Ice cold decks (dealer wins no matter how the deck is cut)

http://www.benjoffe.com/holdem

by jonp • 16 years ago

147 25 16 years ago

benjoffe.com

sp3322• 16 years ago

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

Timothee• 16 years ago

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.

lionhearted2• 16 years ago

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.

sp3322• 16 years ago

This requires a stacked deck, which means it only works once anyway. Unless you have a lot of prepared decks lying around, you won't have to worry about it happening "2-3 times in a game."

heyitsnick4• 16 years ago

You are dealer once an orbit. You have one extra deck that is passed to a 3rd party during the orbit, and the deck is stacked. When it's your turn to deal, the deck is switched with your stacked deck; the previous deck is passed back to the 3rd party to be stacked again. The 2 decks rotate in this way. Deck switching seems trivial; passing the deck to another every orbit seems the tricky bit. But hey, I'm no con artist :)

flipper• 16 years ago

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?

lionhearted2• 16 years ago

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.

mos1• 16 years ago

A solid heads-up player will pretty much never fold A4 from the small blind. small blind is also the button, so they'll have position for the rest of the hand.

Really, the only question is how much to raise, or if to limp. the limp/mini-raise tactics would be there essentially to add some deception to post-flop play.

Ace high is a pretty strong hand heads-up, even with the shit kicker, as the odds are against your opponent flopping any pairs at all.

koepked3• 16 years ago

I was thinking the same thing, a lot of those winning hands are hands that shouldn't win in actual gameplay. It'd be neat to see an algorithm that maximizes the hands that the dealer not only wins, but should win, for small numbers of players. This would be pretty complex I'm thinking.

koepked3• 16 years ago

Could go even further with this: maximize the the number of hands the dealer should win while also maximizing the number of hands the oppenent will likely think they should win to keep the pot big! Now I'm just getting crazy...

heyitsnick4• 16 years ago

This seems impossible. There will always be hand combinations that hero would fold preflop, with the exception of the dealer in heads up fixed limit, less so NL. Even 3-handed with hero on the button (who should be correctly opening say 60% of his hands), it doesn't seem possible to stack the deck so that every hand combination will be opened.

heyitsnick4• 16 years ago

Allow me to offer a HU pros take on the hand :)

In a heads up pot, The dealer is the small blind. In both fixed limit and NL variants, a competent player will raise close to 100% of his range preflop. A4 will be defending by the opponent in fixed limit, unlikely to be 3betting (we need to have some weak aces in our non-3betting range), but actually quite often folded in no limit due to reverse implied odds, assuming stacks were 30BBs deep+.

On 998, a 'donk bet' from A high is very unlikely and a pretty poor, unbalanced play with a weak showdownable hand. Our hero will continuation bet in fixed limit and NL (probably somewhere around 2/3 pot). A checkraise would be a very poor play in NL; possible thin value play in FL but XR will be called by the 3straight anyway.

In all likelyhood the turn will be seen in FL (from whereon the dealer will certainly win the hand), unlikely in NL (button will win without showdown preflop or on the flop)

johnwatson11218• 16 years ago

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• 16 years ago

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• 16 years ago

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.

Tichy• 16 years ago

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

koepked3• 16 years ago

I'm sure with a little creativity one could figure something out. Figuring that out would certainly be easier than figuring out the order to put the cards in ;) Anyway, this is less about actually doing it and more about the coolness factor in knowing it's possible, and in seeing how to figure it out.

tseabrooks• 16 years ago

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.

rrhyne2• 16 years ago

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

ojbyrne• 16 years ago

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.

rrhyne2• 16 years ago

You are completely right. My bad!

Dove• 16 years ago

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.

lotharbot• 16 years ago

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.

Ice cold decks (dealer wins no matter how the deck is cut)

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