when can we get advantage to buy pairs ?

beyondbj · Mar 24, 2009

if 8 decks , the casino cut 1/2 deck pen only

when should we get advantage to buy pairs??

if it offer 1:11 reward

for example , how many the same card is leave in the last deck should we bet pairs

and how to count our advantage ?/

any one can show me a table to show how many player will get an edge for different number of same series of card is leave in the last deck card

ricopuno · Mar 26, 2009

beyondbj said:
if 8 decks , the casino cut 1/2 deck pen only

when should we get advantage to buy pairs??

if it offer 1:11 reward

for example , how many the same card is leave in the last deck should we bet pairs

and how to count our advantage ?/

any one can show me a table to show how many player will get an edge for different number of same series of card is leave in the last deck card

Hi,

The best time to buy pairs is when the TC is = +3 and up , decks remaining
are 2. The probability that pairs of faces cards to come out is high.
Although I don't have math to prove it but I have done it myself and it works.

Sonny · Mar 26, 2009

Don't bother. Even with computer-perfect play you will earn more money by playing a few extra hands per hour on the main bet.

-Sonny-

ricopuno · Mar 26, 2009

I understand what your saying Sonny. The only thing is that the min. bet for pairs is usually $5 which is not a big loss if you did not get pairs but if you get pairs then the pay out is 12 to 1. Also you won't be playing the pairs all the time only when the TC is + 3 and up with 2 decks remaining.

Sonny · Mar 26, 2009

The bet only gets positive about 1.33% of the time, and most of those are near the 1-deck mark, so you won't find many opportunities to play with an advantage. Even when you do find an advantage, it is only about 7% on average so your win rate is very small to begin with (around 0.09%). Using a card counting system will drop that even lower since you are using imperfect information. Also, as you pointed out, the variance is pretty high so you will be "bleeding" your bets most of the time and only occasionally hitting that big bonus payout. The advantages are small, rare and pretty risky. If the casino has a low limit on the bet, like $25 or so, then the profit potential practically disappears. It's beatable, but not really worthwhile.

-Sonny-

ricopuno · Mar 26, 2009

Thanks for the advice Sonny. I agree with you

But I think I will continue playing it.

Sonny · Mar 27, 2009

ricopuno said:
The best time to buy pairs is when the TC is = +3 and up , decks remaining are 2.

Using that strategy in the game beyondbj described will have you betting into an average advantage of -9.5%. For every $5 bet you make you should expect to lose $0.475. Luckily you will only be making the bet about 5.77% of the time so your win rate increases to -0.55%. Basically you're doubling the house edge.

ricopuno said:
The probability that pairs of faces cards to come out is high.

The problem is that most of those face cards are different ranks. They need to be the same rank in order to get the bonus. Hands like (K,Q), (J,K) and (10,Q) are good BJ hands but they lose the side bet. That's why a standard card counting system is not good for this bet.

-Sonny-

FLASH1296 · Mar 27, 2009

I hope that the well-meaning posters here ALL understand that the absurdly awful side bet, generally referred to as "Perfect Pairs",
does NOT consider two face cards to be a pair.
Where I have seen this nightmarish sucker bet offered the payoff is 10 to 1 and 15 to 1 for suited pairs.

Your expectation on this side-bet is truly horrid and the casinos that I saw accepted bets from $5 to $100 on it.

Of course it substantially slows down the game.

ricopuno · Mar 27, 2009

FLASH1296 said:
I hope that the well-meaning posters here ALL understand that the absurdly awful side bet, generally referred to as "Perfect Pairs",
does NOT consider two face cards to be a pair.
Where I have seen this nightmarish sucker bet offered the payoff is 10 to 1 and 15 to 1 for suited pairs.

Your expectation on this side-bet is truly horrid and the casinos that I saw accepted bets from $5 to $100 on it.

Of course it substantially slows down the game.

Thanks for the advice Sonny..

Flash, Actually in the casino I go to the payout for non perfect pairs is 6 to 1
and for perfect pairs it is 12 to 1. I guess in some casino they only pay for perfect pairs.

As I said as long as I have the luck for pairs I will play.
I must admit I combine luck with card counting. The truth is I also play slots.

FLASH1296 · Mar 27, 2009

SIX to ONE !

Pulleeze tell me where this larceny is taking place.

My hunch is that this is on an Indian Reservation, but ...

Sonny · Mar 27, 2009

ricopuno said:
Flash, Actually in the casino I go to the payout for non perfect pairs is 6 to 1 and for perfect pairs it is 12 to 1.

It sounds like you are playing a variation of the Perfect Pairs side bet (House edge of 37.59% as described). The numbers I gave above are for the Dare Any Pair bet (sometimes called the Lucky Pairs bet) that the OP described so they do not apply to your game. With so many different variations (don't forget the Pair Square bet!) it's easy to get confused.

-Sonny-

ortango · Mar 28, 2009

The original poster most likely plays at an Asian casino which has a dumbed down perfect pair side bet: 11 to 1 for any pair. This bet is over 10% for the casino.

Also, please don't tell people +3 is the pivot for advantage, with no proof except that "you have done it". There is no math behind this, as far as I can tell, even the wizard of odds hasn't gotten around to it. I'm sure it comes around to an advantage at really high or low counts, but I don't know where that is.

FLASH1296 · Mar 28, 2009

T T B O M K, it is impossible to gain an advantage at any "pairs" side bet.

I challenge anyone to show us otherwise.

Tarzan · Mar 28, 2009

I saw a perfect pair recently

I'm totally unsure of what the payout may be but I saw multiple perfect pairs at the Borgata last week. Apparently they do not hire any waitresses that do not have... a perfect pair!

Katweezel · Mar 30, 2009

Double distractions

Tarzan said:
I'm totally unsure of what the payout may be but I saw multiple perfect pairs at the Borgata last week. Apparently they do not hire any waitresses that do not have... a perfect pair!

Some stuff has been written here on the subject of distractions.
TZ, you seem to have a real nose for perfect pairs, and my guess is your nose would like to investigate very closely between... those perfect pairs, just to see if one is more perfect than the other... :cat:

beyondbj · Mar 31, 2009

i dont mean use TC to buy pairs

i mean the uneven remaining cards makes pairs become advantage for 1:11 payout

can any one help me to check how to calculate the advantage , if the probability become 1/10 , it will get 10% advantage

for example if we got remaining 50 cards

which are 16 A , 8 J ,8 Q , and all other cards got 2 for each

whats the probability we can get pairs for such case and any other combination

can anyone write a program to show the probability to get pairs for different combination of remaining cards? i believe for uneven numbers for some of the cards , the pairs will come out more often which makes an advantage for players

Sonny · Apr 1, 2009

beyondbj said:
can any one help me to check how to calculate the advantage , if the probability become 1/10 , it will get 10% advantage

As far as side bets go, this is about as easy as it gets. Just find the number of two-card combinations of each rank, add them all up and divide the sum by the total number of two-card combinations. That will give you the probability of winning the bet for any subset of cards. If you want the house edge, just multiply the number of winning combinations by the payout (11 in this case), subtract the number of losing combinations and divide by the total number of two-card combinations. I'll give an example below.

beyondbj said:
for example if we got remaining 50 cards

which are 16 A , 8 J ,8 Q , and all other cards got 2 for each

whats the probability we can get pairs for such case and any other combination

For that example, we get:

Code:

CARD  QUANTITY    COMBINATIONS
A        16           120
J         8            28
Q         8            28
K         2             1
10        2             1
9         2             1
8         2             1
7         2             1
6         2             1
5         2             1
4         2             1
3         2             1
[U]2         2             1[/U]
TOTAL    52           186  

Non-Pair Combinations: 1140
Total Combinations: 1326

Edge = (11*186-1140)/1326 = 68.33%

That’s a huge advantage! Incidentally, it doesn’t matter what the rank of the cards are. The advantage would be the same if we replaced those ranks with other ranks as long as the grouping is the same.

beyondbj said:
can anyone write a program to show the probability to get pairs for different combination of remaining cards?

Let’s do that now. Computers tend to get upset when you try to use straight factorials (like 52!*51!) so we will use a simplified version of the combination formula. Instead of:

C(x,y) = x!/y!(x-y)!

We will simplify the formula for this specific case:

C(x,2) = (x*(x-1))/2
or:
C(x,2) = (x*x-x)/2

The pseudo-code for our program might look like this:

Code:

Pairs=0
for (i=1; i<=ranks; i++){
   Pairs+=(numCardsRank(i)* numCardsRank(i)- numCardsRank(i))/2
}

totalHands=(numCards*numCards-numCards)/2
nonPairs=totalHands-Pairs
Edge=(11*Pairs-nonPairs))/totalHands

I used a similar routine as the heart of my simulation program. This could also be used in a spreadsheet or done by hand if necessary.

-Sonny-

when can we get advantage to buy pairs ?

beyondbj

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ricopuno

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Sonny

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ricopuno

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Sonny

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ricopuno

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Sonny

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FLASH1296

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ricopuno

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FLASH1296

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Sonny

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ortango

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FLASH1296

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Tarzan

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Katweezel

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beyondbj

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Sonny

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