Absence of ACES before the FLOP
 

In Brunson's Super System, Mike Caro gives a ton of statistic. One of these is Absence of Aces before the Flop.
Colume One is "The pobability that no player has an Ace, (including yourself). Colume Three is "If you have no Ace the probability that no other player has an Aces". These two columes seem to be saying the same thing. What am I missing?

Re: Absence of ACES before the FLOP
 

I don't have SS handy but if these are worded exactly this way there is a huge difference.

Column one seems to be tallying the chance that no Ace(s) have been dealt out.

Column two seems to be tallying the chance that none of your opponents have an Ace (or Aces) ASSUMING YOU HAVE NONE YOURSELF.

Obviously there is a higher chance of one [or more] of your opponents having an Ace if you know you do not.



See the difference ?

*

Re: Absence of ACES before the FLOP
 

I must be slow... I don't see the difference. The wording is as show in SS. "The probaility that no player has an Ace (including yourself) "If you have no Ace the probability that no one has an Ace"

In both cases, the hole cards have been dealt and I have no Aces. There is not an ASSUMPTION indicated. Both columes state I don't have an Ace. Why would the proability of an Ace having been dealt to anyone change just because I know I don't have one?

Tell me again the difference.

Re: Absence of ACES before the FLOP
 

Assume 10 handed. In the first example, you have no knowledge whatsoever, including your own cards. In the second example you have knowledge of your cards.

First: C(48,20)/C(52,20)
Second: C(46,18)/C(50,18)

Re: Absence of ACES before the FLOP
 

[ QUOTE ]
Assume 10 handed. In the first example, you have no knowledge whatsoever, including your own cards. In the second example you have knowledge of your cards.

First: C(48,20)/C(52,20)
Second: C(46,18)/C(50,18)

[/ QUOTE ]

Re: Absence of ACES before the FLOP
 

OK, I'm slow but I'm almost there and I'm still missing the obvious.
First: C(48,20)/C(52,20)
Second: C(46,18)/C(50,18)
I see what the 20 and 18 are - unseen dealt cards. Where do the other numbers come from? This newbie thanks you for your patience and help.

Re: Absence of ACES before the FLOP
 

First: Number of combinations you can deal 48 non-aces when you deal 20 cards divided by the number of combinations you can deal 52 cards when you deal 20 cards.
Second: Number of combinations you can deal 46 non-aces when you deal 18 cards divided by the number of combinations you can deal 50 cards when you deal 18 cards.

Re: Absence of ACES before the FLOP
 

Getting closer... can you explain the formula?
c1(48,20)/c2(52,20) I understand that C1 is being divided by C2. What is happening within the parentheses? At a 10 player table, the formulas should equal 13.28% asw shown in Caro's table. Am I really dense?

Re: Absence of ACES before the FLOP
 

The first one is the chance that no one at the table will be dealt aces. (Considers the times you are dealt aces)

The third one is the chance that no one ELSE has aces when you are already not dealt them. (Does not consider the times you are dealt aces)

Re: Absence of ACES before the FLOP
 

In the first case; imagine you are an observer at a table. What is the chance that none of the 10 players you see have an A? This is the general question.

In the second case; imagine you are sitting at a table. You see your two hole cards . . . and you know you that don't have an A (so two less unknown cards). "What are the chances that the ramining 9 players don't have an A either?" is essentially the question.