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

147 pointsby jonpover 15 years ago

7 comments

lionheartedover 15 years ago
The computer part is interesting, but it wouldn't actually fly. Take a look at the first heads up hand:<p>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.<p>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.<p>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.<p>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.
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lotharbotover 15 years ago
What follows is a constructed solution for the six-player problem, wherein the small blind always wins.<p>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.<p>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.<p>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.<p>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.<p>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.
sp332over 15 years ago
Scam School shows you how to pull it off! <a href="http://revision3.com/scamschool/texascheat" rel="nofollow">http://revision3.com/scamschool/texascheat</a>
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johnwatson11218over 15 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.
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Doveover 15 years ago
How many such decks can be expected to exist?<p>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).<p>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:<p>2 3.44e50<p>3 2.40e41<p>4 7.64e34<p>5 6.98e29<p>6 5.33e25<p>7 1.76e22<p>8 1.69e19<p>9 3.71e16<p>...<p>10 1.55e14<p>11 1.09e12<p>12 1.18e10<p>13 1.84e8<p>14 3.90e6<p>15 1.08e5<p>16 3.77e4<p>17 1.61e2<p>18 8.25<p>19 0.49<p>20 3.44e-2<p>21 2.72e-3<p>22 2.42e-4<p>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.<p><i>Finding</i> them, now . . . that might be more work.
Tichyover 15 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?
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rrhyneover 15 years ago
His script is wrong, he didn't burn a card before each of the flop, turn and river.
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