Yes, there's a limited amount of space.

If you hold T9o, then the chance of flopping a straight is 1.31% and the chance of flopping >=quads is 0.01%.

If you hold T9s, then this chance of flopping a straight will remain 1.31%, however in 0.02% of the cases this straight will also be a flush, in other words a straight flush. For that reason the "straight" statistic will be reduced by 0.02% to 1.29%. The >=quads statistic will increase from 0.01% to 0.03%.

But the straight is also a flush in 1.31*0.84/100=0.01% of the cases, so shouldn't it be >=quads 0.02%, straight 1.30%, flush 0.83%?

No, nothing has really changed compared to the T9o situation, except for some of the straights álso being a flush (making them a straight flush). This happens in 0.02% of the cases. Those 0.02% of the straights then move to >=quads.

You can completely ignore the flush statistic here.

Again, it's 0.01%, not 0.02% so I'm wondering where this error comes from.

Are you saying the flush % is not reduced by the straight-flushes % and only the straight % is? Then why the flush % is 0.82 when it should be 0.84?

No, it's 0.02%.

Flopping a straight flush
There's 19600 different ways of dealing a flop with T9s out of the deck (50*49*48/3!=19600).
There's 4 ways of dealing a straight flush with T9s (KQJT9, QJT98, JT987 and T9876).
And 4/19600*100%=0.02%.

Flopping a flush
After removing T9s from the deck there's still 11 remaining cards of that suit.
There's 11*10*9/3!=165 ways of drawing 3 of this suit for the flop, 4 of which are a straight flush.
So, there's 161 ways of drawing a flop that gives T9s a flush.
So there's 161/19600=0.82% chance of flopping a flush

Flopping a straight
Let's look at the chance of flopping a KQJT9 straight.
There's 4 K's in the deck, 4 Q's and 4 J's.
There's 4*4*4 ways of drawing a KQJ flop, with a total of 6 permutations (KQJ, KJQ, QKJ,QJK,JKQ,JQK), making for a total of 4*4*4*6 ways of flopping a straight with KQJ.
There's 4 ways for T9 to flop a straight (KQJT9, QJT98, JT987, T9876), so that makes for a total of 4*4*4*6*4 ways of flopping a straight.
You still need to divide by 3! though to compensate for overlaps between the 6 permutations, so there's 4*4*4*6*4/3!=256 ways of flopping a straight, 4 of which are straight flushes.
So that's 252 ways of flopping a straight.
This makes for a percentage of 252/19600*100%=1.29%.


Straight flushes are not included in the flush statistic.
After all, they're not flushes

Ah, I'm an idiot, I totally missed that there are 4 straight flushes possible, not two.

Scylla, thanks for the detailed explanation! It all makes sense if we just count combos.
What I'm still struggling with though is if P(A AND B)=P(A)*P(B), then why P(straight flush)
is not P(straight)*P(flush) or 1.31*0.84/100=0.01%???

One factor is that if P(straight) then the chance P(flush) actually increases a bit, since if there's a straight there's guaranteed to be no pairs (or 3 of a kind) on the board. So you're fishing the flushes out of a smaller subset of boards, thus increasing their chances. However, this will only increase the chance to flopping a flush to about 1%, which is not enough.

A second factor is that if there's a straight on the board, there's also a guarantee that there's no T or 9 on the board. Since 2-8,J-A each contain a heart and T and 9 don't, this will again increase the chance of each of the cards being of the same+correct suit. In fact, the chance of drawing the correct suit out of each card is 1/4, so the chance of now flopping a flush becomes (1/4)^3)=1.56%.

And if we nów apply the same formula then 1.31%*1.56%=0.02%.

OK, so I blew it once again
The correct formula is, of course, P(A AND B)=P(A)*P(B|A), where we're looking for P(B|A) which is the probability of flopping a flush given that we've flopped a straight.
So in essence we have 256 boards where we've flopped a straight and out of those 4 are flushes, which makes P(B|A)=4/256=1.56%. Did I finally get it right?

Yes, I believe that is correct.
After all, the chance of flopping a straight is 1.29% and 1.29%*1.56%=0.02%, which is indeed the number we're looking for.

It's 1.31 in this context so it's 1.31*1.56=0.02%, you had this right two posts above.

Another question: how do we get P(OESD)=9.71% for T9o?

I get 10.45%.

We have OESD on the following boards:
QJx, where x is not K or 8 (4*4*40=640 boards)
J8x, where x is not Q or 7 (640 boards)
87x, where x is not J or 6 (640 boards)
KJ7 (4*4*4=64 boards)
Q86 (64 boards)

So it's 640*3+64*2=2048 boards, making P(0ESD)=2048/19600=10.45%

Yes, you are correct.
My bad.

The error is in the number of combinations in which QJx, J8x and 87x can be drawn. It's not 4*4*40, although that seems perfectly logical.

It didn't dawn on my what the problem was until I wrote it down (for QJx) category by category on a piece of paper:

Category 1:

QJA: 4*4*4=64
QJ7: 4*4*4=64
QJ6: 4*4*4=64
QJ5: 4*4*4=64
QJ4: 4*4*4=64
QJ3: 4*4*4=64
QJ2: 4*4*4=64

Total: 7x64 combos=448 combos

Category 2:

QJT: 4*4*3=48
QJ9: 4*4*3=48

Total: 2x48 combos=96 combos

Category 3:

QQJ: 4*3*4/2=24
QJJ: 4*3*4/2=24

Total: 2x24 combos=48 combos

Now, the trick is actually in the part where you need to divide 4*3*4 by 2 again. This is because, for example, QhQdJx is the same flop as QdQhJx.

So, that makes for a total of 448+96+48=592 combos.

And J8x and 87x have 592 combos as well.

So that makes for a total of 3*592+2*64=1904 combos.
And therefore a chance of 1904/19600*100%=9.71%.

That makes perfect sense, thanks, Scylla!

OK, a tougher problem, let's say we want to figure out the probability of T9s flopping an OESFD.

We'll need to calculate on how many boards we flop a FD for T9s, given that a board also gives us and OESD. Let's say we have Th9h.

Again, T9 flops an OESD on the following boards:
QJx, where x is not K or 8
J8x, where x is not Q or 7
87x, where x is not J or 6
KJ7
Q86

Let's deal with a simpler one first - KJ7:
We can make 9 FDs on this board (Kh Jh and 3 non-heart 7s, same with Kh 7h J and Jh 7h K). That makes 18 FDs on KJ7 and Q86.

A tougher one now: QJx
QJ [A, 7-2]: 9 * 7 = 63 FDs
QJ [T-9]: 6 FDs (only Qh Jh and 6 non-heart Ts and 9s are possible since we have Th9h)
QQJ: 3 FDs (Qh Jh and 3 non-heart queens)
QJJ: 3 FDs (Qh Jh and 3 non-heart jacks)
That makes 75 FDs on this board texture and 225 FDs total for QJx, J8x, 87x combined.

In total we have 225+18=243 FDs, so P(OESFD)=243/19600=1.24%, and that's what Flopzilla gives me.

What I don't understand though is why it shows 9.60% for P(OESD). You seem to keep P(FD) fixed at 10.9%, in that case I'd expect to see P(OESD) as 9.71% (as you calculated above) - 1.24% = 8.47%, making P(FD OR OESD) 10.94%+8.47%=19.41%, but Flopzilla gives me 19.3%. So it looks like you do substract P(OESFD) out of the sum of P(FD) and P(OESD) which would be another way of calculating P(FD OR OESD), but for some reason you reduce P(OESD) by 0.11%.

A minor question: why is P(FD) rounded to 1 decimal places but P(OESD) to 2?

The reduction of 9.71% to 9.60% is unrelated to the cases where there's a 2 flush board. The reduction is due to the fact that some of the OESD combos are monotone and give T9s a flush, thus nullifying the fact that they are also a draw to a straight.

All numbers consist out of 3 digits; it's an esthetic choice.
It just looks better.
I'm pretty sure I remember trying different variations, including showing all numbers with 2 digits behind the comma, but it looked "weird" that the numbers were different sizes.
3 digits was the variant that just looked best.

Should you want more digits for some reason, press Ctrl+T for text output.
The output in that field is shown in 4 digits (unless a number ends in a 0, such as 9.210 being shown as 9.21).

OK, let me do this one. We're looking for P(OESD AND flush)=P(OESD)*P(flush|OESD)

T9s flops an OESD on the following boards:
QJx, where x is not K or 8
J8x, where x is not Q or 7
87x, where x is not J or 6
KJ7
Q86

Let's see how many are monotone:
QJx, where x is not K or 8: 7 flushes (K, 8 are out and we can't have a flush with Qh Jh [Q-9])
J8x, where x is not Q or 7: 7 flushes
87x, where x is not J or 6: 7 flushes
KJ7: 1 flush
Q86: 1 flush

So we have 23 flushes out of 1904 boards, making P(flush|OESD)=23/1904=1.21%
Then P(OESD AND flush)=9.71%*1.21%=0.12%

Rounding error?

QJx, J8x and 87x contain 75 flushes.
KJ7 contains 9.
Q86 also contains 9.

So that's 3x75+2x9=243 flushes.

Therefore P(flush|OESD)=243/1904*100%=12.76%.

And P(OESD)*P(flush|OESD)=9.71%*12.76%=1.24%.

Scylla, you've counted flush draws, not flushes
Q86 does not contain 9 flushes, there's only one, Qh8h6h.

I asumed you were looking for flushdraws.
After all, you can't have an OESD ánd a flush.
If you already have a flush, there's no straight to draw to that would improve your hand.
Unless you're looking for a draw to a straight flush?

You said that P(OESD) reduction is related to some OESD boards being monotone so I've calculated the probability of that.

Ah, right.
I misunderstood what you meant.
Yes, your math seems fine.
There's indeed 3x7+2x1=23 OEDS+flush combos out there.
If we deduct these from the original 1904 OESD combos we had for T9o then we'll be left with 1881 OESD+fd combos, resulting in a percentage of 1881/19600*100%=9.60%.

The one hundreth of a percent difference you get can be perfectly explained by the fact that you're working with rounded numbers for P(flush|OESD) and P(OESD).


PS:
In fact, given that P(OESD AND flush)=P(OESD)*P(flush|OESD), and P(OESD)=1904/19600 and P(flush|OESD)=23/1904, that makes:
P(OESD)*P(flush|OESD)=1904/19600*23/1904=23/19600.
And P(OESD)-P(flush|OESD)=1904/19600-23/19600=(1904-23)/19600=1881/19600=9.60%.
So, yes, rounding error, due to working with rounded numbers.

Thanks, Scylla! I was doing some research double-checking results with Flopzilla, and I have a complete match, which shows yet again how awesome this software is!

Help! Flopzilla will not run. I tried to open it today and out of the blue it is saying that it can't verify my license. It is saying that it is being blocked by a firewall but I don't even have a firewall on my computer. This just occurred completely out of the blue. Please tell me how to fix this!

If you're not using a firewall, then you're automatically using the Windows Firewall, which can indeed cause all sorts of weird issues. To solve this, please set an exception for all of Flopzilla's .exe files.

If you're having trouble doing this, or need more detailed instructions, please send an e-mail to support.

It says windows firewall is off under domain profile, private profile and public profile. This is all greek to me. How do I make sure the firewall is not blocking flopzilla or CREV?