Beating perfect pairs?

[dacium]

The perfect pairs bet seems to be one of the worst but it is beatable under certain circumstances.

On our 6 deck games,
It pays 30:1 on pair of same card same suit.
10:1 for two black or red cards of same face,
5:1 for one red one black.

This edge I think was about 5% to 7%.
Now there seem to be a number of ways to count. If you are counting high low and it goes either way extremely high or extremely low you can get a small edge because the deck has more cards in a smaller value range.

Also I you are really really lucky you can simply count red and black, and if the deck has way more of 1 colour left than the other, the chance of a perfect pair that pays 30:1 goes up massively.

The problem is of course is that the house edge is 7.3% on the full 6 decks and deteriotes badly after that. Before the shuffle with only 1 deck left obviously the house edge has blown out to like 50%.

Any one know of any easy ways to beat this one? The red/black count need to get past a ratio of about 10:1 (depends on how many cards are left) which is so rare it almost never happens. But bets at the right time can get you about 2 - 4% edge when it does happen. But of course a perfect pair is so rare the variance is huge. Its sweet when you do hit a few perfects in a row though.

[E-town-guy]

Originally Posted by dacium

Any one know of any easy ways to beat this one? The red/black count need to get past a ratio of about 10:1 (depends on how many cards are left) which is so rare it almost never happens. But bets at the right time can get you about 2 - 4% edge when it does happen. But of course a perfect pair is so rare the variance is huge. Its sweet when you do hit a few perfects in a row though.

If you don't know how to beat it how do you know you can get a 2-4% advantage? From what I've read it seems like its not worth it as it would likely require a 2nd count but I don't know for sure.

[dacium]

I was wondering if anyone has any other easy count methods thats all.

I had the spread sheet at work that I did the calculations in and didn't know the exact number I worked out for the edge. I can beat it counting black/red but its extremely rare, likly to get a few pecentage house edge maybe once in about 4,000 shoes. I think I worked out that count would need to be +20 or more if you were doing the count, which is rediculous, so mabye its just not feasable to beat this.

Also in australia its the law that house edge must be below 15%, based on my calculations the average house edge (making a bet randomly in the shoe from 6decks left to 1 deck left) gives an average house edge of 17%, making the bet illegal. However I don't know if I want to point this out because it seems like the casinos here will shut down the tables if they dont have the perfect pair bet, because they don't make much money on the $10 tables.

[Sonny]

Are you sure this was a 6-deck game? I’ve heard of this bet on an 8-deck game with a 3.37% house edge, but those payouts on a 6-deck game shoot the house edge up to 5.79%. That is truly a greedy casino!

Unfortunately, I don’t think this bet can be sufficiently beaten using traditional card counting. The problem is that the red+black bet and the red+red/black+black bet are always fighting against each other. You want an abundance of both colors to get an edge at the red/black bet, but that makes the other bet less likely to occur. Similarly, when the remaining cards are mostly one color you are less likely to win the red/black bet. As soon as one bet starts to turn positive the other one turns negative. The result is that the house edge tends to hover around 6% despite the “ethnicity” of the shoe.

One way to beat this bet would be by tracking the relative number of each rank and waiting for a significantly uneven distribution. That would be very difficult and may not occur very often (as you noted). This bet could also be beaten by using some sort of sequencing or tracking, but that is far beyond the scope of this public forum.

-Sonny-

[dacium]

All the australian ones are 6 deck and have perfect pairs. Pretty much impossible to beat.

[avbj]

I think an even odd count would be much better to beat this than a red/black count. If either even or odd goes high then the probability of getting a perfect pair would go up. I have not done any calculations though

[Automatic Monkey]

Originally Posted by avbj

I think an even odd count would be much better to beat this than a red/black count. If either even or odd goes high then the probability of getting a perfect pair would go up. I have not done any calculations though

If you are going to count even/odd you might as well count high/low. Same thing. But the high/low info is also obviously useful for betting.

You probably need two guys to beat this bet. One is counting red/black and the other is counting high/low. When the count is extreme in both color and rank, you may have an advantage.

If you want to get fancy, you could have a third guy counting the odd/even, or any other way you want to break the ranks up, and when you have a preponderance of any kinds of ranks and a color/suit you would have the advantage.