Ask Mission

My Answer:

I don't know how many decks you're playing, so we'll assume eight. If we look here:

We see that the tie bet is supposed to win 9.5156% of the time and pays eight units on a win. Assuming you bet $10 every single time you play a hand, here are the actual results based on your percentages:

(355 * 80) - (2887 * 10) = -470

Therefore, you would actually still be down with flat betting $10 despite the number of times that the Tie has come up. With that being said, since you are using a system, I would assume that you are not flat betting the Tie every single hand. You're certainly not ahead because of your system, just when it so happens you have been betting more, but the Tie running better than expected has certainly not hurt you.

Also, are you playing somewhere that pays 9-TO-1 on the Tie bet? My understanding is that some online casinos do that and perhaps there is a land casino somewhere that offers that. If that's the case:

(355 * 90) - (2887 * 10) = $3080

So, you would be ahead flat betting in that situation given how well the Tie bet has been coming in.

As far as the probability of your tie bet hitting as much as it has goes, we can use a binomial probability calculator to get an idea how likely that is:

Probability: .095156

Trials: 3242

Successes: 355

And we get: 0.00298 so the probability of the Tie bet coming up this frequently (or more frequently) in that sample is 0.298% or 1 in 335.57

Therefore, we can conclude that the Tie bet has been running extremely well for you. It's running ever so slightly better than three standard deviations to the good. If one less Tie bet had come, then you would be within the third standard deviation.

So, it just depends on how much the Tie bet pays. If it pays 8-TO-1, it just so happens that you're betting it (or betting more) at fortunate times. If it pays 9-TO-1 and you were flat betting it, you would be winning regardless.

ALSO:

For the purposes of advantage play, the Tie Bet at Baccarat is effectively not countable. However, bets in Baccarat do have a variable house edge based upon the composition of the remaining cards. I tend to doubt it, but it is possible that whatever decision-making method you're using to place the tie bets might also correlate with the tie bets having a lower house edge than off of the top of a fresh shoe.

Even then, I would suggest that the ties running 3SD+ to the good is undoubtedly helping you. There's really no way to attribute that to your system, method, whatever you want to call it...whatsoever. That's like having a coin toss calling system that relies on selectively calling heads (never tails), having 65/100 flips (not all of them bet) come up heads and then attributing financial success to the system. Granted, you'd still be down flat betting the tie under this 8-TO-1 scenario, but you certainly woulkd not be down nearly as much as expected given that you are running 3SD+ to the good.

If anything, that should be evidence of how fortunate you've been running on this high house edge bet. Even with this kind of a sample size, if the tie was being flat bet every single hand AND running 3SD+ to the good, it would still be losing. That tells you it's an awful bet.

What you would do is this: (Number of wins on tie * 80) - (Number of losses on tie * 10) = Profit/Loss

Otherwise:

(Number of wins on tie * 80) - ((Number of times tie bet placed - Number of wins on tie) *10) = Profit/Loss

As far as only flat betting the ties is concerned, if ties are running 3SD+ to the good overall in your limited sample size AND you're not betting it every time, then your sample size of tie bets placed is effectively smaller. And, again, your method for choosing when to bet the tie might have a lower house edge than, "Normal," if it correlates somehow with a deck composition that would be, "Good," for the tie bets. Either way, tie bets are overall running far better than expected, so there can be no doubt that is helping you tremendously.

CONCLUSION: If you’re running well on the Baccarat Tie Bet, you’re just getting lucky. It has a HUGE house edge. Again, it is technically possible to count the Baccarat Tie Bet, but any such efforts would be fruitless from a financial standpoint because the deck composition is such that positive opportunities almost never present. There is even a guaranteed way to win, if all of the remaining cards are 0 value cards (tens and faces) the result is guaranteed to be a tie, but that almost never happens. Casinos wouldn’t let you write down results at the table if it did.

Mission146 Answer:

Technically, it's not, "ONLY,' from a new shoe, it just usually is. The reason why it always happens from a new shoe is because the cards start out equally distributed, which means the same amounts of each rank from a new shoe. That's why it happens regardless of the number of decks in a shoe, because the number of each rank is perfectly distributed before you remove the card.

That's why I say it would usually be from the beginning of a shoe. It's possible for the distribution of the ranks to be equal later on in the shoe, as well, it just doesn't happen very often.

Let's look at 32 cards left in the bacarrat deck distributed as follows:

A-3 (3/32 * 2/31) = 0.00604838709

2-3 (3/32 * 2/31) = 0.00604838709

3-3 (3/32 * 2/31) = 0.00604838709

4-3 (3/32 * 2/31) = 0.00604838709

5-3 (3/32 * 2/31) = 0.00604838709

6-3 (3/32 * 2/31) = 0.00604838709

7-1---0

8-2 (2/32*1/31) = 0.00201612903

9-1---0

10-3 (3/32 * 2/31) = 0.00604838709

J-2 (2/32*1/31) = 0.00201612903

Q-2 (2/32*1/31) = 0.00201612903

K-3 (3/32 * 2/31) = 0.00604838709

Total Pair Probability: (0*2) + (0.00604838709*8) + (0.00201612903 * 3) = 0.05443548381

Okay, so the difference between this and a, "Fresh shoe," is that not all of the ranks are equally likely to pair to begin with prior to removing one of the cards. For example, the 7's and 9's cannot pair, because there is only one of each. Therefore, if we remove a 7 or 9, the pair probability is going to go up because we're taking away a card that screws up pairing.

Removing one of the cards that there are only two of (8's, J's, Q's) is going to cause the overall pair probability to go up also, as you will see, because the result is three remaining ranks that cannot pair...but the 8's were not all that likely to pair to begin with. I guess we'll just have to see what happens when we remove one of the cards from the ranks there are three of.

Seven Removed

A-3 (3/31 * 2/30) = 0.0064516129

2-3 (3/31 * 2/30) = 0.0064516129

3-3 (3/31 * 2/30) = 0.0064516129

4-3 (3/31 * 2/30) = 0.0064516129

5-3 (3/31 * 2/30) = 0.0064516129

6-3 (3/31 * 2/30) = 0.0064516129

7-0----0

8-2 (2/31*1/30) = 0.00215053763

9-1---0

10-3 (3/31 * 2/30) = 0.0064516129

J-2 (2/31*1/30) = 0.00215053763

Q-2 (2/31*1/30) = 0.00215053763

K-3 (3/31 * 2/30) = 0.0064516129

Total Pair Probability: (0.0064516129*8) + (0*2) + (3*0.00215053763) = 0.05806451609

As predicted, the total pair probability increased, because we removed a card that could not pair anyway. If you were betting on a pair to come, we took away a, "Bad Card," leaving open more possibilities of pairing.

Eight Removed

Now, we are going to remove an eight from the deck, which will leave only one eight. Because of that, the probability of pairing should drop a little bit because there are now three cards that could come out and make a pair impossible....at least, one would think...

A-3 (3/31* 2/30) = 0.0064516129

2-3 (3/31* 2/30) = 0.0064516129

3-3 (3/31* 2/30) = 0.0064516129

4-3 (3/31* 2/30) = 0.0064516129

5-3 (3/31* 2/30) = 0.0064516129

6-3 (3/31* 2/30) = 0.0064516129

7-1---0

8-1---0

9-1---0

10-3 (3/31* 2/30) = 0.0064516129

J-2 (2/31*1/30) = 0.00215053763

Q-2 (2/31*1/30) = 0.00215053763

K-3 (3/31* 2/30) = 0.0064516129

(0.0064516129*8) + (0*3) + (2*0.00215053763) = 0.05591397846

That's interesting, isn't it? The probabilities of any individual pair remained the same as a seven being removed (except 8's, which became impossible) but our overall probability of pairing went UP compared to before we removed a card. Again, that's because the 8's weren't terribly likely to pair to begin with, as there were only two of them...so the likelihood of the ranks of three each (and remaining ranks of two each) goes up more than enough to compensate.

Ace Removed

A-2 (2/31*1/30) = 0.00215053763

2-3 (3/31* 2/30) = 0.0064516129

3-3 (3/31* 2/30) = 0.0064516129

4-3 (3/31* 2/30) = 0.0064516129

5-3 (3/31* 2/30) = 0.0064516129

6-3 (3/31* 2/30) = 0.0064516129

7-1---0

8-2 (2/31*1/30) = 0.00215053763

9-1---0

10-3 (3/31* 2/30) = 0.0064516129

J-2 (2/31*1/30) = 0.00215053763

Q-2 (2/31*1/30) = 0.00215053763

K-3 (3/31* 2/30) = 0.0064516129

Total Pair Probability: (0.0064516129*7) + (0.00215053763*4) + (2*0) = 0.05376344082

In this case, the overall pair probability went DOWN compared to the starting state. The reason for this is because we drastically reduced the probability (tied for highest) of the aces pairing up and the remaining card pair probabilities simply did not increase enough to compensate.

Conclusion

Anyway, I did this with a more extreme example this time in order to highlight the difference and have the change in overall probability being recognizably meaningful without having to go more than a few decimal places to see the difference.

As you remove the cards, you no longer have a uniform starting state in which all pairs are equally likely. Removing cards that are more likely (composition) based on the number of those cards left to pair will generally cause the overall pair probability to go down. If not always, then generally. Removing cards that cannot pair (if only one card of that rank) or cards that are less likely to pair will generally cause the overall pair probability to go up.***

***It might always cause it, I just didn't feel like doing an eight-deck example to see if there are exceptions.

In Your Example

In your example, you can either remove another king as the second card to be removed, or you can remove a card that is not a king. Let's see what happens in each:

A 32 0.005746 A 32 0.005774

2 32 0.005746 2 32 0.005774

3 32 0.005746 3 32 0.005774

4 32 0.005746 4 32 0.005774

5 32 0.005746 5 32 0.005774

6 32 0.005746 6 32 0.005774

7 32 0.005746 7 32 0.005774

8 32 0.005746 8 32 0.005774

9 32 0.005746 9 32 0.005774

10 32 0.005746 10 32 0.005774

J 32 0.005746 J 32 0.005774

Q 32 0.005746 Q 32 0.005774

K 32 0.005746 K 31 0.005413

416 0.0746988 415 0.0746988

A-31 (31/414 * 30/413) = 0.00543916903

2-32 (32/414 * 31/413) = 0.0058017803

3-32 (32/414 * 31/413) = 0.0058017803

4-32 (32/414 * 31/413) = 0.0058017803

5-32 (32/414 * 31/413) = 0.0058017803

6-32 (32/414 * 31/413) = 0.0058017803

7-32 (32/414 * 31/413) = 0.0058017803

8-32 (32/414 * 31/413) = 0.0058017803

9-32 (32/414 * 31/413) = 0.0058017803

10-32 (32/414 * 31/413) = 0.0058017803

J-32 (32/414 * 31/413) = 0.0058017803

Q-32 (32/414 * 31/413) = 0.0058017803

K-31 (31/414 * 30/413) = 0.00543916903

Total Pair Probability: (0.00543916903*2) + (11* 0.0058017803) = 0.07469792136

Because we took away a card more likely to pair, as opposed to less likely to pair, the any pair probability went down very slightly (because there are so many cards) compared to your 415 card state with only one king removed. Let's see what happens when we return our ace and just remove a second king:

A-32 (32/414 * 31/413) = 0.0058017803

2-32 (32/414 * 31/413) = 0.0058017803

3-32 (32/414 * 31/413) = 0.0058017803

4-32 (32/414 * 31/413) = 0.0058017803

5-32 (32/414 * 31/413) = 0.0058017803

6-32 (32/414 * 31/413) = 0.0058017803

7-32 (32/414 * 31/413) = 0.0058017803

8-32 (32/414 * 31/413) = 0.0058017803

9-32 (32/414 * 31/413) = 0.0058017803

10-32 (32/414 * 31/413) = 0.0058017803

J-32 (32/414 * 31/413) = 0.0058017803

Q-32 (32/414 * 31/413) = 0.0058017803

K-30 (30/414 * 29/413) = 0.0050882549

Total Pair Probability: (0.0050882549) + (0.0058017803*12) = 0.0747096185

It went up! And, again, the reason it went up is because, though the probability of getting a pair of kings was reduced, the probability of any other pair coming went up more than enough to compensate.

On and on forever and ever this will continue...with the only exception being a, "Fresh shoe," or a naturally occurring (though rare) state in the current shoe where all ranks are evenly distributed prior to the removal of a card again.

Final Conclusion

The fresh shoe is NOT the cause. The fact that all ranks are equally distributed prior to one card being removed is the cause. It just happens to be a characteristic of a fresh shoe that all ranks are equally distributed before a card is removed.