Quiz: Draw dead preflip with AA

Inspired by a discussion on this board, I have a quiz question regarding holdem poker:

What is the minimum number of opponents where you can draw dead preflop with AA?
Which cards should each of these players hold?

By drawing dead, I mean that the player should not be able to win the pot or get a split of it no matter what cards hit the board.

[waffle]

I don't know about the minimum. I tried to create a scenario where AA had 0% preflop equity.

AA
vs..
33
44
55
66
77
88
99
TT
JJ
QQ
KK
KK
A2
A2
22

Hi, thanks for your suggestion.
I guess with your given hands, there is a possibility for the board to show a straight flush for a split between all players. Depending on the suits there is also a possibility for AA to draw the highest flush.
There is a solution where AA draws dead and with less than 15 players on the table.

You have to stop:

- Quads and sets
This requires that both other aces be out

- Board four flushes (includes straight flush)

this requires that 9 out of 13 of each Ace's suit are in other player's hands. (total:18 cards)

- Board three trips (e.g. 77749) which gives a full house.

Requires that every card 2-K is present at least once (total: 12 cards) Having every card out also satisfies the requirement that someone will have at least 2 pair to beat AA no matter what comes.

- 10-A and A-5 straight

Requires that all of (K or Q or J or 10) are out AND a 6 is out

So we can see it cannot be done with less than 9 hands (18/2). Putting this all together gives:

You hold AdAh. Opponent's cards are:

AsAc
KsKc
KhKd
QhJd
Th9d
8h7d
6h5d
4h3d

Now we need a 2, 3 more hearts, 3 more diamonds:

Qd2h
Td5h
4d3h

That's 11 players.

AA is not drawing dead in your scenario.
There is still the possibility of flopping a straight flush for a split between all players.
Can you reaarange for the eleven players to prevent that split also?

EDIT:
There is also the possibility for a lot of split pots between the pocket aces for one pair if I see this correct.

[ QUOTE ]
Can you reaarange for the eleven players to prevent that split also?

[/ QUOTE ]
Oops.

For the straight flush you'll need 5 blockers and T blockers of the other two suits. We can remove the KK as it is no longer needed for straight blocking with four 10s out. So the list becomes:

AsAc
KhQd
Jh9d
8h7d
6h5d
4h3d

TsTc
TdTh
5s5c

Now we need a 2, 3 more hearts, 3 more diamonds:

Qd2h
8d5h
Kd3h

Which is 12.

edit:
[ QUOTE ]
EDIT:
There is also the possibility for a lot of split pots between the pocket aces for one pair if I see this correct.

[/ QUOTE ]
No because every other card is out, so there will always be at least 2 pair and AA will always lose. When the board double pair, somone will have trips, etc. It's mentioned in the conditions above.

edit again: actually hold that thought, some further rearrangement may be required to prevent this on non pairing boards. It should be possible with 12 though /images/graemlins/grin.gif

Damn, you're right, K7492 is one such possibility for example. This is tougher than I thought /images/graemlins/smile.gif. Back to the drawing board. I think part of the trick will be to unpair the 5s and 10s and separate the connectors so that a straight becomes inevitable when 2 pairs are missed.

Thanks for trying it out. It seems you have Qd twice in the list. As I have edited into my previous post, I still think that you are missing when the board comes rags like: 2d 4c 6c 9c Qs. Then the Aces will split the pot with top pair.

I have constructed an example where AA is drawing dead that I have verified with http://twodimes.net. I will post the example later today but it might be fun if someone could beat me with the number of opponents I had to use or come up with another combination than I did.

Here's my solution:

You hold Ad Ah. Opponents hold:

As Ac
Kd Qh
Jd Jh
Jc Js
9d Th
8d 8h
6d 6h
Qd 7h
5s 5c
4d 2h
3d 3h

Solution with 11 numbers (http://www.twodimes.net/poker/?g=h&b=&d=&h=Ad+Ah%0D%0AAs+Ac%0D%0AKd+Qh%0D%0AJd+J h%0D%0AJc+Js%0D%0A9d+Th%0D%0A8d+8h%0D%0A6d+6h%0D%0 AQd+7h%0D%0A5s+5c%0D%0A4d+2h%0D%0A3d+3h)

+ any remaining diamond-heart couple = 12 hand total.


I'm guessing you did it in 11? If so, I don't know how. You need:

- 9 diamonds + 9 hearts to block a four flush = 9 hands minimum
- One offsuit high pair (e.g. JJ) to block a high straight & straight flush - 1 extra hand
- One offsuit low pair to block the board doing a straight flush split - 1 extra hand
- One offsuit AA to block quads and a set - 1 extra hand

For a total of 12 hands.

Here is my suggested solution to the quiz with 11 opponents. (http://twodimes.net/h/?z=1292724)

The reason my solution is one less than Phils solution is that I have taken an extra card out for each suit of the aces. This allows the aces to flop a flush, but if that happens one of the opponents has flopped a straigh flush.

I guess it is impossible to find a solution with less opponents but I could stand corrected.

[Yads]

Phil your solution allows the Aces to win 34 times.

[rchandra]

So now we finally have the solution to "when should you muck AA preflop?" When you have a clear read on each opponent (marked cards to get such a read?) and they have those cards. I recommend folding faceup.

Yes, 12-handed games aren't common. But wasn't WSOP 11-handed at the start? Could be 12-handed next year.

[ QUOTE ]
Phil your solution allows the Aces to win 34 times.

[/ QUOTE ]
[ QUOTE ]
+ any remaining diamond-heart couple = 12 hand total.

[/ QUOTE ]

Two dimes won't calculate it with the extra hand included

[wrestler_118]

I counted 22050 different ways to for AA to be drawing dead (against 11 opponents):

AdAh will be drawing dead if:

Opponent 1 = KdKh
Opponent 2 = 7d7h
Opponent 3 = AsAc
Opponents 4,5,6 and 7 = any combo of Qs and 6s (105 ways)
*Opponents 8,9,10 and 11 = any combo of JdTd9d8d5h4h3h2h (105 ways)
OR
Opponents 8,9,10 and 11 = any combo of JhTh9h8h5d4d3d2d (105 ways)

105 x 105 x 2 = 22050

Think I got them all.

Matt

[wrestler_118]

just looked a PlasticHats solution. There are many times more possible combinations. But I am too tired to count.

Matt

[samscott17]

Hi,
For the AA to be drawing dead, you would have to have the other two aces out(stops you hitting the set or four of a kind). KK and KK out as well(stops the top straight). 22 and 22 out(stops the bottom straight).
Then each of your aces suits say hearts and diamonds, all the hearts and all the diamonds bar 3 of each suit have to be dead.(no flushes available for your aces)
This still does not mean you are drawing dead, as say 333 on the board gives you a full house.
So left in the pack(unseen cards)You cannot have any 2 of the same( 4 and 4) you cannot have five of any suit.
I work it out that you could have something like 3c4c5h6d8c9s10hJdQs left, but you are still not finished. As you could hit five on the board to hit the straight.
To stop this happening remove the 9 and 6 or 7 or 10(you choose) by my calculations for the AA to be drawing completely dead you need only the 3c4c5h7h8c10hJdQs left in the pack.
I am probably completely wrong here, but hey I usually am.
It was fun removing almost every card in the pack.
Sam Scott