I’ve been trying to learn probability and other math from books and from online tutorials. It’s a little hard to ask questions because I don’t understand the topic yet. This is my attempt…

Let’s say that there are four cards: *w,x,y,z* but only the exact combination, *w,x,y,z* is winning. (*w,x,z,y* **;** *x,w,z,y* etc. are losing combinations.)

Using the factorial function, (4! = 24) for the total number of possible combinations and (0! = 1) for the number of ‘winning’ combinations, the odds of getting the ‘winning’ combination is 1 : 24.

Now, if one need only get three of the four to be ‘right,’ it would seem the odds would be better. However, the odds are the same. (0! = 1 = 1!) The combinations, *w,x,y,** and * *,x,y,z* are logically equivalent to * w,x,y,z*.

The real complication for me comes if one need only get *two* or fewer cards right. If, by two right you mean: * w,x,*,**, then (2! = 2) there are two winning combinations. Likewise, if one need only get *one* card right then (3! = 6) the odds are 6 : 24. (FWIW, I had to verify all of this for myself with fingers and toes math.)

But, if by two you mean *any* two: (*w,x,*,* **;** *,x,y,* **;*** or **, ,y,z ) or if by one * you mean * any * one: (w,,,* ; ,x,,*** ;**

*,*,y,* * or **,

*,*,z * then the logical equivalents throw a wrench into the factorial machine.

*Any* two gives you 4 : 24. (**,x,y,* * --> 2! = 2 but one of those combinations was already counted in the equation: *w,x,*,* * --> 2! = 2. (i.e.: *w,x,y,z ) Since the same is true for **,,y,z*, the total is (2! + (2! - 1) + (2! - 1)) = 4. ( I’m working on the assumption that while I might have the right answer, I may have brought in the factorials where they don’t belong.)

My question is, what is the ‘generic’ or theoretical equation used to give you the ‘winning’ combinations in the case of *any* one or two? What would really help would be the “a^2 + b^2 = c^2” answer… or better yet, the “pythagorean theorem” answer, not 9 + 16 = 25.

Explaining that you just finished teaching your friend, Joe that the answers are four and fifteen respectively and that your friend, Joe is your 12 year old, paralyzed, lobotomized cocker spaniel who can only communicate by blinking his left eye will not help.

Anything I could use to search wolfram or similar sites to find instruction on this type of problem would really help!