Here's a "French Canadian" english
way to translate this:
First, you need to know that (13 over 1) actually means COMBINATIONS of 1 in 13 and that you can solve it like this:
13!/(13-1)!1!
=13*12*11*10*9*8*7*6*5*4*3*2*1 / 12*11*10*9*8*7*6*5*4*3*2*1*1 =13*........................../..........................*1
=13/1 =13
A better example would be: How many DIFFERENT pairs can you build with Aces?
Heart(h), Spades(s), Diamonds(d) and Clubs(s)
EASY! You can solve (4 over 2), or 2 in 4 now...
4!/(4-2)!2! =
=(4x3x2x1)/(2x1x2x1)
=4x3..../....2*1
=12/2
=6
{HS,HC,HD,SC,SC,DC}
Now, the equation...
First,
out of 13 different values, let's choose one, Ace. (Remember, 1 in 13 = 13)
Second,
out of 4 different Aces, pick 2. (2 in 4 is 6.)
Now, we know that having any PAIR is 13*6 = 78 different Pairs. Thing is, there could be any other three cards with this pair...
Third,
you want your next three card to be: one of 4 suits and they CAN be suited, so 1 in 4 suited POWER 3. This renders the possible suit outcomes for the three remaining cards. one in four is 4... 4 power 3 is 64. there are 64 outcomes of suits possible write them down { HHH, HHC, HHS, HHD, ... DDD}.
Fourth,
but not least, is to decide what can be the value of the three other cards. the last thing you need to consider is that you DO NOT want it to be an Ace NOR ALIKE TOGETHER... you would end up with two pairs, a full or three or four of a kind.
So to solve this it's simple, you need 3 different values out of the 12 remaining (Ace is not possible).3 in 12
12!/(12-3)!3!=
=12*11*10*9*8*7*6*5*4*3*2*1 / 9*8*7*6*5*4*3*2*1 3*2*1
=12*11*10................... / ................. 3*2*1
=1320 / 6
=220
FINALLY
We solved 13 * 6 * 64 * 220 and remember, you can place these numbers in any order... so 13*6*220*64 is just as good. And what's 64's cubic root ? right. It's 4...
13 * 6 * 220 * 4power3
answer: 1098240 hands of 2598960 are "single pair" or approx. 42.2569 percent
Hope that helps. Math, not even once! ;)