Looking for help with odds on a few specific scenarios in Hold 'em.

What's up everybody? This is my first time posting anything, and I just signed up, so I hope I'm doing this right.
So I work in a poker room in Washington State, and for a promotion I'd like us to get going, I need to know the numbers on a few things:

What are the odds of flopping a flush in one specific suit (let's use clubs), with both holes cards being suited?
The odds of the flop being all clubs?
The odds of the whole board (flop-river) being all clubs?
The odds of hitting a flush in each suit, with both hole cards being suited, in a 8 hour session?

Thank you very much to anyone deciding to help!

Quote:

What are the odds of flopping a flush in one specific suit (let's use clubs), with both holes cards being suited?

If you already have suited hole cards (but before the flop), probability to flush on the flop is:
c(11,3) / c(50,3)
= 0.008418
= 1 in 118.79

If you have yet to receive hole cards (so before the deal), the probability to flush in a specific suit on the flop is:
c(13,5) / c(52,5)
= 0.0004952
= 1 in 2019.39

If the suit doesn’t matter, it would simply be four times the last probability.

Quote:

The odds of the flop being all clubs?

Without knowing any other cards, the probability of an all clubs flop is:
C(13,3) / c(52,3)
= 0.1294
= 1 in 77.27

If the suit doesn’t matter, again, it would just be four times this.

Quote:

The odds of the whole board (flop-river) being all clubs?

Without knowing any other cards, this would be the same as the 1 in 2019 probability I gave above.

If your hole cards are clubs and you want to know the probability the board will be all clubs:
C(11,5) / c(50,5)
= 0.0002181
= 1 in 4586.06

If you want to know before the deal, what’s the probability your hole cards and the entire community will be all clubs:
C(13,7) / c(52,7)
= 1.28E-05
= 1 in 77963

Quote:

The odds of hitting a flush in each suit, with both hole cards being suited, in a 8 hour session?

The probability to hit a flush in any suit, with the hole cards being of the flush suit :
[(11,3)*c(39,2) + c(11,4)*c(39,1) + c(11,5)] * [4*c(13,2)/c(52,2)]
= 19371 / 1286390
= 0.015058

Now we can find the number of flushes in n-number of hands and the probability there are at least one of each suit. The math is a little messy, but it’s the summation from i=4 to i=infinity of the following:
c(h,i)pⁱ(1-p)^(h-i)[1-4(3/4)ⁱ+6(1/2)ⁱ-4(1/4)ⁱ]

Where,
p = the probability to hit flush, 0.015058
h = number of hands played

So it depends on the number of hands you play in 8 hours. Here are the probabilities for a few different number of hands played:

h = 150; probability = 0.03408
h = 200; probability = 0.077653
h = 250; probability = 0.137716
h = 300; probability = 0.209405
h = 350; probability = 0.28736
h = 400; probability = 0.36686


All the above probabilities assumed any kind of flush (so Royal or Straight are included). Also, c(x,y) is the combination function where y-number of items are chosen from a pool of x-number items.