Some players at a full, 10 person limit holdem table will play any ace at any time.
Those who have read Sklansky know that Ace-little unsuited does not make it into
his 8 hand groupings and often times should not be played from early position.
Suppose you are in first position, that is the seat directly to the left of the big
blind and you are dealt Ace of clubs and 2 of diamonds (Ac,2d).
Deciding to play the hand and go against Sklansky's advice could cost you.
For this example let us suppose that you call the big blind and several players behind you also
call but no one raises behind you. In addition the small blind calls and the big blind checks.
The flop comes 3h,6s,Ad. Now what? You are third to act after the blinds. Do you bet because you have
top pair on board?
It is obvious that if another player is holding an ace you could be in big trouble.
Hopefully he is holding ace deuce like you so that a split pot is possible.
If another player has an ace with any card besides a 2 you are behind at this
point even though you hit top pair. Intuitively one could argue that since 2 of the 4 aces
are already accounted for there is a good chance that no one else has another ace.
However, the math shows that there is a 62 percent chance
that at least one other player was dealt an ace in our example.
In this example you hit top pair yet this may be a bad thing in that you may have
the second best hand all the way to the river. It is a classic example
of a scenario where as Mike Sexton says, "you don't know what you're rooting for."
If you decide to fold the hand after the ace comes on board then why were you playing
it in the first place? Were you hoping for an ace not to come on board?
Did you think the flop would have both an ace and a deuce?
Are you going to stick around to try and get a deuce on board
or a 4,5 for a bicycle straight?
Hopefully the above example showed you some of the problems that can be encountered
when playing ace-little offsuit from early position. Ace-little suited is another story
because you have a better chance of getting the nut flush. For example, if you
have Ad,2d and the flop has 2 more diamonds then you have a 35 percent chance
of getting the flush by the river. If the flush in your suit is the best
hand then you have the nut flush because you have the ace(unless of course there is a straight flush out there).
The Math:
How is it that we know you have a 35 percent chance of completing your flush
when 2 of the flop cards are the same suit as your hole cards?
The math works by showing the odds of you not getting your flush and then subtracting that
percent from 1. The odds of the turn card not helping with your flush are equal to the
number of cards left minus your outs divided by the number of cards left
or (47-9)/47. The odds of you not getting the flush card on the river are a similiar number
but there is now one less card so the odds are (46-9)/46.
This means the odds of completing the flush are:
=1-(((47-9)/47)*((46-9)/46))
=1-((38/47)*(37/46))
=1-(.808511*.804348)
=1-.650324
=.349676
=35 percent
My friend Tony Mueller helped me get the odds for at least one other player holding an ace using
logic similar to the logic for computing flush completion odds.
We solve the problem by first seeing the odds of no one else getting an ace and then we
subtract that percent from 1.
The odds of no one else being dealt an ace are:
The first of the 18 cards times the second times the third and so on.
The odds of the first card not being one of the 2 other aces are 45/47.
The odds of the second card not being one of the other 2 aces are 44/46 and so on to the 18th card.
Thus the chances of no one else getting one of the 2 remainin aces are:
=(45/47)*(44/46)*...*(28/30)
=.375578
Therefore the chances that at least one other player got another ace are 62 percent.
It is easy to get this number in excel using the product function to multiply the 18 fractions together.
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