Beating Perfect Pairs by Red/Black Count
Some people are selling a red/black count system for perfect pairs bet. This side bet pays on 6 deck games. 30:1 is paid if your two first cards are identical (same value and suit). 10:1 is paid if they are same value same colour. 5:1 is paid for same value. KQ does not consitute a pair of tens etc it has to be a proper pair. The payout system is slightly different for the 21 variant of blackjack where there are no ten cards.
The count systems claim that because perfect and coloured pairs pay the most and are colour dependant, keeping track of the red to black ratio of the shoe allows you to bet at the right time and make money.
Anyway if counting red/black I have determined the ratios needed to achieve 0% EV (break even point), this shows that a red/black count cannot beat the side bet pratically, so those count systems they are selling must be fake:
The basic probabilities are:
Perfect pair: 5/311 = 1.61% Pays 30
Color pair: 6/311 = 1.93% Pays 10
Mixed Pair: 12/311 = 3.86% Pays 5
Loose bet: 100 - (1.61+1.93+3.86) = 92.60% Pays -1
EV = -5.79%
I then adjust the probabilities based on the ratios of red to black and find what ratio is needed to acieve EV = 0%. Here are the results:
Decks Left: Ratio Needed: Chance: Card outs:
4.25 (infinity) 2.170E-36% 0
4 16.3:1 1.482E-31% 6
3.5 4.9:1 6.64E-24% 22
3 3.3:1 5.24E-22% 36
2.5 2.5:1 1.16E-19% 52
2 2:1 1.11E-18% 68
1.5 1.7:1 4.86E-17% 86
1 1.5:1 2.65E-18% 104
[Ratio is ratio of red to black, or black to red (which ever is greater) that have come out of the shoe]
[Cards out is the number of red or black cards that have now come out (which ever color is fewer). Fewer than this number means a player Advantage exists.]
It is clear all those situations are next to impossible. With 4 decks left, you would need to have seen just 6 cards of 1 colour! With 1 deck left you pratically need every remaining card to be one colour! Accumulated Hyper geometric probability distribution was used to calculate the chance that such rare situations (or better situations) occur in a shoe. As seen in the chance values they are nearly impossible to occur.
In conclusion a red/black count like a high/low count cannot beat the perfect pairs side bet practically since you would need to play zillions of shoes before you see a profitable situation.
Also because perfect pair is 5/311 and colored is 6/311 its unlikly that an 'suit count' could beat the perfect pairs either (practically of course).
Also some people are saying High/Low count can effect the EV since if its all high, there is more chance at getting a pair of high cards, and if more low there is more chance at getting a low pair. A high card will increase its presence in the deck by factor of (1+(C/20*1/5) where is C is true count. A low card by (1-(C/20*1/5). Any card in general would be (1+-(C/20*0/13)) = 1. So the high/low count has no effect on the EV of the perfect pairs bet. This is obvious as if there are too many high cards, then chance of high pair has not gone up, because high cards of other values have just replaced the low values, and the probabilitiy of getting a high pair is offset by the propability of not getting a low pair.
Some people are selling a red/black count system for perfect pairs bet. This side bet pays on 6 deck games. 30:1 is paid if your two first cards are identical (same value and suit). 10:1 is paid if they are same value same colour. 5:1 is paid for same value. KQ does not consitute a pair of tens etc it has to be a proper pair. The payout system is slightly different for the 21 variant of blackjack where there are no ten cards.
The count systems claim that because perfect and coloured pairs pay the most and are colour dependant, keeping track of the red to black ratio of the shoe allows you to bet at the right time and make money.
Anyway if counting red/black I have determined the ratios needed to achieve 0% EV (break even point), this shows that a red/black count cannot beat the side bet pratically, so those count systems they are selling must be fake:
The basic probabilities are:
Perfect pair: 5/311 = 1.61% Pays 30
Color pair: 6/311 = 1.93% Pays 10
Mixed Pair: 12/311 = 3.86% Pays 5
Loose bet: 100 - (1.61+1.93+3.86) = 92.60% Pays -1
EV = -5.79%
I then adjust the probabilities based on the ratios of red to black and find what ratio is needed to acieve EV = 0%. Here are the results:
Decks Left: Ratio Needed: Chance: Card outs:
4.25 (infinity) 2.170E-36% 0
4 16.3:1 1.482E-31% 6
3.5 4.9:1 6.64E-24% 22
3 3.3:1 5.24E-22% 36
2.5 2.5:1 1.16E-19% 52
2 2:1 1.11E-18% 68
1.5 1.7:1 4.86E-17% 86
1 1.5:1 2.65E-18% 104
[Ratio is ratio of red to black, or black to red (which ever is greater) that have come out of the shoe]
[Cards out is the number of red or black cards that have now come out (which ever color is fewer). Fewer than this number means a player Advantage exists.]
It is clear all those situations are next to impossible. With 4 decks left, you would need to have seen just 6 cards of 1 colour! With 1 deck left you pratically need every remaining card to be one colour! Accumulated Hyper geometric probability distribution was used to calculate the chance that such rare situations (or better situations) occur in a shoe. As seen in the chance values they are nearly impossible to occur.
In conclusion a red/black count like a high/low count cannot beat the perfect pairs side bet practically since you would need to play zillions of shoes before you see a profitable situation.
Also because perfect pair is 5/311 and colored is 6/311 its unlikly that an 'suit count' could beat the perfect pairs either (practically of course).
Also some people are saying High/Low count can effect the EV since if its all high, there is more chance at getting a pair of high cards, and if more low there is more chance at getting a low pair. A high card will increase its presence in the deck by factor of (1+(C/20*1/5) where is C is true count. A low card by (1-(C/20*1/5). Any card in general would be (1+-(C/20*0/13)) = 1. So the high/low count has no effect on the EV of the perfect pairs bet. This is obvious as if there are too many high cards, then chance of high pair has not gone up, because high cards of other values have just replaced the low values, and the probabilitiy of getting a high pair is offset by the propability of not getting a low pair.
