Poker puzzle challenge - Fundamental theorem

[CurryLover]

I was thinking about that bit in Theory of Poker where Sklansky talks about how the Fundamental Theorem of Poker sometimes needs to be qualified in multi-way pots (p.25-26). Then I imagined a situation that could possibly arise to illustrate this concept in an odd way:

Imagine you've got the current nuts on the flop. You go all-in for a huge overbet of the pot. The situation is this: against every individual hand you are a substantial favourite and thus they should not call your bet. However - and this is the important point - because of a freak distribution of the cards, if every player calls your bet and sees the river it is impossible for you to win.

The point would be this:
Every individual player would be making a mistake according to the fundamental theorem of poker if they called your bet - since they have only a small individual chance to outdraw you. So you'd think that you would be rooting for every individual to call, since they'd all be making a mistake. However, since you could not possibly win the hand if every individual player called, you would not, in fact, be rooting for everyone to call - although you'd be happy if only one or two of them called.

I hope I've explained the imaginary situation well enough so far. Now here is the puzzle to work out:

Can you construct a Hold'em deal that illustrates this concept? i.e. The current nuts is unable to win the hand if all 9 other players see the river, but is a massive favourite against each opposing hand individually.

Now, I remember Ciaffone once gave an example of a possible hand where the current nuts in a multi way pot would not be able to win. However, this was in Omaha not Hold'em. Can you do the same for Hold'em?

If anyone is able to contruct a hand like this then great. If you're not bothered then fair enough - I just made this post because I was thinking about the concept last night and have not been able to construct such a hand myself.

Another thought. Imagine that all hands were turned face up on the flop after the player with the nuts goes all-in. Now each player sees that he can't call the bet because he knows he does not have the odds. However, one of them notices that, if they all call, then collectively they take away any chance that the player with the current nuts has of winning the hand. He says to his opponents/collaborators, "Hey, we're all huge underdogs to this guy. In fact, I'd say that we've each got only a 10% chance of outdrawing him. But let's all agree to call him. Then - regardless of which of us actually beats him - we'll just carve the pot up between us. So what we'll be doing, in effect, is just dividing that guy's money between us. We can't lose - what about it guys?"

This could be a situation in which the Fundamental Theorem of Poker is totally turned on its head.

I know this is not particularly important from a practical point of view since this situation is not going to happen. But the more I think about it, the more interesting I'm finding this whole idea from a theoretical point of view.

Final thing: In this situation you obviously don't want them all to call because then you can't possibly win. You'd obviously like one caller, however, since you'd be making Sklansky bucks. You'd probably like 2 callers as well. But what would be the optimum number of callers in this situation - how many callers would make you the most Sklansky bucks? And at what point would it turn into a break even proposition? For example, you might have the highest EV+ with, say 3 callers. With 7 callers the situation might become EV=. With 8 callers it might be EV-. And with all 9 callers it is obviously EV- because you cannot possibly win. Can this even be worked out before a hand has been constructed?

Anwyay, inane ramblings over now.

[Jazza]

as far as i can see this is one:

you have Q[img]/images/graemlins/heart.gif[/img]Q[img]/images/graemlins/spade.gif[/img], and the board after the turn is:

Q[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/club.gif[/img]6[img]/images/graemlins/diamond.gif[/img]3[img]/images/graemlins/diamond.gif[/img]

you are up against 9 opponents who hold:

99
66
33
4[img]/images/graemlins/club.gif[/img]5[img]/images/graemlins/club.gif[/img]
7[img]/images/graemlins/diamond.gif[/img]8[img]/images/graemlins/diamond.gif[/img]
J[img]/images/graemlins/spade.gif[/img]T
A[img]/images/graemlins/heart.gif[/img]A[img]/images/graemlins/spade.gif[/img]
57
QJ[img]/images/graemlins/heart.gif[/img]

[CurryLover]

Yes, that is an example. Nice one m8!

After I made my initial post I thought of an example myself. It's not as good as yours and it is a pre-flop example, but here it is:

You are in a 15(!) handed NL Hold'em game with 2 Black Aces. You have the current nuts, but cannot possibly win if all the hands see the river because they are:
A [img]/images/graemlins/spade.gif[/img] A [img]/images/graemlins/club.gif[/img]
A [img]/images/graemlins/heart.gif[/img] K [img]/images/graemlins/diamond.gif[/img]
A [img]/images/graemlins/diamond.gif[/img] K [img]/images/graemlins/heart.gif[/img]
K [img]/images/graemlins/spade.gif[/img] K [img]/images/graemlins/club.gif[/img]
Q [img]/images/graemlins/spade.gif[/img] Q [img]/images/graemlins/club.gif[/img]
J [img]/images/graemlins/spade.gif[/img] J [img]/images/graemlins/club.gif[/img]
T [img]/images/graemlins/spade.gif[/img] T [img]/images/graemlins/club.gif[/img]
9 [img]/images/graemlins/spade.gif[/img] 9 [img]/images/graemlins/club.gif[/img]
8 [img]/images/graemlins/spade.gif[/img] 8 [img]/images/graemlins/club.gif[/img]
7 [img]/images/graemlins/spade.gif[/img] 7 [img]/images/graemlins/club.gif[/img]
6 [img]/images/graemlins/spade.gif[/img] 6 [img]/images/graemlins/club.gif[/img]
5 [img]/images/graemlins/spade.gif[/img] 5 [img]/images/graemlins/club.gif[/img]
4 [img]/images/graemlins/spade.gif[/img] 4 [img]/images/graemlins/club.gif[/img]
3 [img]/images/graemlins/spade.gif[/img] 3 [img]/images/graemlins/club.gif[/img]
2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img]

Let's say you are UTG. The pot is $0 (okay, there will be blinds in there, but to keep the maths simple we'll call it a pot of zero). You decide to go all-in for $1,000.

How many hands do you want to call your bet? I am not so good at the maths involved, but can work a few things out. For example:

Everyone passes: EV = $0

Only 2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] calls: EV = $600
(Rounding the percentages, you will win 4 times out of 5. So 4 times you win £1,000 and one time you lose $1,000. That means your EV is $3,000/5 = $600)

Only 2 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] and 3 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/club.gif[/img] calls: EV = $1,100
(You will win about 7 times out of 10. So your EV = ($14,000 - £3,000)/10 = $1,100)

4 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,200
(You will win about 11 times out of 20. So your EV = ($33,000 - $9,000)/20 = £1,200)

5 [img]/images/graemlins/spade.gif[/img]5 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,250
(You will win about 9 times out of 20. So EV = ($36,000 - $11,000)/20 = $1,250)

6 [img]/images/graemlins/spade.gif[/img]6 [img]/images/graemlins/club.gif[/img] also calls: EV = $1,000
(You will win about 15 times out of 40. So EV = ($75,000 - $35,000)/40 = $1,000)

So, it seems like the optimum number of callers for you in this coup would be 4 since that is the highest EV. Any more than 4 callers and your EV starts to decline. Of course, this is not exact because different combinations of callers might be slightly better or slightly worse for you - but it is a reasonable estimate I think.

For those of you who are good at maths (I am not) can you check my working and my conclusion for me in case I've botched it up? Also, is there a method or formula that would allow me to work out the above in a quicker, more accurate way?

I am not sure what conclusions are to be reached from thinking about this type of theoretical idea. However, I think there might be some understanding to be gleaned somewhere in there about the inner structure of the game. What do others think?

[Anders]

[ QUOTE ]
as far as i can see this is one:

you have Q[img]/images/graemlins/heart.gif[/img]Q[img]/images/graemlins/spade.gif[/img], and the board after the turn is:

Q[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/club.gif[/img]6[img]/images/graemlins/diamond.gif[/img]3[img]/images/graemlins/diamond.gif[/img]

you are up against 9 opponents who hold:

99
66
33
4[img]/images/graemlins/club.gif[/img]5[img]/images/graemlins/club.gif[/img]
7[img]/images/graemlins/diamond.gif[/img]8[img]/images/graemlins/diamond.gif[/img]
J[img]/images/graemlins/spade.gif[/img]T
A[img]/images/graemlins/heart.gif[/img]A[img]/images/graemlins/spade.gif[/img]
57
QJ[img]/images/graemlins/heart.gif[/img]

[/ QUOTE ]

Running Jacks win you the pot. No?

[Jazza]

if there was a sixth street it would [img]/images/graemlins/wink.gif[/img]

[Anders]

[ QUOTE ]
if there was a sixth street it would [img]/images/graemlins/wink.gif[/img]

[/ QUOTE ]

Ah, I am retarded. nh [img]/images/graemlins/smile.gif[/img]

[FishAndChips]

[ QUOTE ]
as far as i can see this is one:

you have Q[img]/images/graemlins/heart.gif[/img]Q[img]/images/graemlins/spade.gif[/img], and the board after the turn is:

Q[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/club.gif[/img]6[img]/images/graemlins/diamond.gif[/img]3[img]/images/graemlins/diamond.gif[/img]

you are up against 9 opponents who hold:

99
66
33
4[img]/images/graemlins/club.gif[/img]5[img]/images/graemlins/club.gif[/img]
7[img]/images/graemlins/diamond.gif[/img]8[img]/images/graemlins/diamond.gif[/img]
J[img]/images/graemlins/spade.gif[/img]T
A[img]/images/graemlins/heart.gif[/img]A[img]/images/graemlins/spade.gif[/img]
57
QJ[img]/images/graemlins/heart.gif[/img]

[/ QUOTE ]


Nice example! The only problem that I see is that a few of the hands (8 [img]/images/graemlins/diamond.gif[/img]7 [img]/images/graemlins/diamond.gif[/img], 4 [img]/images/graemlins/club.gif[/img]5 [img]/images/graemlins/club.gif[/img]) have about 30% equity against your hand, and thus depending on the size of the bet and the pot, it would not be a "mistake" to call. In limit with nine callers going to the turn, the pot would be quite large and they definitely have the correct odds to call. I guess it could work in no limit if you can bet enough to price out those hands, and also negate their implied odds.

[TomCollins]

Except You go all-in for a huge overbet of the pot.

I usually like to do that in limit.

[M.B.E.]

Cross-reference:

http://archiveserver.twoplustwo.com/...?Number=546354

[Louie Landale]

Caro contrived an example where Hero flopped a set of nines and its the nuts, yet has perhaps a 6% chance of winning (got to snag quads or get running 22, 33, or 44 (IIRC). But lets suppose it really is possible.

The situation you described is definately NOT a FTOP issue. If everyone calls then someone is the favorite: someone or two or three are going to win more than their 10% fair share. In fact, if hero cannot win then its possible that All the others are favorites; at least theoretically. These folks are NOT "making a mistake" by calling your bet. Now the first guy to call MAY be "making a mistake" when he has only a 12% chance of calling, but just because he doesn't know who else will call, but that's a different issue.

FTOP doesn't always apply to multi-way pots because there are certainly times where what's good for one person is also good for someone else. If the pair bets and the flush draw calls, the straight draw may be getting the right odds to call (good for him, bad the for pair), but its also good for the flush draw since the straight draw cannot beat the flush if he makes it. Likewise its a mistake for the straight draw to raise as a big underdog, but that's raising is also bad for the flush draw. In hi-lo split games, the best high hand almost always WANTS the best low hand to raise, trapping the other players.

- Louie