
Ice cold decks (dealer wins no matter how the deck is cut) - jonp
http://www.benjoffe.com/holdem
======
lionhearted
The computer part is interesting, but it wouldn't actually fly. Take a look at
the first heads up hand:

Small blind (opponent) gets dealt Ace-Four, a very strong heads up hand,
because ace-high will be good lots of time heads up. Dealer gets Ten-two,
which is a very bad hand that makes bottom pair, or a pair with a bad "kicker"
(if two hands match, you go to the high card - a 2 can never be high enough to
play). Ten-two is junk. Anyway, a good heads-up player will probably raise
Ace-four 3-5 times the value of the pot.

Okay, so then the board comes down 989 on the flop. At this point, opponent
has ace high on a pair of nines. He's feeling quite good. He probably bets
out, maybe half the pot. At this point, any sane player folds the ten-two if
they called to begin with.

If you call, and the catch the pair of tens on the turn, the person will
probably think you're an idiot. If this happens 2-3 times in a game, and
you're winning more than probability suggests, they'll probably nab you for
cheating. Which, depending on where you're at, winds up in you getting thrown
out of the game politely, thrown out of the game impolitely, or treated...
well, really impolitely.

Long story short - interesting computer usage, real life application low-ish
even if you wanted to try something like this, which I wouldn't recommend
anyways.

~~~
flipper
Yeah, I wouldn't recommend it either, but in the situation you describe
couldn't the cheater just raise on the flop? If they get looked up by their
opponent, when they catch a lucky Ten on the turn they can just say they were
bluffing with air.

I'm not a heads-up player so I'm not sure how often you could get away with
such gambits. Also, in this situation the opponent might suspect the cheater
holds JT, T9, 98, 87, 76, A9, A8 or any number of other hands that could have
hit the flop. Wouldn't a solid heads-up player fold A4 frequently enough in
that situation that they never realize they're playing against a stacked deck?

~~~
lionhearted
Maybe, but I think first action heads up is on the small blind (opponent).
He'll be raising that hand. You play most hands heads up, but you toss 10-2
offsuit normally to a raise. After the flop comes down, then I think first
action is the dealer. If he bluffs 10-2 there, that's a _complete_ bluff - and
the paired board makes it less likely he hit something, though JT would be an
option for someone who calls. If he bets the flop, and the turn, maybe the
Ace-Four folds at some point. But if he shows it down (and some people will),
it'll look like a completely unnatural sequence of events. You usually don't
call significant raises with 10-2 offsuit, and then you're pure bluffing when
the flop hits - you have almost no chance of making a hand, except the very
small chance you hit a ten and they don't have an overpair or hit an overcard
or have a straight themselves.

~~~
heyitsnick
The dealer always has position by definition (we deal to our left and finish
with ourselves, and action starts to the left and finishes with us).

The page is actually mistaken in this aspect; it has out opponent labelled as
"SB" whereas in fact hero, the dealer, will be SB and out opponent is BB.

------
lotharbot
What follows is a constructed solution for the six-player problem, wherein the
small blind always wins.

With six players, the small blind will receive the first and seventh cards of
the deck, which I will denote (1,7). The remaining hands will be (2,8), (3,9),
(4,10), (5,11), and (6,12). The burn cards are 13, 17, and 19. The board is
cards 14, 15, 16, 18, and 20. The first and last board cards are exactly 13
cards after the small blind's two cards, so by simply permuting the 13
possible ranks and then repeating the permutation, one can guarantee the small
blind will hit two pair via the first flop card and the river card. What
remains is to guarantee that two pair will be the winning hand.

Because cards of the same rank will be exactly 13 spaces apart, no player will
be dealt a pocket pair, and the board will never pair. Therefore, larger
multiples (trips, quads, boats) will never come up. Furthermore, no other
player will hit two pair.

We can eliminate the possibility of a flush (including a straight flush) by
guaranteeing that, of the seven cards from 14 and 20, no three are of the same
suit. This is simple -- cycle the cards in arbitrary suit order, say, spades -
diamonds - clubs - hearts.

Now we simply need to eliminate the possibility of a straight. The permutation
A Q T 8 6 4 2 K J 9 7 5 3 does this. Each player gets connected cards like
A-2, 5-6, or J-Q, but the board hits the first, third, and fifth cards of a
straight (or doesn't put one out at all), meaning one or two players have a
gutshot draw but nobody ever hits.

It even appears realistic for a player in the small blind to play this hand
regardless of which of the possibilities they're dealt, unless two other
players both put out big raises. With a strong hand (say, pairing an ace or
king on the flop) they can simply act as though they knew they were in the
lead the whole time. With a weak hand (say, 3-4 offsuit) they can act like
they were playing position with a Gus Hansen hand, semi-bluffed with bottom
pair, and sucked out on the river.

------
sp332
Scam School shows you how to pull it off!
<http://revision3.com/scamschool/texascheat>

~~~
Timothee
I saw it there too. One thing to note though is that you can only play one
game at a time since the desk is stacked.

In the video, Brian Brushwood plays two games but there's a cut (no pun
intended) in the video in between the two.

------
johnwatson11218
Didn't see it in the article but it might be nice to see the total number of
solutions as n increases. This might be a good application for genetic
algorithms to search this large space.

~~~
leif
I'm interested in this function as well. I'd like to try to analyze his script
to see if there is such a bound, or at least some theoretical results to be
pulled out of it, but I didn't see the guy's script on the page; has anyone
found it?

~~~
jonp
The page <http://www.benjoffe.com/holdem_files/heuristic.html> is running the
two-player case. It seems to be using a Javascript hill-climbing algorithm:
pick a scoring function which measures how good the solution is, then start
from a random deck and try small changes, keeping them only if the score
improves.

------
Dove
How many such decks can be expected to exist?

Well, there are 52 possble cuts, including "no cut", giving us 52 possible
scenarios the dealer must win. Assuming naively that winning one scenario does
not affect the odds of winning another scenario, the probability of the dealer
winning all scenarios is (1/n)^52, where n is the number of players. There are
51! possible decks (given that the deck is cyclic under cutting and we may
hence assume the first card).

Thus a naive expectation for the number of ice cold decks that exist for any n
is 51!/n^52. This has the following values:

2 3.44e50

3 2.40e41

4 7.64e34

5 6.98e29

6 5.33e25

7 1.76e22

8 1.69e19

9 3.71e16

...

10 1.55e14

11 1.09e12

12 1.18e10

13 1.84e8

14 3.90e6

15 1.08e5

16 3.77e4

17 1.61e2

18 8.25

19 0.49

20 3.44e-2

21 2.72e-3

22 2.42e-4

So the odds of such a deck existing are very strongly in our favor in all of
the non-silly cases. Indeed, it appears likely that we can get almost all of
the silly cases, too, especially if that assumption about the scenarios being
independent was wrong.

 _Finding_ them, now . . . that might be more work.

------
Tichy
How does one stack a deck? Having a second deck prepared and somehow
exchanging it after shuffling the cards of the original deck?

~~~
tseabrooks
I can't imagine how someone would stack all 52 cards in a deck without passing
it to another person. A few years ago I taught myself to stack maybe the top 5
cards and the bottom 5 cards and deal either the top or bottom card very
quickly depending on what I wanted someone to have. That was mostly a matter
of memorizing which card was which and practicing shuffling enough so that I
could move a card to the top and then keep it there.

------
rrhyne
His script is wrong, he didn't burn a card before each of the flop, turn and
river.

~~~
ojbyrne
I think you're mistaken, because he said that he did take that into account,
and the few random checks I did confirmed it.

For example, starting with the first card in the first example he has

A 10 4 2 K 9 8 9 8 10 8 9

A 4 is one hand, 10 2 is the second, K is burned

989 is the flop, 8 gets burned

10 is the turn, 8 gets burned,

9 is the river.

~~~
rrhyne
You are completely right. My bad!

