In fooling around with the probability of flopping top pair, by accident I came across an interesting relationship (perhaps only interesting to me).

Here it is posed as a problem, which I think is not very easy.

f(k) =C(50-4*k,3) - C(47-4*k,3), k=0,1,2, … ,10

Find a much simpler combinatorial function, g(k) such that

g(k) = f(k) - 1

First correct answer gets a BruZee!

You gave a very strong hint, I came up with:

g(k) = 6*(k−12)*(4*k−47)

EDIT: As you asked for combinatorial expression the above is from: C(3,1)*C(47- 4*k) +C(3,2)*C(47-4*k, 1)

Sorry missed a 2 in the formula: C(3,1)*C(47- 4*k, 2) +C(3,2)*C(47-4*k, 1)

You get a BruZee. Maybe it wasn't as hard as I thought. Your result is not what I got, but obviously we can convert one to the other with some combinatorial algebra.

My result is

3*C(4*(12-k),2)

There's no mathematical "magic" going on here.

Anyone with a little bit more than a passing familiarity with Pascal's triangle knows that the 1st diagonal is the constant 1, the 2nd is linear, the 3rd is quadratic, the fourth is cubic, etc. And that the differences of any 2 separated by some constant along any of theses diagonals will be of a degree of 1 less. So for the 1st diagonal it is just zero, the 2nd is some constant, the 3rd is linear, the fourth is a quadratic, etc.

So for the "puzzle" you posted, you should expect a quadratic.

With your experience, you should have realized this instantly and not bothered to post. Now go put on your dunce cap and sit in the corner for an hour. Go on! I'm not kidding! Go!

You don't need algebra. Even David's cockatoo understands that combinations obey the recurrence

C(n,k) = C(n-1,k-1) + C(n-1,k).

That's how Pascal's triangle is constructed. Each entry is the sum of the 2 above it.

That's one smart cockatoo!