5 ms·

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

147 points by jonp 17y ago

johnwatson11218 17y 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 17y 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 17y ago
    The page http://www.benjoffe.com/holdem_files/heuristic.html 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.
sp332 17y ago
Scam School shows you how to pull it off! http://revision3.com/scamschool/texascheat http://revision3.com/scamschool/texascheat
- Timothee 17y 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.
lionhearted 17y 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.
- sp332 17y 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."
  - heyitsnick 17y 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 17y 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?
Tichy 17y ago
How does one stack a deck? Having a second deck prepared and somehow exchanging it after shuffling the cards of the original deck?
- koepked 17y 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 17y 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.
rrhyne 17y ago
His script is wrong, he didn't burn a card before each of the flop, turn and river.
- ojbyrne 17y 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.
  - rrhyne 17y ago
    You are completely right. My bad!
Dove 17y 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 17y 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.

147 points by jonp 17y 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 17y 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 17y ago
  The page http://www.benjoffe.com/holdem_files/heuristic.html 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.

sp332 17y ago

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

Timothee 17y 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.

lionhearted 17y 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.

sp332 17y 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."
- heyitsnick 17y 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 17y 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?

Tichy 17y ago

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

koepked 17y 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 17y 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.

rrhyne 17y ago

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

ojbyrne 17y 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.
- rrhyne 17y ago
  You are completely right. My bad!

Dove 17y 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 17y 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.