Suppose Hero, sitting on the button, holds
A
, K
, 5
, 4.
This is a starting hand some would consider premium. It has no middle cards, a suited ace with two wheel cards, and the king (along with the ace) provides enough high card strength. Hero decides to see the flop, and the flop is
8
, 7, 2.
Hero has missed his club flush, and has not much going for high at this point, but Hero has flopped the second nut low with counterfeit protection. Hero has an outside chance to win high or scoop, but at this point (immediately after the flop) the second nut low is mainly what this hand has going for it.
Three opponents see the flop with Hero and before there is any action, Hero wonders if any of the three opponents has flopped the nut low. (Hero has 8742A for low and the only better low at this point is 8732A).
In order to have the nut low, an opponent would need a starting hand with at least one ace plus at least one three.
Making a Chart
The plan is to include in the chart all of the possible ways an opponent could have a hand with the nut low. In other words, we want to list all the possible hands containing at least one ace and at least one trey that an opponent could be holding.
The possibilities for an opponent to have the nut low are for that opponent to hold
AAA3
AA33
A333
AA3*
A33*, or
A3**
Before proceeding, you should verify for yourself that we have included all the possibilities.
AAA3
Hero has one ace. There are four ways an opponent could be holding AAA3:
A, A , A, 3,
A, A , A, 3,
A, A , A, 3, and.
A, A , A, 3
These are all playable and there are no other possibilities for AAA3.
AA33
There are three ways an opponent could be holding two aces:
A, A,
A, A, and
A, A
(Or think of removing one ace from three available… there are three ways to do this).
There are six ways an opponent could be holding two treys:
3, 3,
3, 3,
3, 3,
3, 3,
3, 3, and
3, 3
Thus there are a total of 18 ways (3*6) an opponent could be holding two aces plus two treys.
These are all playable and there are no other possibilities for AA33.
A333
Some opponents would fold a hand with trip treys, but others would play A333, especially if the ace were the same suit as one of the treys.
Since there are, from Hero’s perspective, three missing aces, any opponent could have any of the three missing aces. Thus there are three ways an opponent could have an ace. And since there are four treys in all, there are four ways one trey could be removed, leaving three treys.
Thus there are 3*4=12 ways an opponent could have been dealt A333. And three quarters of them (nine) would have a suited ace.
AA3*
As explained above, there are three ways an opponent could be holding two aces. And since an opponent could be dealt any one of the four missing (from Hero’s perspective) treys, there are four ways an opponent could have a trey. Finally, since Hero can see 4+3=7 cards after the flop, there are 45 missing cards, of which 45-3-4=38 are something other than aces or treys. Thus there are 38 possibilities for *.
All in all then, there are 3*4*38=456 possible ways an opponent could be dealt AA3*. These are all playable.
A33*
Since from Hero’s point of view, there are three missing aces, and since an opponent could hold any one of the missing aces, there are thus three ways an opponent could have a missing ace.
As explained above, there are six ways an opponent could be holding two treys.
Also as explained above, there are 38 possibilities for *.
All in all then, there are 3*6*38=684 possible ways an opponent could be dealt A33*. Most of these are playable. The pair of treys is more of a liability than an asset, but over half of the possible hands A33* hands have a suited ace, and would be considered playable by a non-nit.
A3**
This will be the most numerous possibility for an opponent to be holding at least one ace plus at least one trey.
There are three missing aces, four missing treys, and 38 missing “other” cards.
There are 38 ways to choose one of the 38 missing “other” cards. Then after that card is chosen, there are 37 ways to choose another of the “other” cards. But since the order in which the cards are selected to be dealt doesn’t really matter to us, whichever two “other” cards are selected, there are actually two ways to select the same two cards.
Let’s call the first of the “other” cards X and the second Y. Then the two “other” cards would be X+Y. But if we chose the “other” cards in the reverse order, Y+X, they would be the same two cards. Since there are two ways we could choose these two cards, we’ll divide the product by 2. Thus the number of ways to select the other two cards if we’re not concerned about the order in which they’re selected is
38*37/2=703.
Finally 3*4*38*37/2=8,436. That’s the number of ways to choose one of three aces, one of four treys, and two of 38 other cards. Most of these A3** hands are playable. (Some without suited aces and with middle cards are poor starting hands but might be played anyhow).
Let’s put this in a chart for easy tabulation.
Hand type |
method |
number of hands |
AAA3 |
1*4 |
4 |
AA33 |
3*6 |
18 |
A333 |
3*4 |
12 |
AA3* |
3*4*38 |
456 |
A33* |
3*6*38 |
684 |
A3** |
3*4*38*37/2 |
8,436 |
Total |
|
9,610 |
Inclusion-Exclusion Principle
Using this method, as a check on the number of hands containing ace-trey combinations when Hero has AK54 and the flop is 872 (suits don’t matter for low),
C(45,4)-C(42,4)-C(41,4)+C(38,4)=
45*44*43*42/24
-42*41*40*39/24
-41*40*39*38/24
+38*37*36*35/24=
9610.
This is the same result as the result from the chart method (but obtained more quickly).
There often is more than one way to view and solve a probability problem.
Playable Hands
Of these 9,610 possible starting hands probably five sixths are playable by a tight (but not too tight) player. Some starting hands containing an ace plus a trey that would be considered unplayable by a fairly tight opponent would be played by looser opponents.
The number of total possible hands for an opponent when Hero can see seven cards is 148,995.
And then 9,610/148,995=0.0645. In other words, 9,610 hands amounts to about 6.45% of the possible hands dealt.
But a discerning (but not ultra-tight) player might only play five sixths (~8,000) of those 9,610 hands.
Thus somewhere between five and six percent of hands an opponent would have been dealt that would contain at least one ace plus at least one trey would likely be considered playable (depending on the tightness of the individual).
Since Hero has three opponents, and any one (or two or three) of them could have been dealt a hand with at least one ace plus at least one deuce, Hero should expect that at least one of the three might have flopped the nut low in the neighborhood of 15%-19% or the time, roughly one time in five or six.
However, sensible opponents are not playing any hand randomly dealt to them. If someone is only playing the better half of the hands dealt, then he’s only playing half of the 148,995 possible hands, only ~74,500 or so of the possible hands dealt to him. And in that case, Hero should expect at least one of his three opponents might have flopped the nut low in the neighborhood of 30% to 38% of the time.
Or if Hero’s opponents are only playing the top quarter of hands dealt them, then at least one of the three has flopped the nut low roughly between three times out of five and three times out of four.
Opponents vary greatly in how tightly they play. Some play very loosely while others play more tightly. Some mix it up more than others.
Number of Opponents Originally Dealt Cards
Perhaps a more realistic way to approach the dilemma of how likely is an opponent to have flopped the nut low is to consider the number of opponents who were originally dealt cards. Assume anyone would play any of the hands shown in the chart. That’s not true of everybody, but it’s in the ballpark for a typical group of opponents.
Ten-handed Table
At a ten handed table (nine opponents) when Hero is dealt AK54, at least one opponent will be dealt a hand with at least one ace plus at least one trey roughly 45.1% of the time. That’s from simulation data.
When Hero sees the 872 flop with AK54 in a ten handed game, if we assume anyone who was dealt a hand from the above chart would see the flop, then roughly half of the time, Hero will be up against someone who has flopped the nut low.
Nine-handed Table
At a nine handed table (eight opponents) when Hero is dealt AK54, at least one opponent will be dealt a hand with at least one ace plus at least one trey roughly two times in five. That’s from ~42.8% simulation data.
When Hero sees the 872 flop with AK54 in a nine handed game, if we assume anyone who was dealt a hand from the above chart would see the flop, then roughly two times in five, Hero will be up against someone who has flopped the nut low.
Six-handed Table
At a six handed table (five opponents) when Hero is dealt AK54, at least one opponent will be dealt a hand with at least one ace plus at least one trey roughly one time in four. That’s from ~27.3% simulation data.
When Hero sees the 872 flop with AK54 in a six handed game, if we assume anyone who was dealt a hand from the above chart would see the flop, then roughly one time out of four, Hero will be up against someone who has flopped the nut low.
Hand Reading
After this 8, 7, 2 flop, seen by Hero and three opponents, it’s difficult to know why an opponent is betting. In other words, does the bet represent a low hand, a high hand, or both?
If the bet is primarily for low, does the bettor have a better low, the same low, or a worse low than Hero? From Hero’s perspective, the extent to which any of these are feasible depends on how the bettor has played past hands.
If the bet is primarily for high, does it represent a flopped set, a flush draw, two pairs, a straight draw, an over-pair, or some combination of these, perhaps along with a decent low.
In general, as the game size becomes smaller, the play becomes looser more aggressive. In any case when an opponent bets, Hero has to consider the possibility the bet is from the nut low.
