Poker Riddle

I found an interesting poker riddle yesterday. Don't google it for the answer, because that's just unfair.

Somewhere in Vegas, there's a crazy Texas Hold'em tournament going on. It's a 1 table tournament, and they allow an unlimited number of people to sit at that table (well, unlimited until the deck runs out of cards obviously). There are two television commentators who can see everybody's hole cards. Both of them will tell the truth about what they see, but one is very dumb and the other is very smart.

As the tournament begins, the dumb one says, "look! The guy under the gun has pocket aces! What a good start for him!"

The smart one says, "Yes, but unless at least one other person folds, he's drawing dead."

How is this possible? What is the fewest number of people you need at the table for this to happen, and what are their cards?

sounds fun, but im just too lazy right now to think

so the game is like 21 handed or some shit? lolll

i would like to think that you can't have more than ~9 players to the hand or you'd be instantly fucked.

so i propose an open jam or raise 1/3 to 1/2 your stack!

actually the answer is + Show Spoiler +

On December 02 2009 16:26 YoMeR wrote: so i propose an open jam or raise 1/3 to 1/2 your stack!

Let's just assume that this is a flipament

I can't comprehend this. Are we supposed to assume that unless one person folds, everyone else is going to go all-in or something? So we see a flop or we see all five cards?

Cause I mean if the smart commentator means he's drawing dead and pointing out AA won't hold up in a billion-way pot, him pointing that out would kind of tilt me.

"HAY YOUR EQUITY ISN'T THAT GREAT WITH AA SUCKS WHEN THE WHOLE TABLE IS IN THE POT"

He's just saying that, if every single person goes to showdown, then the guy with AA has literally 0 equity. Of course it's unrealistic, it's just a riddle, not a real poker situation.

it's easy Oo
opponents have two aces
the 4 "2"
the 4 "K"
and 10 spade and 10 clubs if you have AsAc

so 22 cards => 11 opponents
11+1 = 12 GG me

On December 02 2009 16:35 waga wrote: it's easy Oo opponents have two aces the 4 "2" the 4 "K" and 10 spade and 10 clubs if you have AsAc

board is 4568x spades?

edited

hmmm, I'd think you need:

2x KK
2x 22
4x the one suit
4x the other suit
1x AA

then a few undoubled pocket pairs... I still don't get how you exclude quads for the chop

On December 02 2009 16:26 luddite wrote: actually the answer is + Show Spoiler + 12 people

xplain?

edited

On December 02 2009 16:39 Sicks Macks wrote: hmmm, I'd think you need: 2x KK 2x 22 4x the one suit 4x the other suit 1x AA then a few undoubled pocket pairs... I still don't get how you exclude quads for the chop

That won't prevent the aces from hitting a flush, or the bottom straight. Or they could win in just nobody hits lol.

On December 02 2009 16:41 waga wrote: spades ands clubs are SC and my comment is good flush A hi don't win

You'll have to explain in more detail, because I don't really understand what you're saying.

lol
it's easy Oo
opponents have two aces
the 4 "2"
the 4 "K"
and 10 spade and 10 clubs if you have AsAc

This ...

AA can't make quads or set
AA cant make str8
AA can't hit a flush (you need 4 club or 4 spade , 13-10 =3)
My first tough was good ...
no need SC or smth...

OK now I see what you mean. You still haven't accounted for the possibility that nobody hits anything though, so he could win it with just the one pair of aces.

On December 02 2009 16:42 luddite wrote: Show nested quote + On December 02 2009 16:39 Sicks Macks wrote: hmmm, I'd think you need: 2x KK 2x 22 4x the one suit 4x the other suit 1x AA then a few undoubled pocket pairs... I still don't get how you exclude quads for the chop That won't prevent the aces from hitting a flush, or the bottom straight. Or they could win in just nobody hits lol.

can't hit the bottom straight because the twos are covered, I meant 5x each suit.

On December 02 2009 16:26 luddite wrote: actually the answer is + Show Spoiler + 12 people

Just to clarify, this solution has aces actually drawing dead not to a chop?

Yeah, there's a way you can arrange it so that they can't even chop.

On December 02 2009 16:55 Sicks Macks wrote: Show nested quote + On December 02 2009 16:26 luddite wrote: actually the answer is + Show Spoiler + 12 people Just to clarify, this solution has aces actually drawing dead not to a chop?

i thought the same...drawing dead to a WINNING showdown. since the refs see the holecards, do they know someone else has AA aswell??

Yeah they know someone else has AA. That person is also drawing dead (0 equity).

ok:

Seat 1: You have AhAs

2: KhQh
3: JhTh
4: 3h2h
5: 5h4h
6: KsQs
7: JsTs
8: 3s2s
9: 5s4s
10: 2c2d
11: KcKd
12: 66
13: 77
14: 88
15: 99
16: AcAd

ehhhh lemme double check this. 12 seems hard.

Yeah I think that works

PM 12 person solution, pls?

You'll need to at least give somebody else AA, to prevent the possibility of you getting a set/quads.

Sicks how the hell does that work? off the top of my head board come QQAxx we win

oops forgot to put the other AA in. Thought of it but didn't write it.

Im just taking a shot in the dark here, but if 24 cards are in play then the remaining 28 in the deck will guarantee someone to hit 2 pair.

9itrs something aobut pairs, and I'm drunk as fuck and still figured that out. OTherwise it's avboubtt 1 cdard straoitgngts.-

On December 02 2009 17:37 NewbSaibot wrote: Im just taking a shot in the dark here, but if 24 cards are in play then the remaining 28 in the deck will guarantee someone to hit 2 pair.

but you have to block the following things too:

sets
both straights
both flushes
quads with Ace kicker (by blocking this you also block FHs and Aces up that can win)

whoops I forgot that they can see everyone's cards, not just the guy with AA.

Luddite just PM'ed me the answer and it's cool. I was missing something. Refining my answer within the logical framework I constructed would only get me to 13.

Well since the answer is 12 it probably is something like the first guy has 2 red aces, the two black aces are taken, and everyone has like every pocket pair (red to take flush draws away) except one or two dudes has a suited connector of black cards. Probably something like that; gimme a second to think about it unless someone else gets it first.

Also note every straight requires a Ten or a Five so that might have something to do with it.

There's probably a 76s to kill the wheel draw in there somewhere

If you leave 4 consecutive of one suit that's okay too because the other pps will make a straight flush. I have one that is 13 but I can't think of 12...

I have a guess try this:

AA red (first guy's hand)
AA black
KK red
QQ red
JJ red
TT red
TT black
98 diamonds
76 diamonds
54 hearts
33 red
22 red

red aces can't win with hearts because TT gets straight flush, there aren't enough diamonds in the deck to win with the diamond flush either.

No straights can split if I'm right. You can't have a straight higher than 98765 because all the straights with a Ten are blocked.

Originally had 55 red and 44 red instead of 54hh for 13 people but I THINK any board that doesn't hit the pps will give the suited connectors two pair ftw.

ARGH, AMIRITE? THE SUSPENSE IS KILLING ME. I LOVE RIDDLES

AcAs

vs
AKc
AKs
KK
QsQc
2s2c
2s2c
4s4c
5s5c
6s6c
7s7c
8s8c
9s9c
TsTc
JsJc
3sXs

he can't straight, flush or quad, and regardless of what cards come out with the remaining 333445566778899TTJJQQ someone hits a better hand ~_~ so now if the person with the 3 folds and 333 comes out he can still win

so 16 total?

AA black
AA red
3s2s
5s4c
66 black
77 black
88 black
99 black
TT black
JJ black
QcT
KcT

there are only 3 spades and 3 clubs left in the deck (KsQs4s, 5c3c2c) so no 4flush board
no 4kind board possible
wheel loses to 66
all 4 tens are out so no broadway

AA lose to a set if any J,T,9,8,7,6 fall, leaving only K,Q,5,4,3,2 as possibilities to check
but you can only have one of (5,4) and (3,2) otherwise 3s2s or 5s4c hits 2pair

the board can then have KQ53 safely, but any card left in the deck as the river will improve someone to better than AA

never mind

hero = AsAc

2= AdAh
3= KsKc
4= QsQc
5= JsJc
6= 10s10c
7= 9s9c
8= 8s8c
9= 7s7c
10= 6s6c
11= 5s5c
12= 4s4c
13= 3s3c
14= 2s2c
15= 2h2d
16= KhKd


Thats 32 cards, we cant make flush, str8 , set, and guarantees someone a set in the remaining players or something better than AA. Ballllaaaaa!!

On December 02 2009 18:56 rgfdxm wrote: AA black AA red 3s2s 5s4c 66 black 77 black 88 black 99 black TT black JJ black QcT KcT there are only 3 spades and 3 clubs left in the deck (KsQs4s, 5c3c2c) so no 4flush board no 4kind board possible wheel loses to 66 all 4 tens are out so no broadway AA lose to a set if any J,T,9,8,7,6 fall, leaving only K,Q,5,4,3,2 as possibilities to check but you can only have one of (5,4) and (3,2) otherwise 3s2s or 5s4c hits 2pair the board can then have KQ53 safely, but any card left in the deck as the river will improve someone to better than AA

Board can make a straight flush e.g. 34567 all diamond for split

good point. I hadn't thought of that. Doesn't yours have the same problem with e.g. 23456 of spades?

Ah, I think your solution can be modified to win. Change your 76dd into 7s6c. Now the 4 hearts left in the deck are 9876 so TT wins with a straight flush, the four diamonds left are 7654 so 98dd would win with a straight flush.

On December 02 2009 18:43 edzwoo wrote: I have a guess try this: AA red (first guy's hand) AA black KK red QQ red JJ red TT red TT black 98 diamonds 76 diamonds 54 hearts 33 red 22 red red aces can't win with hearts because TT gets straight flush, there aren't enough diamonds in the deck to win with the diamond flush either. No straights can split if I'm right. You can't have a straight higher than 98765 because all the straights with a Ten are blocked. Originally had 55 red and 44 red instead of 54hh for 13 people but I THINK any board that doesn't hit the pps will give the suited connectors two pair ftw.

23456 of spades ?

On December 02 2009 19:24 rgfdxm wrote: Ah, I think your solution can be modified to win. Change your 76dd into 7s6c. Now the 4 hearts left in the deck are 9876 so TT wins with a straight flush, the four diamonds left are 7654 so 98dd would win with a straight flush.

Edit : Win ?

Edit 2 : mmmm no. 2345 of spades and x =/= 6 of spades ? ie 2345s9h or 2345s 8d etc etc.

I don't understand your objection. In edzwoo's solution, the player is holding AA red. If the board is 4 spades then he loses to a flush. The point of the 7s6c hand is to prevent a complete straight flush on the board from causing a split. If the board is 65432 spades, then 7s6c wins with the 7 high straight flush.

I know but what about 2345 of spade and 9/8/7/6 of heart for example ? The other guy with AA would win the pot with straight flush 5 high

I think i got it.

+ Show Spoiler +

On December 02 2009 18:43 edzwoo wrote: I have a guess try this: AA red (first guy's hand) AA black KK red QQ red JJ red TT red TT black 98 diamonds 76 diamonds 54 hearts 33 red 22 red red aces can't win with hearts because TT gets straight flush, there aren't enough diamonds in the deck to win with the diamond flush either. No straights can split if I'm right. You can't have a straight higher than 98765 because all the straights with a Ten are blocked. Originally had 55 red and 44 red instead of 54hh for 13 people but I THINK any board that doesn't hit the pps will give the suited connectors two pair ftw.

very close, but not quite. There's still a possibility that the board could be 23456 spades or clubs, which gives everyone the straight flush for the chop.

I think you have to give one person + Show Spoiler +

On December 02 2009 21:22 freezes wrote: I think i got it. + Show Spoiler + UTG holds AdAh (red) vs KdKh (red) QdQh (red) JdJh (red) TdTh (red) 9d9h (red) 8d8h (red) 7d7h (red) 6d6h (red) Tc5c Ts4s As3s Ac2c (12 opponents) In order for utg to get a flush there must be 2345 of the suit which gives 6d6h the straight flush. 6d6h also blocks its low straights. AKQJT is out of the question since all 4 T are used. Tc5c and Ts4s blocks chops by straight flushes of the other suits. As3s and Ac3c block quads and such. There are no 5 card combos that don't improve the opponents hands.

you can do it with one less opponent

i like candy

Does this work?

+ Show Spoiler +

On December 03 2009 00:03 Elite00 wrote: Does this work? + Show Spoiler + KdKh QdQh JdJh TdTh 8h9h 7d7h 6d6h Tc5c Ts4s As3s Ac2c

no 8d9d2d3dQc gives the aces the best flush

also god damn i was off by one guy. The math major in me must solve this.

Ok final answer


AdAh (red)

vs

AcAs (black)
KdKh (red)
QdQh (red)
JdJh (red)
TdTh (red)
TcTs (black)
8d9h
6c7s
4d5h
3d3h (red)
2d2h (red)

11 opponents