[Buzz]

[ QUOTE ]
If you can please provide calculations for the numerator of those last three percentages, I'd much appreciate it. I still can't make my numbers work. Thanks.

[/ QUOTE ]

Domester - Numerator for the last three percentages? Sure. I didn’t save my scratchwork after I computed and posted. Not sure I’ll do it the same way I did last time. But in any event, what you do is make a chart showing the various possibilities.

You hold A24Q. The low cards in the stub are
3 aces, 3 deuces, 3 fours,
4 treys, 4 fives, 4 sixes, 4 sevens and 4 eights.
The other 19 cards in the stub are all high cards.

After taking out (mentally) of the deck your four card hand there are three cards left in the stub for each of the three low ranks in your hand, and four cards left in the stub for each of the five low ranks not in your hand. There are also nineteen high cards left in the stub.
-------
• 1 rank of low cards on flop. In this case, there must be either one card, two cards, or three cards of a single low rank. The other flop cards must be high cards.
For a single ace, deuce, or four plus two high cards: 3*19*18/2
For a pair of aces, deuces, or fours plus one high card: 3*19
For trip aces, deuces, or fours: 1

For a single trey, five, six, seven or eight plus two high cards: 4*19*18/2
For a pair of treys, fives, sixes, sevens or eights plus one high card: 6*19
For trip treys, fives, sixes, sevens or eights: 4*1

Computing:
3*(513+57+1) = 1713 ways to have a flop with a single rank of aces, deuces, or treys.
5(684+114+4) = 4010 way to have a flop with a single rank of treys, fives, sixes, sevens or eights.
Then 1713+4010 = 5723.

What I just did seems totally strange to me. I must have thought of the solution differently when I answered your prefious post. But in any event, the numerator I’m getting is the same.

Instead of my spending the rest of the night trying to figure out how to make the above response clearer to you, ask about what you don’t understand and maybe I (or somebody else) can explain it to you. Basically, I’m listing the number of possible different combinations that satisfy the requirement of exactly one low rank on the flop, and I’m doing that for each of the different low ranks.
-------
• 2 ranks of low cards on flop. In this case, there must be either a single card of each of two low ranks plus a high card, or there must be a pair of one low rank and a single card of another low rank.

There are two groups of low ranks.

*first group: 3 aces, 3 deuces, 3 fours,
*second group: 4 treys, 4 fives, 4 sixes, 4 sevens and 4 eights.
*The other 19 cards in the stub are all high cards.

Here are all the possibilities:
(d.r. means different rank)
(s.c. means single card)
first group pair + d.r. first group s.c.,
first group pair + second group s.c.,
second group pair + first group s.c.,
second group pair + d.r. second group s.c.,
high card + first group s.c. + d.r. first group s.c.
high card + first group s.c. + second group s.c.
high card + second group s.c. + d.r. second group s.c.

That’s a chart, of sorts, a seven line chart. Honestly it looks contorted to me. last time I did it, as I recall, it was neater. Oh well.
Now I’ll write the number of possibilities for each line. Here goes:
3*3*6 = 54
3*3*20 = 180
5*6*9 = 270
5*6*16 = 480
19*9*6/2 = 513
19*9*20 = 3420
19*20*16/2 = 3040

Then 54+180+270+480+513+3420+3040 = 7957.

That sure looks contorted to me, but lo and behold, it’s the same answer I got before. You can see it’s a bitch doing these. Lots of places to go wrong.
-------
• 3 ranks of low cards on flop.
O.K. here comes the last one. This is actually going to be pretty easy. All three different ranks from group one or group two, then two ranks from one group and one rank from the other.
This one’s sort of a four line chart.
I’ll just write the numbers that lead to an answer without an explanation. Ask if you don’t understand and I’ll try to explain.

20*16*12/6
9*6*3/6
9*20*16/2
20*9*6/2
Computing, 640+27+1440+540 = 2647.

Anyone wants to explain more clearly, be my guest.

Buzz