Background
I have been giving much thought to the question as to whether it is possible to gain an edge at VBJ by maintaining a “pseudo” card-count on a machine which uses a virtual multi-deck shoe with an unknown shuffle point. The “pseudo” count would begin when the players first sits down at the machine, though of course the player would have no idea if or when a shuffle has taken place during the count.

My initial thoughts were as follows:

• A pseudo-count contains some information, and thus should allow a player to perform better than with basic strategy alone.
• Conversely the pseudo-count does not contain as much information as a genuine count, and thus a player cannot do as well as if he knew the genuine count
• The optimum strategy may not be the same as that for one using a genuine count

The first question is whether it is possible for a pseudo count to contain enough information to make play +EV. The conclusion I have come to is that, at least in extreme circumstances, then yes it can.

Definitions
Before I start here are a few definitions. I am considering a 6-deck shoe where the shuffle occurs after four out of six decks have been dealt. I am considering the standard hi/lo count where a card of 2, 3, 4, 5 or 6 is assigned a value of +1, and a card of T, J, Q, K, A is assigned a value of -1. You maintain a running count, which is expressed as what is known as the “true count” by dividing by the number of decks currently undealt/unseen. This “true count” corresponds to the density of high cards left in the deck. I will be using the term “genuine true count” to refer to this quantity, and the term “pseudo true count” to refer to the equivalent quantity calculated from a shoe with unknown shuffle point.

Roughly speaking (depending on precise rules played) a genuine true count of zero corresponds to a house advantage of 0.5%, and each one point change in true count swings the advantage up or down by 0.5%. Thus if 6 decks remain to be dealt then play will be +EV provided the running count is greater than +6 (and thus the true count is greater than +1).

Example
So on to the example, and as I indicated it is going to be an extreme example. Suppose you had been maintaining a pseudo running count for 10 cards, and so far every card had been a low card. You currently have a running pseudo-count of +10, equating to a pseudo true count of around +1.7.

Now what could the genuine counts be? It is important to observe that if there has not been a shuffle during these 10 cards then your pseudo count is mathematically equivalent to a genuine count starting on a new shoe – it does not matter whether you have “missed” earlier cards or if they are still in the shoe.

Now the chances of there being a shuffle somewhere during the deal of 10 cards will only be something like 10/208, or around 5%. So it is actually pretty likely* this count can be considered genuine and thus playing the next hand will be +EV.

And what if there had been a shuffle during those 10 cards? Well, it might have occurred during the first two or three of those 10 cards, in which case the genuine running count should only be perhaps 7 or 8. But this corresponds to a genuine true count of 1.2 or 1.3 which means the next hand is still +EV.

The break-even point would come if the shuffle had occurred after around the 4th card (as the genuine running count would only be +6 or less) . The worse case would be if the shuffle came immediately after the 10th card, in which case the genuine true count would be zero. In this case the next hand would have a 0.5% house advantage.

However overall it works out that the chance of the next hand being +EV for the player heavily outweighs the chance it is -EV. Overall it would be a +EV move.

So this gives us a simple but highly impractical system. Walk up to a VBJ machine. If the first 10 cards dealt are all 6 or lower, then play the next hand.

Where next?
Now of course this is highly impractical because the chances of the first 10 cards all being a 6 or lower is extremely remote, so you would not be playing the machine very often. But I think it establishes the principle that it is possible to come up with a +EV system on these machines.

So the debate has moved on from “is it possible to make a +EV play?” to “what is the optimum strategy for +EV play?”.

And remember, it is not like there will be a dealer watching you closely so the system could be pretty wacky.


* The probability of there having been a shuffle within the last 10 cards conditional upon seeing a count of +10 is not quite the same thing as the basic (unconditional) probability calculated here, but the point is illustrative.

Very well thought out outline here. You could have one hell of a discussion with TBC sometime about it...I'd truly be interested in hearing how that turns out.

I have questions.

At what point would someone using this system reset the count? If you are seeing 10 cards per hand and there is a 5% chance of a reshuffle during that 10 cards. after just 5 hands there is a 25% chance there was a reshuffle at some point in those hands.

Since you don't know your starting point. If I come along and see 10 low cards come out and assume therefore I am playing with an advantage I may in fact be incorrect for a number of reasons.

1) there may have been a reshuffle.
2) The count may not have originally been 0 when i started counting (lets assume that in fact the prior hand which I did not see was 10 high cards. Now I see 10 low cards believe that I am playing with with an advantage but in fact the count is now actually 0). Since we don't know the shuffle point we don't know the starting point.


If we assume you are correct .... can you quantify the advantage a player could get (factoring in the effect of the times you get whacked by the reshuffle).

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If liberty means anything at all, it means the right to tell people what they do not want to hear. -- George Orwell

I do not have the answer to this, but I think it is the next thing to work out.

As you start to play you can assume your count is correct, and at first you will probably be right . Provided you set your criterion for playing a hand high enough, then the number of times you are right will outweigh the number of times you are lead astray by a misleading count.

However, the longer you play without resetting, the less reliable your count becomes (as the probability increases that there has been a shuffle). Eventually there comes a point when the penalty/frequency of the count being wrong outweighs the benefit of the times it is correct. I don't know yet where this point comes, but it is something that can be calculated.

There is a trade-off between running the count for long enough to maximize playing opportunities, but not running it too long until the count become too unreliable.

In fact, the idealized "optimum" system would probably use some kind of rolling count, where greater weight was given to the cards seen more recently as they are more likely to have been dealt since the last shuffle. Maybe impossible to put into practice without some kind of rainman-like abilities, though.



Your first issue (i.e. the reshuffle) is outlined above, but actually your second point is not a problem.

Suppose the shuffle had occurred 20 cards before you started counting. Then of course things would be better if you had known what the cards were - but of course you don't. This is actually mathematically no different to those 20 cards still being in the shoe. Provided no shuffle has occurred during the count then your expected values work out the same as if you had started counting from the start of a new shoe.

I'm not following why this does not make a difference.

Your calculation of when you have an advantage is based from starting at a count of 0. If you are in fact not starting at a true count of 0 how do you know when you have an advantage?

_________________
If liberty means anything at all, it means the right to tell people what they do not want to hear. -- George Orwell

I don't play VBJ, I have 3 questions about the player's knowledge of the "game mechanics", specifically on any machine does the player know any or all of the following:

1) How many decks are in the shoe?
2) How deep in the shoe does the machine go before shuffling?
3) When the shuffle occurs?

I would think for a counting system to provide a significant advantage, you would have to know all 3, just like in a live game. If you do know all 3 then you can not accurate reset you count to zero and would almost certainly be resetting your count at the wrong times. Also, since much of the advantage of a count comes deep in the shoe and you have no idea when you would be in that position, you lose a significant portion of the benefits of counting and could actually bet large at inappropriate times.

Take an extreme example. Say you know there are 6 decks and the machine reshuffles 4 decks in. Your reset your count 3.5 decks ago and after a recent run of low cards, now have a pseudo true count of +20, so you decide to bet huge since the count is so heavily in your favor. However, suppose that the machine actually reshuffled after the last deal. The actual count is zero and you would be betting large with no advantage. While you would gain some advantage while your count was escalating at the same time as the real count, much if not all of that advantage would be lost when you bet big with no advantage or even worse with an unfavorable count.

Dave

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The opinions in this post are my own and do not necessarily reflect the views of The Poker Atlas, AVP or PokerTrip Enterprises.

The problem is VBJ is the unknown shuffle point (# of decks and penetration before shuffle is known). Hence this post. TBC was counting the VBJ machines which seeminly made no sense. There have been some debates whether a system can be devised to increase the EV against the machine. I took a crack and came up with a reset strategy after every 2 decks, which theoretically should give the player true counts 2 times out of 3.

AXB, I liked the way you framed the problem here. It is often useful to look at the extremes of the solution space, and then try to figure out the optimized solution and quantify (in this case the change in EV) of the solution. Excellent approach.

I'm thinking out loud here...

Equate the EV shift, resetting after X cards as: counting a deck with X/312 % penetration
Estimate the EV shift (positive or negative) for the count cycle where shuffle occurs.

Does this turn into a polynomial equation that can be solved? If not, we can simply solve the solution by working out the overall EV for each scenario, varing the count reset from 1 to 208. Probably a quick spreadsheet solution can be constructed.

We can futher increase the EV because there is no restriction from starting play mid-shoe on a machine. As you alluded in the OP, you can choose to jump in after a really positive count shows up. Not sure how to quantify such a strategy.

Let me try another example.

You have a jar with 2 red beads and 98 black beads.

a. What is the probability a bead drawn at random is red?
b. You draw a bead out at random, and it is red. You do not replace the bead. What is the probability that the next bead drawn at random is also red?

The answers to a. should be 1 in 50 (there are 100 possible beans to draw, 2 of which are red). The answer to b. should be 1 in 99 (there are 99 beads to choose from, 1 of which is red) . The reason these answers are different is akin to card counting - once red beads have been drawn they are less likely to appear again.

Now lets start again with the full jar. This time you take out 50 beads at random one at a time without looking at them, and throw them away.
c. What is the probability that the next bead drawn at random is red?
d. You draw a bead out at random, and it is red. You do not replace the bead. What is the probability that the next bead drawn at random is also red?

The answers here are still 2 in 100 (1 in 50) and 1 in 99 respectively. If you don't know what the first 50 beads out of the jar are, then the probability of the 51st bead out of the jar being red is still 2 in 100 (1 in 50). And if you know the 51st bead is red, but not the color of any of the others, then the probability that the 52nd bead is red is still 1 in 99.

Finally we re-fill the jar. Now you take out 98 beads out of the jar one-by-one, without looking at any of them, and throw them away.
e. What is the probability that the next bead drawn at random is red?
f. You draw a bead out at random, and it is red. You do not replace the bead. What is the probability that the next bead drawn at random (i.e. the last bead in the jar) is also red?

Yes, you've guessed it, the answers are 1 in 50 and 1 in 99 again.

This is why I say, provided no shuffle has occurred during the count and you do not know the cards that have already gone, then starting your count part-way through a shoe is mathematically equivalent to starting a fresh shoe.

So I have a jar filled with with 100 marbles 2 are red 98 are black. I take out the two red marbles and discard them. what is the chance that the next marble I will draw ill be red. 0%.

Isn't the value of counting coming form knowing the composition of the remaining deck?

If we were counting marbles then I would be able to determine the actual chances of drawing a red marble next only if I knew the original number of red and black marbles and then counted the number of red and black marbles that were removed.

certainly withjout counting I agree that I can say there is a 1 in 50 chance of that marble being red. But the value of the counting process comes from determining when the odds are actually better than 1 in 50 that we start with.

_________________
If liberty means anything at all, it means the right to tell people what they do not want to hear. -- George Orwell

If you took out the two red marbles, then you are right the probability of drawing a red marble from the next draw is 0%.

If you took out two marbles but don't know the colour, then the probability of drawing a red marble from the next draw is still 2%.



That is not true. For example, if someone had randomly removed 50 marbles in your absence. When you draw from the remaining 50 marbles, the chances of drawing a red marble is still 1 in 50. You are probably thinking, what if there are no red marbles in the jar anymore? It is easy to work this out mathematically. I'm too lazy to do that right now, but think about the 3 possible cases (0, 1, or 2 red marbles remain in the jar). You can work out the probability for each case. Now you can figure out the probability of pulling a red marble for each case (0%, 2%, or 4%). Now multiply the probability for each case by the probability of pulling a red marble, add them up, and you end up with exactly 2% (1 in 50).



I think what you are really saying is the value of counting is diminished if you didn't keep track of the marbles removed prior to the start of your count. This is not true. The removed marbles is irrelevant to the probability in the remaining marbles. For all intents and purposes, it is as if the marbles were still in the jar. If you start the count from the 51st marble onwards, you can use the count to figure out the probability of a red one in the remaining marbles.