Micro-Counting - Denting the House Edge?

Hi BlackJack enthusiasts,

I wanted to clarify that the use of basic strategy assumes that the casino is using an infinite deck?

I have this hypothesis on Micro Counting, whereby you use the info that you can see on the table to play accordingly, and I want to see whether this theory is plausible? An exaggerated example would clarify what I mean.

Assume the casino uses CSM to prevent counting and ace tracking. There are seven spots on the table, and you play on all of them. Same bet sizes. Cards come out and your cards are four 2's, four 3's, four 4's, two 5's, and the dealer has a 6. With Basic Strategy you should hit until you get anything above a 12 in most rules. But because the cards dealt are all small cards, the probability of getting an 8 or above (two 8's or above = bust for dealer) increases, especially compared with an infinite deck. Of course the more decks used, the smaller the difference in odds. However, given that the casino uses 4 or less decks, would it be wise to stand on all hands (even when hitting is riskless) because the probability of getting 8's or above is high for the dealer than compared to normal situations, thus creating a much higher likelihood for the dealer to bust?

I was thinking during these statistical deviances it'd be wiser to stand in most cases than hit, and by taking advantage of this situation we'd effectively reduced the house edge.

Thank you for your help!

First, your use of the word 'infinite' is incorrect. I think you meant to say just the opposite, a finite number of decks, whether it be 1 or 4.

Answer to your question depends on just what the finite number of decks is. If it was say 4 or greater, your running count of +15 translates to a TC of 4 or less, which isn't going to make a big enough difference to consider what you are thinking. Now if it was a single deck, it that would mean a TC of +15 (actually TC about 20 because 1/4 deck has been used), what you are proposing might have some merit. I am wondering though if the TC was great enough that the best play is standing on your 2 cards no matter what, because of a higher chance of dealer break, then wouldn't your best play be to double? I mean if it is +EV to stand, wouldn't it be nearly double the +EV to double? And then of course it gets even fuzzier if any of those four 2's, 3's, or 4's where paired together, giving an option of splitting.

Quote: ttactics

Hi BlackJack enthusiasts,

I wanted to clarify that the use of basic strategy assumes that the casino is using an infinite deck?

Nope

Quote:

I have this hypothesis on Micro Counting, whereby you use the info that you can see on the table to play accordingly, and I want to see whether this theory is plausible? An exaggerated example would clarify what I mean.

Assume the casino uses CSM to prevent counting and ace tracking. There are seven spots on the table, and you play on all of them. Same bet sizes. Cards come out and your cards are four 2's, four 3's, four 4's, two 5's, and the dealer has a 6. With Basic Strategy you should hit until you get anything above a 12 in most rules. But because the cards dealt are all small cards, the probability of getting an 8 or above (two 8's or above = bust for dealer) increases, especially compared with an infinite deck. Of course the more decks used, the smaller the difference in odds. However, given that the casino uses 4 or less decks, would it be wise to stand on all hands (even when hitting is riskless)

Why would you stand if hitting is riskless? If hitting is riskless then it is strictly better than standing in all situations.

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because the probability of getting 8's or above is high for the dealer than compared to normal situations, thus creating a much higher likelihood for the dealer to bust?

Sure, it's higher than usual. It remains higher than usual if you hit.

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I was thinking during these statistical deviances it'd be wiser to stand in most cases than hit, and by taking advantage of this situation we'd effectively reduced the house edge

You're increasing the house edge if you stand when hitting is riskless.

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Thank you for your help!

No problem.

@ kewlj,

Hmm sorry I am very new to this +EV thing so let me give you a bit of my thought process instead.

Let's say single deck, and the cards come out like I illustrated in the example. In this optimistic example, the odds of hitting an 8 or greater become 24/37 for the dealer. The problem with doubling and splitting is because we may risk reducing the dealer's chances of getting an 8 or above consecutively. So if we doubled all 7 hands and received a ten for each of the hand, we've effectively reduced the odds of the dealer drawing an 8 or above to 17/30. The aim of this concept is to ensure that the dealer busts. The question that needs to be asked is - does doubling each hand in this situation, given that these hands are more than likely going to result in less than 17 in most cases, justify the 8.1% reduction in the dealer getting an 8 or greater?

Of course, the other way can be true. If we happen to draw all the cards that are small instead, we've raised the chances of the dealer busting to 24/30.

And what I was thinking was that maybe it is optimal to double only in certain sums (5+6 = 11) and not other sums that has a high probability of resulting less than 17 (such as sums of 5's and 6's). Or perhaps you can keep doubling / splitting each hand until the % of the dealer getting an 8 or above is below 60%.

Thank you for your feedback.

Quote: ttactics

Let's say single deck, and the cards come out like I illustrated in the example. In this optimistic example, the odds of hitting an 8 or greater become 24/37 for the dealer. The problem with doubling and splitting is because we may risk reducing the dealer's chances of getting an 8 or above consecutively.

Unseen cards are unseen cards. You haven't seen the burn card or the dealer's hole card, and you have two 5's unaccounted for. Impressive that you found a 7 spot single deck face up table, though - I thought most were 6 spot and face down, specifically to prevent this type of play.

A 5 under the dealer's 6, and the dealer drawing a ten isn't a great situation.

I believe Thorp (and probably several others) mentioned this sort of play (look up "end decking" or "end play"). My understanding is the first few hands are definitely draw (or split), so you can see a few more cards. If you find the two previously unseen 5's, your odds are better for doubling the later hands.

You've still got 6's and 7's and Aces in play. 6,7,7 = 20, which is going to be rough against a lot of your hands.

Quote: ttactics

Let's say single deck, and the cards come out like I illustrated in the example. In this optimistic example, the odds of hitting an 8 or greater become 24/37 for the dealer. The problem with doubling and splitting is because we may risk reducing the dealer's chances of getting an 8 or above consecutively. So if we doubled all 7 hands and received a ten for each of the hand, we've effectively reduced the odds of the dealer drawing an 8 or above to 17/30. The aim of this concept is to ensure that the dealer busts.

That is incorrect logic. The "bust cards" are just as likely to be 5 cards down as they are to be coming up next. Hitting does not change the dealer's a priori chances of busting. Of course after you hit the probability will be different, but that doesn't matter.

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The question that needs to be asked is - does doubling each hand in this situation, given that these hands are more than likely going to result in less than 17 in most cases, justify the 8.1% reduction in the dealer getting an 8 or greater?

There is no reduction. Sometimes when you take a card the dealer's chances of busting go up, and sometimes they go down. In the end it all evens out. If the dealer busts 24/37 of the time when you don't hit, he will bust 24/37 of the time when you do hit.

Consider a much simpler problem. Imagine that there are two tickets, a losing ticket and a winning ticket. We will each choose a ticket out of a hat. You see that your chances of winning are the same (50%) whether you pick first or 2nd, right? If you pick 2nd, then, at the time of your pick, sometimes you have a 100% chance of winning, and sometimes 0%, but it still averages out to 50%.

@AxiomOfChoice

Thanks for the feedback.

I can understand in games like Roulette where the last 5 colors were red, the odds of hitting the red or black remain equal (the past doesn't affect the future). In the long run, they both will average slightly less than 50%, but will have equal %.

But for this example, this is how I see the problem...
because the 15 cards dealt are all smaller than 8, we are left with 24/37 cards that are 8 or bigger. In this situation, the past (the cards that are dealt) does affect the future (the ongoing play). I'll try to explain it in a simple example, simplifying to 24/36, so this is equivalent to 2/3.

Explanation for more likelihood of "bust cards" to appear next.
You have 3 cards. Two of them are black, one of them is red. You shuffle them and write down each shuffle whether the top card is black or red. Sure there are times when the top card is red, but in the long run the probability that the top card is black is 66.7%, while the red is 33.3%.

Why reduction happens:
Very simple game where a friend has to pick a card. Again we have 3 cards, 2 are black and 1 is red. If he picks a black he loses. You have the option to randomly pick and take away a card first. If you don't pick a card, the odds of your friend choosing a black card is 66.7%. Reduction does occur when you do randomly pick a card and it happens to be black. Your friend now has better odds, red and black are both 50%.

Quote: ttactics

Why reduction happens:
Very simple game where a friend has to pick a card. Again we have 3 cards, 2 are black and 1 is red. If he picks a black he loses. You have the option to randomly pick and take away a card first. If you don't pick a card, the odds of your friend choosing a black card is 66.7%. Reduction does occur when you do randomly pick a card and it happens to be black. Your friend now has better odds, red and black are both 50%.

So you think that if he picks the 2nd card from the top instead of the top card, he has a better chance of winning? That is ridiculous.

@Dieter

Sorry I didn't clarify the scenario enough. The casino does not have dealer's hole card. After dealing the first two cards for each player's spot on the table and the dealer's card, there is no card underneath the dealer's single card. So effectively the dealer's second card is only dealt after all the players have acted.

If the table has a dealer's hole card involved, this strategy instantly does not work, this I understand as you have demonstrated.

Yes, and you are right that there are 6's and 7's and Aces involved, this is not an arbitrage. But I believe that the probability of getting an 8 or above consecutively is greater than the probability that the dealer will get a sum that will equal to 17-21 (if someone can mathematically prove this whether to be correct or not that would be great!) given that the remaining cards are so heavily weighted towards 8 or above. So given an infinite times that this situation occurs, in the long run this will be advantageous to the player who stands all his hands rather than play by basic strategy, giving a higher than normal odds for the dealer to bust.

Thanks for the references. I think with using the first few hands to see more cards, that is also a good strategy for some situations. However for my situation, because we understand that the deck is about 2/3 stacked with cards worth 8 or higher, we have enough information as we already do.

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In fact I want to continue on with what kewlj about doubling and such.

The problem with doubling and splitting, as I mentioned in my previous reply to him/her, is that we are reducing the dealer's odds of getting an 8 or above (bust cards). The odds that we are dealt with a bust card is 2/3. So my question is, does doubling or splitting justify the reduction of odds for the dealer of getting an 8 or above when we know that the chances of getting a bust card for ourselves is 2/3?

Quote: ttactics

If the table has a dealer's hole card involved, this strategy instantly does not work, this I understand as you have demonstrated.

The strategy does not work anyway. It does not matter whether the dealer starts with the top card, the 2nd card, or the 3rd card. His chances of busting are the same.

Quote: ttactics

Why reduction happens:
Very simple game where a friend has to pick a card. Again we have 3 cards, 2 are black and 1 is red. If he picks a black he loses. You have the option to randomly pick and take away a card first. If you don't pick a card, the odds of your friend choosing a black card is 66.7%. Reduction does occur when you do randomly pick a card and it happens to be black. Your friend now has better odds, red and black are both 50%.

Let's just work through the math of this.

If you don't remove a card, his chances of winning are 1/3.

If you do remove a card:

1/3 of the time you remove his only winner and he wins 0% of the time.
2/3 of the time you remove one of his losers and he wins 50% of the time.

So his chances of winning are (1/3 * 0%) + (2/3 * 50%) = 0 + 1/3 = 1/3, which are the same as his chances of winning if you don't remove a card.

It's obvious that it has to work out this way, because it doesn't matter whether he picks the top card (which is what happens when you don't remove one) or the 2nd card (which is what happens when you do remove one). Since the cards are shuffled, both the top card and the 2nd card have the same probability of being black.