Adjusting outs for low draws in Limit O8

[L0QTiS]

I had a notion this weekend I'd like to run past the this group.

Given the following scenario:

You have A2XX unsuited in late position (XX cards unimportant)

Assume 3 called + yourself + BB who checked (5 to the flop)

Flop is 48Q rainbow

Assume you have no high draw

Given this scenario we have 16 clean outs to the nut low, but can we realistically assume that all 16 outs are unused by our opponents? Probably not (or at least, this would be somewhat unlikely). So, based on that assumption, can we make a mathematical estimation of how many outs may be in use (or, at least most likely in use)? Based on that alone, it would seem impossible because those who folded may have also folded our outs. But, can we perhaps assume that most players who entered the pot probably have some of our outs and adjust accordingly?

Where this may factor in is in the above scenario if it's checked through on the flop and then non-low brick falls on the turn. It may be a single bet to you, or a bet and a raise. I would think if you low draw decision is close, it should probably be folded for this very reason (not to mention the fact that you might be getting quartered).

Is this worthy of working through, or perhaps a fools errand?

[cielo]

Hey L0QTiS,

I saw a similar question regarding a flush draw once. The idea is basically that on any average hand (or over the long run) the ratio of your outs that are already used up vs. your outs that are still in the deck should remain constant.

For example, in the case of the nut low draw, you have, as you said, 16 outs to the nut low, out of 45 unknown cards.
So, you are a 16 to 29 dog to hit the nut low on the turn. But, of course it is highly likely that someone else has some of those outs and that even more of your outs have already been folded. However, we must not neglect the cards that your opponents are playing (or have folded) that would NOT complete your draw. So, I guess the point is that in the long run, your opponents will have 1.8 non-outs for every out they have, so the odds remain the same.

As a side note, I was recently looking at CardPlayer's OMAHA Calculator and was surprised to see how few cards remain after a whole table was dealt (obviously I always knew this, but never reflected on it explicitly).

So, I hope I answered your question.

cielo

[Buzz]

[ QUOTE ]
can we realistically assume that all 16 outs are unused by our opponents?

[/ QUOTE ]

L0QTiS - It is almost certain that some of your 16 outs have been dealt to your opponents. It is also almost certain that some of your 29 non-outs have been dealt to your opponents.

Indeed, with nine opponents who have been dealt 36 cards, and with only 9 cards left in the stub, 16*(36/45) of your outs figure to have been dealt to your opponents and 29*(36/45) of your non-outs figure to have been dealt to your opponents.

Meanwhile, 16*(9/45) of your outs figure to be in the stub and 29*(9/45) of your non-outs figure to be in the stub.

Thus the odds against catching one of your outs on the turn are actually 29*(9/45) to 16*(9/45).

Notice this reduces to 29 to 16.

Seems so straightforward to me. I hope what I have written above makes it clear to you also.

[ QUOTE ]
based on that assumption, can we make a mathematical estimation of how many outs may be in use (or, at least most likely in use)?

[/ QUOTE ]

Yes. We can use a factor to estimate how many outs are in any given group of cards. But it doesn't matter because the factor cancels when we're figuring odds for outs.

[ QUOTE ]
Based on that alone, it would seem impossible because those who folded may have also folded our outs.

[/ QUOTE ]

It's not at all impossible to determine the probability. Some of our outs may have been folded, but some of our non-outs may also have been folded. The proportion stays the same.

[ QUOTE ]
But, can we perhaps assume that most players who entered the pot probably have some of our outs and adjust accordingly?

[/ QUOTE ]

Depends on what the outs are. Your opponents probably are more likely to play hands with aces than any other cards. Next, for most but not all opponents, come wheel cards, probably in order 2>3>4>5. Next, for most but not all opponents, come honor cards, probably in order K>Q>J>T. Finally for most opponents come middle cards, probably in order 6>7>8>9. Your opponents don't all favor the same starting cards. For example, some of your opponents like kings and queens better than deuces and treys.

At any rate, if your outs are aces, you're more likely to find some of them in the hands of active opponents - and if your outs are nines, you're more likely to find some of them in the muck.

In terms of encountering one of your outs in the hand of a still active opponent, it clearly depends on what rank the card is. But that doesn't seem to be what you are asking about.

In terms of catching one of your outs on the turn, once one of your opponents has been dealt a hand with one of your outs, it's the same whether the opponent plays the hand or mucks it. Either way, you're not going to catch that particular out on the turn. But you're also not going to catch any non-outs that opponent has been dealt, whether the opponent has mucked or not.

Buzz

[chaos]

That was a nice, clear explanation Buzz.

[dilly]

Didn't see addressed in the original post or responses what effect if an Ace or Deuce counterfeits the hand. What does this do to the math?

dilly [ QUOTE ]


[/ QUOTE ]

[Buzz]

[ QUOTE ]
Didn't see addressed in the original post or responses what effect if an Ace or Deuce counterfeits the hand. What does this do to the math?

[/ QUOTE ]

Dilly - Nothing. An ace or a deuce is one of the 29 non-outs.

Buzz

[4_Outs]

Let me take a broader approach and say that in absence of other information what previous posters have said is correct (i.e. that the probability of hitting your outs does not change).

But what if you “knew something”? E.g. that everyone in the pot had a low hand? Or that no one in the pot had a low hand? Then because you know something, the chances that the low cards you need are/aren’t already in your opponents hands alters the probability of them coming on the turn.

I would guess that in the situation you proposed (5 players see a 48Q flop) that you don’t know much different than normal. So for this specific situation I would guess you’d ignore this approach.

Practically speaking, you may not be able to realize much of an advantage from this approach most of the time. Perhaps the best sort of situation to use this inference is if you have A2xx, many players see the flop, and the flop is 349T now the likelihood that there are a “more than usual” number of 5,6,7,8s is probably higher than otherwise.

All cards are not created equal – consider how many aces are waiting to be played if many players are in the pot vs. if few players are in. In the former case I’d say it’s less likely that aces are going to fall on the turn/river because they’re more likely to be in someone’s hand. And the likelihood of unattractive starting cards (5, 6, 7, 8) hitting would be increased.

Additionally, the question about being counterfeited (e.g. what if you hit your out, then the river comes with an A or 2) seems like it may also benefit from your approach. If there are several people actively contesting the low, then you have a greater chance that those cards which would hurt you on the river are already in someone else’s hand.

BTW, I’m a O8 rookie (so I’m more of a reader than contributor), but I’m well grounded in probability/statistics (that’s part of what I do for a living). I’d be interested in other folks’ reactions.