texas shootout: A beatable game??

[RiverDood]

Here's the acid test:

You have Ah6h.
The house has Ad3c

Who is favored to win??

Set fire to your wallet if you think you're ahead. I just ran the numbers on twodimes.net. You have a 37.7% chance of winning. The house has a 28.55% chance of winning outright and a 33.76% chance of winning by tieing -- which adds up to a 62.3% chance of taking the pot.

You are almost a two-to-one dog on this hand, even though you have the "better" cards.

I tested some other hands, and the risk of a tie/house win is harrowing. We don't see that in regular holdem play because mediocre hands don't play all the way to the river, heads up. But in this game, they do.

Try 76 vs. 72. Try K7 vs. K3.

It's hideous.

[RiverDood]

OK, here's the ultimate horror

First hand: 8c 3s
Second hand: 8h 2d

You can pick whichever hand you want -- and put up just $1. I'll put up $2 and take the other hand plus the ties. Sounds tempting. But if you run the odds, you'll see that you're on the wrong end of a sucker bet.

Or, if you prefer, you can take BOTH hands and put up $1. If either one wins, you get paid. I'll take only the ties. You pay me $1 for a tie. It's much closer to an even proposition, but if we play a million hands, I'll have at least $3,000 of your money. These hands tie slightly more than 50% of the time.

Ugggh! I think this game is being brought to you by the same folks who run the $2 arcade games at the carnival -- where you throw three baseballs and try to knock down the dancing clown.

[Piz0wn0reD!!!!!!]

will someone please run some real math on this?

[Siegmund]

A 3% house edge vs. skillful play is a number I have heard kicked around several times on the net. I vaguely recall seeing a reputable analyst post this number -- about a year ago I considered analyzing the game myself and gave up after reading something that convinced me it was impossible to beat -- but I can't find it again.

The 3% number can be found in various articles discussing the game's manufacturer peddling it at trade shows, for instance http://grochowski.casinocitytimes.co...les/16348.html - but can't find the source for the actual number.

[AaronBrown]

I don't think it's quite this bad. The dealer will never play 83o, the rules for picking the hand preclude it. Knowing the dealer's distribution of hands, the player can pick hands that maximize the chance of winning outright, rather than the chance of winning plus tying/2. Moreover, there is important information in previous cards seen and other cards on this hand.

My guess is that the casino figured out what happens if the player follows the house strategy, and maybe put a little effort into considering other strategies. But I doubt they figured out the optimal player strategy. Almost no one's going to play that way, the house makes the profit on the other 99.999% of players.

My guess is that this is beatable in theory, but it's a lot of work to analyze. Even if I'm right, that doesn't make it a road to instant riches. Like blackjack card counting, it would take a lot of play to get a reasonable standard deviation. Your pay per hour, especially considering the practice time, would not be impressive, unless you had a huge bankroll to absorb losing streaks. The only point to this is if you think it's fun to play in a casino with an edge, and you make your real profit on comps.

[RiverDood]

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I don't think it's quite this bad. The dealer will never play 83o, the rules for picking the hand preclude it.

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Fair enough. You're right that 83 is at the outer bounds of possibility. I picked it just to see how bad the tie-hand problem could get. But even on more plausible hands, ties are more of a risk than one might think at first blush. Especially if everyone gravitates to high-card hands, including the dealer, and a lot of matchups involve the dealer and the customers playing Ax, Kx or Qx against one another.

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Knowing the dealer's distribution of hands, the player can pick hands that maximize the chance of winning outright, rather than the chance of winning plus tying/2.

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Actually, if the players have to act first and don't get to see the dealer's cards, this is hard. If you have A744r, do you play 44, A7 or both? . . . If the dealer has AQ82, only 44 is playable. If the dealer has 8662, only A7 is playable. There's probably a simulation that can tell you what the best prospects are in such cases against "any four," but it's a lot more intricate than blackjack and the edges are likely to be smaller.

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Moreover, there is important information in previous cards seen and other cards on this hand.

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Yes in a one-deck shoe, but probably not in a six-deck shoe. If the deck is very heart rich, you might keep two hearts over a slightly better offsuit hand. But if they're shuffling with two or more decks still left in the shoe, such skews seldom are going to be sizable enough to justify meaningful changes in play.


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My guess is that this is beatable in theory, but it's a lot of work to analyze. Even if I'm right, that doesn't make it a road to instant riches. Like blackjack card counting, it would take a lot of play to get a reasonable standard deviation. Your pay per hour, especially considering the practice time, would not be impressive, unless you had a huge bankroll to absorb losing streaks. The only point to this is if you think it's fun to play in a casino with an edge, and you make your real profit on comps.

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Well said. I'll sometimes agree to be someone's math jockey if the problem feels like a potential money-maker or is an interesting exercise in classic probability. I'm more dubious on this one than you are, but even if you're right, the rules seem goofy and the $$/hour return on this one doesn't feel worth it. Sorry

[Piz0wn0reD!!!!!!]

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will someone please run some real math on this?

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[fnord_too]

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will someone please run some real math on this?

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You understand that running real math on this is real time consuming and not trivial, no? Not to sound harsh, but this is purely for your benefit, so why don't you do the real math on it? I am certain enough it is not beatable to not care.

To the person who brought up the BJ counter example: Players don't have an edge with continuous shuffle machines or if the house reshuffles frequently enough. On any random hand, the house has an edge, and it is only due to ill conceived processes for assigning random hands (i.e. dealing from a deck that does not have the normal proportions of each rank card) that palyers can get an edge. Read Sklansky's article on WPT showdown from the magazine if it is still available (I don't think it is, unfortunately). He comments on exactly what you need to go through to get a game approved by the Nevada Gaming Commission. I suspect most gaming commissions in this country are also quite strict.

[Piz0wn0reD!!!!!!]

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will someone please run some real math on this?

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You understand that running real math on this is real time consuming and not trivial, no? Not to sound harsh, but this is purely for your benefit, so why don't you do the real math on it? I am certain enough it is not beatable to not care.

To the person who brought up the BJ counter example: Players don't have an edge with continuous shuffle machines or if the house reshuffles frequently enough. On any random hand, the house has an edge, and it is only due to ill conceived processes for assigning random hands (i.e. dealing from a deck that does not have the normal proportions of each rank card) that palyers can get an edge. Read Sklansky's article on WPT showdown from the magazine if it is still available (I don't think it is, unfortunately). He comments on exactly what you need to go through to get a game approved by the Nevada Gaming Commission. I suspect most gaming commissions in this country are also quite strict.

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i havent done any real math since 1998.

if i could do the math, i wouldnt have posted this here. I see posts here all the time with crazy hard math and assumed someone would jump at this. also its a 6 deck game, so i woulnt know where to begin.

Do you think that if you could perfectly count all the A-T you still couldnt push an edge? I see how it wouldnt help like in BJ cause the dealer cant "bust".

[AaronBrown]

If someone wants to do the math, I suggest starting out by assuming infinite decks and ignoring straights and flushes. You can add those refinements later. The dealer can either have a pair or not. If he does, you have have a higher pair, the same pair, a lower pair, two unpaired cards higher than him, one card higher than his and one equal, one card higher and one lower, one equal and one lower or two unpaired lower. If he has two unpaired cards you can have two higher than his highest, one higher than his highest and one higher than his lowest, one higher than his highest and one lower than his lowest, two between his two cards, one between his two cards and one lower than his lowest, or two lower than his lowest.

Given a strategy for picking your two cards, it's easy to figure out the probability of each of these things happening, and what your average odds of winning are in each case. That should allow you to pick an optimal strategy for this reduced game.

Next, I'd adjust that strategy by favoring suited hands and connectors, and set the weights for that by simulation. If I couldn't get close to even this way, I would give up. If I could, I would start working on counting algorithms and do a more exhaustive strategy search.