Who wants to figure this out for me?

[Dynasty]

A post in the "Poker Theory" forum in which the merits of KT vs. 76 on a K65 rainbow flop is being debated made me think of something.

We know that AK will flop top pair about 1/3 of the time. In fact, any non-pair hand will flop a pair about 1/3 of the time. However, all hands besides AK won't always flop top pair. For example, When AQ flops a pair of Queens, the pair will not be top pair if a King is on the flop.

So, how often do the big offsuit hands flop top pair? AK? AQ? AJ? AT? KQ? KJ? KT? QJ? QT? JT?

I don't remember every seeing this kind of data in anything I've read. I think it could be useful.

[BruceZ]

[ QUOTE ]
So, how often do the big offsuit hands flop top pair? AK? AQ? AJ? AT? KQ? KJ? KT? QJ? QT? JT?

[/ QUOTE ]

That would be real easy to crank out. Do you just want top pair, or do you want to include 2-pair, trips, full houses, and quads, or some combination of those?

[ QUOTE ]

I don't remember every seeing this kind of data in anything I've read. I think it could be useful.

[/ QUOTE ]

And useful to you means it's worth money. Just how useful are we talking? /images/graemlins/wink.gif

[Senor Choppy]

The most useful would be top pair alone, since all unpaired hands are going to flop 2 pair, trips, etc. equally, unless you were to also look at how often they flopped top 2.

[BruceZ]

[ QUOTE ]
The most useful would be top pair alone, since all unpaired hands are going to flop 2 pair, trips, etc. equally, unless you were to also look at how often they flopped top 2.

[/ QUOTE ]

I was referring only to hands which paired the top card, but also made 2-pair, trips, full house, or quads. Like 2-pair with the top pair, but not necessarily the top 2 pair. Or top trips, top quads, or full house using the top card on the flop. This would be a different number for different hole cards just as top pair is different.

You bring up a third possibility, which is simply top pair or any hand ranked higher than top pair, even if it doesn't contain top pair. As you say, this would simply add a constant amount to every top pair hand, but we may still want these numbers.

Let me know what you want to include. I have developed a general formula and a spreadsheet which allows you to easily enter any preflop hand, and I can configure it to include any of these variations.

[Dynasty]

Clarkmeister sent me a MP saying a chart in Carson's book has this info. So, nobody do any serious work. If you want to post that chart though... /images/graemlins/smile.gif .

[Dynasty]

If it truly is easy, I was intersted only in flopping top pair.

[aloiz]

I took a stab...

Assuming that we hold two offsuit cards, and that the only hand we make is top pair (no two pair, trips, or boats).

AKo = 2 * (3 * C(40,2) / C(50,3) - 11 * 6 / C(50,3)) = .283
AQo = .1414 (odds for ace making only top pair)+ [3 * C(40,2) / C(50,3) - 10 * 6 / C(50,3)] = .258
AJo = .1414 + 3 * C(36,2) / C(50,3) - 9 * 6 / C(50,3) = .235
ATo = .1414 + 3 * C(32,2) / C(50,3) - 8 * 6 / C(50,3) = .215

If I'm doing this right, the calculations get pretty repetitive.

KQ = .233
KJ = .21 (exactly)
KT = .190
QJ = .187
QT = .167
JT = .147

aloiz

[Senor Choppy]

Wow, those are pretty eye-opening. AK being almost twice as likely to flop TP than JT is nuts. I knew there was a reason that hand sucked so badly.

[BruceZ]

My general formula is:

[ 3*(44 - 4*N)*(40 - 4*N)*3 + 3*(44 - 4*M)*(40 - 4*M)*3 ] / (50*49*48)

where N and M are the numbers of ranks higher than each of your hole cards which you don't already hold. For example, for KJ, N = 1 for the A, and M = 2 for A,Q. For the first hole card, there are 3 ways to pair it, 44 - 4*N ways to deal the second card so it doesn't pair either hole card and is not an overcard to the pair, and 40 - 4N ways to deal the third card so it doesn't pair either hole card, doesn't pair the board, and is not an overcard to the pair. Then multiply by 3 since the pair can come in any position. This is all repeated for the other hole card.

<font class="small">Code:</font><hr /><pre>
hand overcards (N) overcards (M) top pair

AK 0 0 26.9%
AQ 0 1 24.5%
AJ 0 2 22.3%
AT 0 3 20.3%
KQ 1 1 22.0%
KJ 1 2 19.8%
KT 1 3 17.9%
QJ 2 2 17.6%
QT 2 3 15.7%
JT 3 3 13.7%
</pre><hr />

Note that we can verify AK as 6*44*40*3/(50*49*48) = 26.9%. The rest are the same except for different values of N and M.

I can email you the Excel file that does this if you want.

[MushashiAce]

Hey bruce, thanx for responding to all these % questions, I've been coming across alot of your odds questions lately, real good stuff /images/graemlins/wink.gif

[BruceZ]

[ QUOTE ]
I took a stab...

Assuming that we hold two offsuit cards, and that the only hand we make is top pair (no two pair, trips, or boats).

AKo = 2 * (3 * C(40,2) / C(50,3) - 11 * 6 / C(50,3)) = .283
AQo = .1414 (odds for ace making only top pair)+ [3 * C(40,2) / C(50,3) - 10 * 6 / C(50,3)] = .258
AJo = .1414 + 3 * C(36,2) / C(50,3) - 9 * 6 / C(50,3) = .235
ATo = .1414 + 3 * C(32,2) / C(50,3) - 8 * 6 / C(50,3) = .215

If I'm doing this right, the calculations get pretty repetitive.

KQ = .233
KJ = .21 (exactly)
KT = .190
QJ = .187
QT = .167
JT = .147

aloiz

[/ QUOTE ]

Refer to my solutions for differences. It seems your AKo is off (should be 6*44*40*3/(50*49*48) = 26.9%) and that affects all the others. If you change your C(40,2) to C(44,2) and make the 3 multiply your 11*6, then you'd have it.

[digdeep]

BruceZ,

I'd be curious to see the excel file if you don't mind. If it's no trouble my email is jastainb@ouray.cudenver.edu.

Thanks

[BruceZ]

[ QUOTE ]
If it truly is easy, I was intersted only in flopping top pair.

[/ QUOTE ]

That is what I computed (see "My solutions" in this thread). Let me know how these compare to Gary Carson's numbers. They should be the same unless Gary is computing something slightly different from what you asked for.

[aloiz]

The C(40,2) was a typo, but yea I did forget to multiply the second half by three.

thanks,
aloiz

[uuDevil]

[ QUOTE ]
Let me know how these compare to Gary Carson's numbers. They should be the same unless Gary is computing something slightly different from what you asked for.

[/ QUOTE ]
For those following along, if I found the right table, Carson is answering a somewhat different question.

From p. 62, table "Chances of Flopping the Top Pair":
<font class="small">Code:</font><hr /><pre>
If you hold an unpaired % time it will be highest card on flop
A 16.6%
K 13.9%
Q 11.3%
.
.
.
</pre><hr />