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johnny005
04-10-2005, 10:59 PM
Hi just wondering if I should be sticking around in pots drawing only to the nut low? lets say i have A-2xx, and the board is 5-6-k should I be sticking around in this pot and if so how many players do i need to stay?

johnny005
04-10-2005, 11:22 PM
oh yeah another good question if anyone knows it... what are the odds in a 10 handed game that another player has a-2 in their hand? how do you do this math?

slavic
04-10-2005, 11:31 PM
[ QUOTE ]
Hi just wondering if I should be sticking around in pots drawing only to the nut low? lets say i have A-2xx, and the board is 5-6-k should I be sticking around in this pot and if so how many players do i need to stay?

[/ QUOTE ]

there are 16 cards that improve you to winning half the pot (or maybe only a quarter but for simplicity we'll say half).

So 57% of the time you'll make your low. The pot doesn't have to be very large at all for you to continue playing, and if the field is large multiway you should put in a raise.

Buzz
04-11-2005, 01:07 AM
[ QUOTE ]
lets say i have A-2xx, and the board is 5-6-k should I be sticking around in this pot and if so how many players do i need to stay?

[/ QUOTE ]

Johnny - Your other two cards matter when you have A2XX and the board is 56K.

But in general, I'll probably usually see another card with a bare ace-deuce and two low cards on the flop.

How many other players? I don't know how to answer. I wouldn't be thinking that way. (And those other two cards matter).

Buzz

johnny005
04-11-2005, 02:24 AM
/images/graemlins/smile.gif lets use this as an example scince in this case there are no other cards.. Probably shouldnt even be playing this hand

Party Poker (9 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)

Preflop: Hero is MP3 with 2/images/graemlins/spade.gif, 2/images/graemlins/club.gif, 2/images/graemlins/diamond.gif, A/images/graemlins/heart.gif.
<font color="#666666">1 fold</font>, MP1 checks, <font color="#666666">1 fold</font>, BB checks.

Flop: 5/images/graemlins/diamond.gif, K/images/graemlins/diamond.gif, 6/images/graemlins/spade.gif <font color="#0000FF">(7 players)</font>
BB folds, UTG folds, MP1 folds.

Turn: K/images/graemlins/heart.gif <font color="#0000FF">(4 players)</font>
Button folds.

River: 7/images/graemlins/heart.gif <font color="#0000FF">(3 players)</font>

Final Pot:

GooperMC
04-11-2005, 02:33 AM
Here are the exact numbers

With A/2 probability of someone else having A/2: 29.8
With A/2 the probability of 2 others having A/2: 2.4
With A/2 the probability of 3 others having A/2: .06

Which means your overall pot equity after making the low is 41.75

With A278 the probability of making the nut low is: 48.8
With A2810 the probability of making the nut low is: 51.6
With A2910 the probability of making the nut low is: 54.2

Which means
With A278 you total draw equity is 20.4
With A2810 you total draw equity is 21.5
With A2910 you total draw equity is 21.5

Basically if you can get 5:1 on your combined flop and turn calls then you should continue.

In a game where there are lots of pople seeing every flop (the type of game you should be playing) it is hard to not get 5:1 unless you have to cold call a raise.

Hope that helped.

GooperMC
04-11-2005, 02:36 AM
In limit I am ok playing A222 because there is less chence of getting counterfeit and less chance of getting 1/4ed.

A222 will be a little bit better then the A2910 numbers I gave in my other post because of the smaller chance of getting counterfeit.

Buzz
04-11-2005, 04:07 AM
[ QUOTE ]
what are the odds in a 10 handed game that another player has a-2

[/ QUOTE ]

Johnny - About three to two against.

In a ten handed game when you are dealt a hand with exactly one ace and exactly one deuce, I think:
• the probability one other player was dealt a hand with at least one ace and at least one deuce is about 0.377,
• the probability two other players were dealt hands with at least one ace and at least one deuce is about 0.053,
• and about one time in a thousand, four of you will have been dealt A2XX.

Looking at it another way, the probability in a ten handed game, when you are dealt a hand with exactly one ace and exactly one deuce, that nobody else was also dealt a hand with at least one ace plus at least one deuce is 0.569, or about 57%.

You want that in odds? O.K. It's about three to two that you won't get quartered or sixthed with A2XX in a ten handed game.

[ QUOTE ]
how do you do this math?

[/ QUOTE ]

Wow! (My wife wonders why I'm laughing). Are you sure you want this? O.K., I'll try to give you the general idea with some specifics.

Not easy for me to explain. My solution involves a whole series of cases and then tabulating the results of each case. The numbers involved, the results of factoral computations, are huge and there are lots of places to go wrong.

But the results shown above seem to be validated by simulations.

Start by giving the player in seat #1 one ace, one deuce, and two other cards, let's say kings, although it doesn't matter what the other two cards are as long as they're not aces or deuces. (That would be a different problem).

Then there are three aces and three deuces to possibly distribute to the nine other players. The other nine players get 36 cards altogether. Just from the standpoint of aces and deuces, here are the possible distributions of cards to the other nine players:

• 3 aces + 3 deuces + 30 other cards.
• 3 aces + 2 deuces + 31 other cards.
• 3 aces + 1 deuces + 32 other cards.
• 2 aces + 3 deuces + 31 other cards.
• 2 aces + 2 deuces + 32 other cards.
• 2 aces + 1 deuces + 33 other cards.
• 1 ace + 3 deuces + 32 other cards.
• 1 ace + 2 deuces + 33 other cards.
• 1 ace + 1 deuce + 34 other cards.
• 1 ace + 35 other cards.
• 1 deuce + 35 other cards.
• 36 other cards (no aces or deuces).

Then it's a matter of calculating the probability of each of the possible distributions shown above. For example, when you hold A2KK, two aces plus one deuce plus 33 other cards (case #6 above) will be dealt collectively to nine opponents 3*3*C(42,33) times.

3*3*C(42,33) = 3*3*42!/33!/9! = 4013026290. That's out of a grand total of 69668534468 times, from C(48,36). Thus the probability of that particular distribution of aces and deuces, two aces and one deuce, being dealt to your nine opponents collectively is about 0.0576.

Then you deal with each of the above distribution cases separately. The bottom three cases don't matter, except maybe as a check. When you get to the sixth case (two aces and one deuce with 33 other cards), call the seat to which the deuce is dealt seat #2, and figure the probability of that hand also being dealt one of the aces.

It's one different problem after another to somehow solve as you go up to more aces and deuces, and they get increasingly complex. The numbers involved are huge, results of factoral calculations. I probably still have my solution stored away in some file on my computer, but it's been a while since I looked at the solution itself. I can't remember if I posted it or not. Probably not because the solution, as I recall, took pages and pages.

For example, when the nine opponents are collectively dealt two aces and one deuce, what is the probability the opponent who has the deuce also has an ace or two?

(That one's easy. Do you see it?)

1-C(33,3)/C(35,3) = 1-5456/6545 = 0.166

The probability is 0.166.

Then what's the probability one of your nine opponents will end up with a deuce and at least one acea when your nine opponents are collectively dealt two aces and one deuce?
P = 0.0576*0.166 = 0.009584. You save this result to be tabulated with others.

Whatever. Maybe you get the idea.

There's a nifty way to approximate some results, but that's not easy to explain either.

And there are ways to simulate the results. Very handy.

I'm not a mathematician. No guarantees about my method or numbers.

Buzz

Buzz
04-11-2005, 07:33 AM
[ QUOTE ]
lets use this as an example scince in this case there are no other cards.. Probably shouldnt even be playing this hand

[/ QUOTE ]

Your example hand is A222-rainbow.

I think A222-suited is marginal, but I'd probably want to see the flop with it in a loose ring game.

But A222-rainbow? I'd call the hand sub-marginal. I don't know what to write except that playing sub-marginal hands is probably pushing your luck.

However, it's more fun to play than to fold, and any hand with an ace plus a deuce is not an absolutely horrid hand.

At any rate, you see the flop with the hand, along with seven opponents, the flop is 5-6-K, and you wonder if you should continue.

The answer, I think, is yes.

With A222/56K, I think all you need to call a single bet after the flop are three opponents who have seen the flop with you and two who will probably stay for the showdown if you make your low.

This is complicated because
• 4/45 you'll get counterfeited on the turn, fold, and lose one small bet,
• (25/45)*(28/44) you'll miss on both the turn and river, fold, and lose three small bets,
• (25/45)*(16/44) you'll make low on the river, winning an average of 5.61 small bets,
• (16/45)*(40/44) you'll make low on the turn and won't get counterfeited on the river, winning an average of 5.61 small bets, and
• 16/45*(4/44) you'll make low on the turn but will get counterfeited on the river, and lose three small bets.

As a check, the above should add up to 1.00 and it looks to me as though it does.

Nobody could do all that math in a game.

Actually, I think your e.v. with A222 after this flop is +1.69 small bets.

Pretty interesting. Just about the crummiest ace-deuce hand you could come up with is playable after a flop with two low cards. But keep in mind that I did the calculations after a 5-6-K flop. Holding A222, you'll only get (I think) an aceless, deuceless flop with two or three low cards about 44% of the time.

Buzz

GooperMC
04-11-2005, 12:40 PM
[ QUOTE ]
But A222-rainbow? I'd call the hand sub-marginal.

[/ QUOTE ]
But you calculation shows that A222 is indeed +EV just to play for the low, and there is a small amount of high EV even if they aren't suited.

With A222 you are going to flop the low around 10% of the time and you will flop a low draw about 30% of the time. That means
- 60% of the time you will lose 1 SB
- 30% of the time you will gain 1.69 SB
- That means you would only have to win .93 SB every time that you flop the low which I think is reasonable.

Strange our calculations seem to be a a small amount different for both A/2 flop and the probability of getting quartered with A/2. I am going to have to check my math when I get home.

gergery
04-11-2005, 03:27 PM
<a href="http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Board=probability&amp;Number=1886580 &amp;Forum=" target="_blank">http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Board=probability&amp;Number=1886580 &amp;Forum= (link),,,,All_Forums,,,,&amp;Words=&amp;Searchpage=3&amp;Limit =25&amp;Main=1859270&amp;Search=true&amp;where=&amp;Name=11039&amp;dat erange=&amp;newerval=&amp;newertype=&amp;olderval=&amp;oldertype=&amp; bodyprev=#Post1886580 </a>

Cooker
04-11-2005, 04:07 PM
Let me clarify a few points: a 20% bet only needs 4 to 1 pot odds to break even in a non split pot game. I am not sure if you said 5 to 1 because of the reasons I give below, or if you made 2 mistakes that roughly cancelled (neglecting what follows plus thinking 20% is 5 to 1 against while it is only 4 to 1 against).

However, in a split pot game you have to significantly beat 4 to 1 in order to make a profit on a 20% winner for half the pot. This is because when you win your half, you only get half your bets back as well. This is an important concept in calculating whether you have odds to draw in high-low split games. Let me illustrate. Suppose your equity is 20% and the pot gives you exactly 4 to 1 on your flop and turn bets and no betting happens on the river (for simplicity):

EV = [( 4 * 1.5BB - 1.5BB/2) - 4 * 1.5BB]/5 = -.15 BB

The extra -1.5BB/2 in the pot size is to compensate the fact that you only got half your bets returned (on a typical high game hand 4 to 1 odds means you get your whole 1 plus 4 extra back when you win, so you lose 1 when you lose and gain 4 when you win). So what odds do you need to continue? To break dead even, you need a pot odds of 4.5 to 1 on flop and turn bets.

I suspect you knew this and that is why you said 5 to 1, but I thought it is a point that should frequently be pointed out for inexperienced players to pick up.

Buzz
04-12-2005, 09:06 AM
[ QUOTE ]
But you calculation shows that A222 is indeed +EV just to play for the low, and there is a small amount of high EV even if they aren't suited.

[/ QUOTE ]

Gooper - Yes, but that's after you see this particular (fairly favorable) flop.

Before you see the flop, I think the hand has a negative EV.

[ QUOTE ]
With A222 you are going to flop the low around 10% of the time and you will flop a low draw about 30% of the time. That means
- 60% of the time you will lose 1 SB
- 30% of the time you will gain 1.69 SB
- That means you would only have to win .93 SB every time that you flop the low which I think is reasonable.

[/ QUOTE ]

I don't have time to get my head into this right now. I'm playing today and I need some sleep. Maybe I'll get to it tomorrow.

I did have my head into it (yesterday or the day before, whatever) when I figured the hand had a negative expectation before the flop. That wasn't a guess; I did the math. But I could be wrong.

[ QUOTE ]
Strange our calculations seem to be a a small amount different for both A/2 flop and the probability of getting quartered with A/2. I am going to have to check my math when I get home.

[/ QUOTE ]

I wondered about that myself. I actually have a lot of confidence in myself and in those particular calculations - but at the same time I recognize I'm capable of making mistakes.

Buzz

GooperMC
04-12-2005, 12:40 PM
For the A/2 flop calcluations I was estimating based on calculations with an A2910 hand (I was at work) and it looks like I estimated a little low. After running through the actual calculations my numbers match up to yours.

Here are the precise calculations:
With A222 you are going to flop the low around 11.7% of the time and you will flop a low draw about 31.9% of the time. That means
- 56.4% of the time you will lose 1 SB
- 31.9% of the time you will gain 1.69 SB
- That means you would only have to win .21 SB every time that you flop the low which I think is more then reasonable.

I will check my probability of quarting calculations later (that one always hurts my head).

GooperMC
04-12-2005, 08:31 PM
I was getting my numbers from here:
http://www.math.sfu.ca/~alspach/mag7/
http://www.math.sfu.ca/~alspach/mag8/

My head hurt too much to try and sort this out, can someone who is smarter then me verify these calculations.

gergery
04-12-2005, 08:52 PM
[ QUOTE ]
I was getting my numbers from here:
http://www.math.sfu.ca/~alspach/mag7/
http://www.math.sfu.ca/~alspach/mag8/

My head hurt too much to try and sort this out, can someone who is smarter then me verify these calculations.

[/ QUOTE ]


try this instead (http://www.math.sfu.ca/~alspach/mag24/)

The first sentence at that website says the following

"The reader is advised that this article and the succeeding article are nonsense and should be ignored. See Low Board Blues: Reprise and Coda for corrections"

woodguy
04-12-2005, 09:05 PM
Greg,

C'mon man /images/graemlins/grin.gif

http://img.photobucket.com/albums/v247/razor27/InstantURLCode.jpg

Regards,
Woodguy

GooperMC
04-12-2005, 11:23 PM
Thanks gergery. I feel a little bit stupud, but at least I now have correct numbers.