Tie bet in baccarat

How common is the "tie pays 9-to-1" version (house edge = 4.844%) of the baccarat tie bet is? My understanding is that the "tie pays 8-to-1" version (house edge = 14.360%) is the usual choice.

8 to 1 is all over the place, though here were I live I dont think there are ANY 9 to 1 games being offered

I say very very uncommon.

maybe more popular with online casinos

Binions was the last I ever saw 9:1 and that was maybe over 10 years ago.
In the 1990s I dealt in casinos and taught Dealers School in Reno and can not remember seeing a 9:1.
Maybe Carson City had one.

The tie bet is susceptible to advantage play, correct? How much more profitable is AP play on this bet with a 9-1 vs 8-1 payout? I assume this is where you are headed teliot.

Quote: Paradigm

The tie bet is susceptible to advantage play, correct?

In the 9-to-1 version, the distribution shows huge potential, I just haven't quite figured out how to exploit all of it. Certainly, John May's method falls short -- he looks for an end-play with a shoe rich in even cards.

The most significant part of my work is the discovery that if you fix the number of cards in the hand, then the player has an edge (off-the-top) of 3.026% if it is known in advance that the hand will require 6 cards. The next question is what cards are most likely to lead to a hand that requires 6 cards. For example, 8's and 9's are clearly bad cards. Shoes rich in the cards that are most likely to lead to six-card hands will lead to a player edge. The answer to this question points to a card counting system I simulated already. This seems to be an entirely new approach.

In practice, however, it looks like the bulk of the tie bet is not susceptible directly to a card counting system, but computer perfect play can thoroughly crush the 9-to-1 tie bet. The main game of baccarat is safe against a device, but the tie bet is not. This is extremely important for casino security who may be getting hit by a device.

None of this pertains to the 8-to-1 version. That's where I'm going.

Just one minor digression please.
This is away from Payout for the Tie Bet momentarily and to the actual making of the Tie Bet and another bet.

If I want to bet Tie and also bet Banker, and a TIE takes place, my understanding is that the Banker Bet remains since although it has not won, it has not lost.

Many electronic simulation programs simply add the dollar amounts of the two bets together for a payoff which is incorrect.

What is correct in a casino. If a bet is on TIE and also on BANKER but a TIE happens to take place. All these darn Indian dealers want to take the "losing" Banker bet even though it did not "lose".

The Banker bet most certainly is not lost on when a tie occurs

Quote: teliot

In practice, however, it looks like the bulk of the tie bet is not susceptible directly to a card counting system, but computer perfect play can thoroughly crush the 9-to-1 tie bet. The main game of baccarat is safe against a device, but the tie bet is not. This is extremely important for casino security who may be getting hit by a device.

Of course if you introduced dice into the game of baccarat, the Tie bet would not be countable :-)

Quote: teliot

In the 9-to-1 version, the distribution shows huge potential, I just haven't quite figured out how to exploit all of it. Certainly, John May's method falls short -- he looks for an end-play with a shoe rich in even cards.

I recommended a number of different heuristics for different subsets in a regular game, the assumption being 8 to 1.

I'm not normally in the habit of feeding casino people info for the sake of it, but in this case I'll give you enough rope to strangle yourself with. Where the 9 to 1 payout is offered linear systems do a pretty good job of capturing the EV. Its not a spectacular play but you can make money with it. There are so many other profitable angles with baccarat casinos should probably completely forget about this altogether in the unlikely event they ever offer the proposition.

Any one who brings a device into a game wants big money fast: the card counting scenario you describe is likely to occur only with very clueless quasi-criminal types.

Quote: teliot

. . . None of this pertains to the 8-to-1 version.

I submit that it DOES pertain to the 8-to-1 version.

It's just that in the 8-to-1 version, the house edge can be reduced but not eliminated. In the 9-to-1 version, the edge can be reduced to the point of being a positive expectation.