable. I would guess only about $3\%$ or $1000$ if them are possible. You can then count up how many have a given number on the first slip, which will be the probability you seek. Then as slips are revealed, remove the draws that are now impossible and compute the chance for each remaining slip.