Is Red Dog Beatable?

It seems like the fewer middle value cards the more likely the player is to win.

Also, does anyone still offer Red Dog?

It always appeared to me that the house edge was too high to overcome.

Yes it is certainly offered at my home base - It is sometimes offered under other names like "Acey-Deucey."

Isn't this the same game as "Faro" or "Faro Bank" - the popular game from the Old Wild West ?

The house edge is indeed prohibitive. That wouldn't neccessarily be a problem is the effects of removal were significantly stronger than blackjack, but they aren't. The problem is certain cards are good or bad for different plays.

As with baccarat, the game begins to become beatable if you use something stronger than a linear count, eg a computer, in a game with deep penetration. Unlike baccarat, the chance of ever encountering a deeply-dealt red dog game is very close to zero.

Since posting I built a quick excel model and am running some simulations. My preliminary results seem to confirm your observation. Hard to overcome the -3% EV.

My simulations show that there is a greater than 1% player advantage about 1% of the time. Assuming a six deck shoe dealt out to the end of the fifth deck. Most of these situations occur very deep in the deck.

It doesn't seem worth the work to devise a count to identify these situation.

One more observation. I had it backwards. The advantageous situations seem to occur when there are lots of mid-range cards and few high and low cards.

You stated: "The advantageous situations seem to occur when there are lots of mid-range cards and few high and low cards."

Please explain this, as it seems counter-intuitive to me.

But I am not going to waste my time researching it since the game can't be beat.

But think of it this way. Take the extreme example. If all the cards were of the same value, EV would be +1100% because three of a kind pays 11:1. If all cards were of just two adjacent values, you still couldn't lose because pairs and adjacents push. So as the card values become more spread out, the EV goes down. The fewer of the edge cards, 2,3 K,A, the more condensed the distribution and the higher the EV.

Just a guess.