low cards in the flop (O8/B )

[domester]

Anybody ever calculate the probability of flopping 0,1,2 or 3 low cards (A-8)? I'm reading Cappelletti's O/8B book and he says that those probabilities, when you hold A24Q, are 5.6%, 33.1%, 46% and 15.3%. I agree with the first number (19 choose 3 = 5.6%) but can't figure on the last three. Buzz, Felicia or anybody else out there, can you help?

[Buzz]

[ QUOTE ]
Anybody ever calculate the probability of flopping 0,1,2 or 3 low cards (A-8)? I'm reading Cappelletti's O/8B book and he says that those probabilities, when you hold A24Q, are 5.6%, 33.1%, 46% and 15.3%.

[/ QUOTE ]

Domester - We don't exactly want the probability of flopping 0, 1, 2, or 3 low <font color="white">_</font>cards. Of more interest is the probability of flopping 0, 1, 2, or 3 low <font color="white">_</font>ranks

The answer to the number of low cards is very easy:
• 0 low cards on flop: P = 0.0560
• 1 low card on flop: P = 0.2867
• 2 low cards on flop: P = 0.4460
• 3 low cards on flop: P = 0.2113

But this is misleading because when there are two low cards on the flop, the two low cards could be a pair, together amounting to only one rank of low card.
Similarly when there are three low cards on the flop we could have a pair or trips.

The answer to the number of low ranks is more tedious:
• 0 different ranked low cards on flop: P = 969/17296 = 0.0560
• 1 rank of low cards on flop: P = 5723/17296 = 0.3309
• 2 different ranks of low cards on flop: P = 7957/17296 = 0.4600
• 3 different ranks of low cards on flop: P = 2647/17296 = 0.1530

Those are evidently the numbers Cappelletti is using.

But, since you’ve brought it up, aren’t we really interested in knowing when we have two or three low cards on the flop that are of benefit to us?

For example, there are some flops with three different ranks of low cards that don’t make a low for us when we hold A24Q. (eg. 2-4-8).
And there are other flops three different ranks of low cards that make a low, but not the nut low. (eg. 2-7-8).

Does he later somehow make a correction for this?

Buzz

[(my name it is) Sam Hall]

[ QUOTE ]
Anybody ever calculate the probability of flopping 0,1,2 or 3 low cards (A-8)?

[/ QUOTE ]

Cappelletti says "What percent of the time do three cards contain zero, one, two, or three different low cards?" "Different" is the key word here. Use that and you should get his numbers.

Sam

[(my name it is) Sam Hall]

[ QUOTE ]
But, since you’ve brought it up, aren’t we really interested in knowing when we have two or three low cards on the flop that are of benefit to us?

[/ QUOTE ]

He talks about a similar situation is the beginning of Chapter 4, although I'm still trying to reconcile all the numbers he throws around.

Sam

[domester]

Buzz,

He doesn't make a later correction for cards that can help. It just says three low cards. 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.

[(my name it is) Sam Hall]

[ QUOTE ]
But, since you’ve brought it up, aren’t we really interested in knowing when we have two or three low cards on the flop that are of benefit to us?

[/ QUOTE ]

Buzz, can you check my numbers?

Holding AQ42 as an example of any low-low-low-high hand:

The flop contains exactly two low cards that are not A, 2, or 4 AND are not themselves paired 17.6% of the time.

The flop contains exactly two low cards, one of which IS an A, 2, or 4, and one of which is not A, 2, or 4 9.9% of the time.

Sam

[domester]

Sam,

I guess my issue is the uncertain meaning of "different". Is it different from A24, or simply different ranks from each other? For example, for one low card, my math is as follows... there are 29 of them left in the deck and there are (29 choose 2 =) 171 ways to pick two high cards, so 29*171/17,296 = 28.7% chance of flopping one low card. If you exclude the As, 2s and 4s, the calc is 20*171/17,296 = 19.8% chance of flopping one low card. Neither one is close to Cappelletti's answer. Am I smoking the wacky weed on this one?

[gergery]

For one low card, there are 29 left in the deck. But once you pick that low card (let’s say it’s the 3c), the other low cards of that rank (3s,3d,3h) become high cards for purposes of this calculation. So odds of 1 low card would be C(29,1) * C(22,2). But that doesn’t take into account the slight adjustment you have to make for picking a low that is in your hand vs. not in your hand. And once you make that calculation, you need to adjust out since picking 3s,3c, T is the same as 3c, 3s, T. Alternatively, calculate C(29,1) * C(19,2) then add in times you choose the same ranked low card with the 2nd or 3rd selection.

[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.
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• 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.
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• 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.
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• 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

[Buzz]

[ QUOTE ]
Buzz, can you check my numbers?

[/ QUOTE ]

Sure, Sam.

[ QUOTE ]
Holding AQ42 as an example of any low-low-low-high hand:

The flop contains exactly two low cards that are not A, 2, or 4 AND are not themselves paired 17.6% of the time.

[/ QUOTE ]

(19*20*16/2)/(48*47*46/6) =
3040/17296 = 0.1758

Looks like we agree on the first one.

[ QUOTE ]
The flop contains exactly two low cards, one of which IS an A, 2, or 4, and one of which is not A, 2, or 4 9.9% of the time.

[/ QUOTE ]

But I get a different answer (19.77%) for the second one. (as shown below):

9*20*19/17296 = 3420/17296 = 0.1977

Buzz

[(my name it is) Sam Hall]

[ QUOTE ]
But I get a different answer (19.77%) for the second one. (as shown below):

9*20*19/17296 = 3420/17296 = 0.1977

[/ QUOTE ]

You are correct. The degeneracy is different since the high, unpaired low, and paired low are distinguishable and I didn't correct for that. Thanks.

6[(19/48)*(9/47)*(20/46)]=0.1977

OK, so the conclusion so far is this: AQ42 flops an uncounterfeited nut low draw 17.6% of the time, and a once-couterfeited low draw 19.8% of the time. In that 19.8% of the time in which the draw is once-counterfeited, 6.6% of the time it will be the 4, and the draw will still be to the nut low. The other 13.2% of the time it will pair the A or 2, making the draw to the second-best low. Given all of that, AQ42 flops a nut low draw 24.2% of the time and a second-nut low draw 13.2% of the time. Is this all correct?

Sam

[(my name it is) Sam Hall]

[ QUOTE ]
But, since you’ve brought it up, aren’t we really interested in knowing when we have two or three low cards on the flop that are of benefit to us?

[/ QUOTE ]

A second amateurish hack attempt at this...

3 low cards, not paired and not paring a hand card flop, making a protected low
1[(20/48)*(16/47)*(12/46)]=3.70%

3 low cards not themselves pair but pairing one hand card flop
3[(20/48)*(16/47)*(9/46)]=8.33%
1/3 of the time (2.78%) it pairs the 4, leaving you with an unprotected nut low. 2/3 of the time (5.56%) it is an unprotected second-nut low.

Three lows cards flop, pairing two hand cards {ex. 24T}
3[(20/48)*(9/47)*(6/46)]=3.12%

AQ42 flops three low split pair
1[9/48*6/47*3/36]=0.16%

This adds up to 15.31%. Buzz shows 15.30% for flopping three low ranks. The rest is probably my rounding error.

Sam

[domester]

Thanks Buzz,

Followed your steps and it all makes sense now. I guess the important question to answer, and my next step, is "what's the chance of having two way post-flop possibilities (nut low or nut low draw w/ a set, top two pair, straight, straight draw, flush or flush draw) with a given hand?"

[Buzz]

[ QUOTE ]
The degeneracy is different

[/ QUOTE ]

Sam - I don't know what that statement means. I don't know what "degeneracy" used in that context means.

[ QUOTE ]
6[(19/48)*(9/47)*(20/46)]=0.1977

[/ QUOTE ]

That's correct, and concise, but not exactly how I probably would have thought of it. (But I'm not suggesting that my path of reasoning is better than yours. Indeed, yours looks more direct).

[ QUOTE ]
OK, so the conclusion so far is this: AQ42 flops an uncounterfeited nut low draw 17.6% of the time, and a once-couterfeited low draw 19.8% of the time. In that 19.8% of the time in which the draw is once-counterfeited, 6.6% of the time it will be the 4, and the draw will still be to the nut low. The other 13.2% of the time it will pair the A or 2, making the draw to the second-best low. Given all of that, AQ42 flops a nut low draw 24.2% of the time and a second-nut low draw 13.2% of the time. Is this all correct?

[/ QUOTE ]

Somehow I have the feeling that you know more about this stuff than I do and that I'm like a country bumpkin struggling along with something he's pretty unsophisticated about, but O.K., I'll bite.

[ QUOTE ]
AQ42 flops an uncounterfeited nut low draw 17.6% of the time

[/ QUOTE ]

10*(2*4*6+4*4*19) = 3520

or 20*16*19/2 + 5*6*16 = 3520

3520 is the number of three-card combinations (not permutations) that would result in a flop where you had an uncounterfeited low draw, if holding A24Q. All of those are nut low draws.

The number of possible flops is C(48,3), or 17296.

3520/17296 = 0.2035. That's the probability of flopping an uncounterfeited low draw. Expressed as a percentage that would be 20.35%.

I had previously written:
(19*20*16/2)/(48*47*46/6) =
3040/17296 = 0.1758

As I look at it now, it seems I left paired flops out of that computation.

My error. Sorry.

It should have read:
(19*20*16/2)+(5*6*16)/(48*47*46/6) =
(3040+480)/17296 =
3520/17296 = 0.2035

Therefore AQ42 flops an uncounterfeited nut low draw 20.35% of the time. I think I've got it right this time.

[ QUOTE ]
and a once-couterfeited low draw 19.8% of the time.

[/ QUOTE ]

•One flop card would be an ace, deuce of four (any of nine cards from your perspective, holding A24Q),
•a second flop card would be a trey, five, six, seven or eight (any of 20 cards from your perspective, holding A24Q),
•and the third flop card would be a high card (any of 19 cards from your perspective, holding A24Q).

O.K., here goes:

(9*20*19)/17296 = 3420/17296 = 0.1977

So, yeah that one looks correct.

[ QUOTE ]
In that 19.8% of the time in which the draw is once-counterfeited, 6.6% of the time it will be the 4, and the draw will still be to the nut low. The other 13.2% of the time it will pair the A or 2, making the draw to the second-best low.

[/ QUOTE ]

Yes. Looks like 6.6 and 13.2 is the correct partitioning of 19.8 to get a 3 to 6 ratio (or a 1 to 2 ratio).

[ QUOTE ]
Given all of that, AQ42 flops a nut low draw 24.2% of the time and a second-nut low draw 13.2% of the time. Is this all correct?

[/ QUOTE ]

No. If the 0.1758 was correct, it would be. (17.58+6.6 = 24.2)

However, the 0.1758 is incorrect. (Sorry, my error).

Instead, 20.35+19.77/3 = 26.94

A24Q flops a nut low draw 26.94% of the time and a second-nut low draw 13.18% of the time.

It also directly flops (no draw) the nut low 6.48% and the second nut low 5.55% of the time.

It also flops various other favorable hands and draws. (I'm not sure where you're going with this).

I usually check my numbers backwards an forwards, as many ways as I can check them. I didn't this time because I got the same result as Mike Cappelletti on my first try. Also, my numnbers added up to 100%. Rats! What a can of worms! There must be something else wrong with that post too, since the numbers added up to 100% using the (erroneous) 17.58%. I'll go back and look at that post when I get back from the game tonight. I'll probably post a correction.

I guess Mike and I both omitted flops such as 3-3-5 from our tabulations. Mike's a nice guy and tries to be helpful in his writings. I find his stuff very readable and interesting. Don't fault him if there's a small glitch here or there. It's easy to go wrong in making these tedious tabulations. Lots of places to make errors.

Just my opinion.

Buzz

P.S. Below is the scratch math for the 6.48% and 5.55%.
---
10*4*4*4 = 640
Another way to think of it is:
20*16*12/6 = 640 (checks)

10*3*4*4 = 480
Another way to think of it is:
3*20*16/2 = 480 (checks)

(640+480)/17296 = 1120/17296 = 0.0648
---
6*10*4*4 = 960
Another way to think of it is:
6*20*16/2 = 960 (checks)
then 960/17296 = 0.555






Whew. How did I get myself into this? Gets tedious after a while. Easy to make a mistake by leaving something out.

[(my name it is) Sam Hall]

Wow Buzz. You've sure taken humoring the newbie to new heights. I will of course check my numbers again and if I can elucidate my reasoning I'll do that. My background includes math, but is not math, so sometimes I get to thinking I know more than I do.

[ QUOTE ]
Sam - I don't know what that statement means. I don't know what "degeneracy" used in that context means.

[/ QUOTE ]

A flop of 35T is the same as 53T, 5T3, 3T5, T53, and T35. It just means I don't really know what I'm talking about. It's the 6 (3!) in (48*47*46)/6=17296. Thanks so much for taking this on.

Sam