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Originally Posted by Overkill
Don and James, thanks for taking the time to reply. What I mean is that whenever (absolutely whenever) a 6 is on the felt, the very next card dealt (or about to be dealt out of the shoe as the first card of the next round if a 6 card was the last card of a round) will be an Ace. I can be playing alone at the table, or with any number of people.

For example, walking up to a table with 4 players, I am talking about seeing the 6s on the felt and looking at the card on the felt that the dealer set down (or is about to set down if she is dealing the current round) that follows each of those 6s and predicting it will be an Ace. We know that in the long run, the math tell us that we have a 1/13 chance of seeing an Ace after each 6 (without regard for card composition that has already been dealt). I am hoping to be able to predict an Ace with slightly greater than a 1/13 likelihood.

Hope this clarifies. A simpler way to look at it would be to thumb through 2 decks with the cards not face up (like a double deck shoe waiting to be dealt). First card is a 3, next card is a 4, next card is a 6, hopefully next card is an Ace, next card is a Q, next card is a 3, next card is a 6, hopefully next card is an Ace, next card is a 9, next card is a K, next card is a Q, etc. Hope I didn't make it more confusing just now.

Agree with GMAN, I guess you're playing a S17 game with "WEAK" cards shuffling procedures, so you can predict the ACE when the last card is "6" or "T".

In an ideal situation, suppose the last card of the last round was "6", then what is the probability that the next card(1st card of first hand in next round) is an ACE ? higher than 1/13 ? 1/9 ? 1/8 ?

Assuming probability is 1/9, meaning that ACE will be WITHIN next 9 cards of the shoe ? So you should bet 4 hands in next round in order to catch that ACE ?

I think it can be simulated if you know the correct probability.

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He stated that the probability that the VERY next card would be an ace is 15.1%.

Don

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Originally Posted by DSchles
He stated that the probability that the VERY next card would be an ace is 15.1%.

Don
So 1/7? In order to lower the variance, just bet 3 box in next round, you will be able to catch it 85.7% of the time ? Huge edge ?

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Sorry to not be clear.

I will try to simplify.

Let's say that I am sitting with 5 other players at first base and the final card of the previous round was a 6. Also assume that I believe I am 15.1% certain that the first card of the next round (and the first card of my hand) will be an Ace (because the 6 card was the last card played).

I would like to know my advantage.

We know that, given the double deck rules I specified, if I were 100% certain that my first card were to be an Ace, then I will have a 51.4% (I referenced Eliot Jacobson for this 51.4% figure) advantage for this round if I am seated at first base.

But instead of being 100% certain, I am 'only' 15.1% certain (which is better than we can expect due to chance, which is 1/13 or 7.69%).

Don, you provided me with the explanation some years ago, but I cannot locate it. The solution for my player advantage was straightforward: Something like (.514 X .151) minus (the probability of my first card being something other than an Ace). But I forget how to express that second part mathmatically.

Please disregard the important consideration that the dealer instead of I may receive 'my' Ace. (Hopefully with 5 players this possibility is sufficiently small.).

1) for the scenario of the final card of the last round being a 6 and my belief that I know with 15.1% certainty that the next card (my first card of the new round) will be an Ace and

2) for the scenario of the final card of the last round being a 10 and my belief that I know with 15.1% certainty that the next card (my first card of the new round) will be an Ace. I was thinking this scenario is trickier than the '6' value card scenario because there are four times as many 10s as Aces - does this fact confound or interfere with one using the same method to calculate the advantage that she/he used to calculate the player advantage for the '6' value card scenario because the probability of having an Ace follow by chance a card that appears four times as often (10) is greater?

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Perhaps this is too simplistic... If you KNOW your first card to be delivered is an ace, you edge is 52%. If you are 15% certain, multiply the two and your edge is 7.8%.

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Originally Posted by 21forme
Perhaps this is too simplistic... If you KNOW your first card to be delivered is an ace, you edge is 52%. If you are 15% certain, multiply the two and your edge is 7.8%.
BJ accounting for the largest part of that 52% at 3 times the normal rate essentially is the same info as above. However, the unknown factor is KNOWING that an ace will follow.

Now, the basic strategist KNOWING that an ace will follow would essentially have the 7.8% edge as shown, however, the edge must be (much) greater for the counter KNOWING that an ace will follow - I think that’s the point Don was trying to make - the point is how much.

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So, there are a couple of points to be made. Obviously, we all understand that, if an ace were certain, the edge on that round would be 51%. That's always the case: if your first card is an ace, your edge is 51%. But, as Overkill mentioned, usually, that probability is only 1/13 (7.69%), and it balances out with all the other first-card possibilities to give whatever the BS house edge is, say, 0.5%. So now, we have to rejigger the edges and the probabilities. Without any special knowledge, and to get to -0.5%, we would have 1/13(51%) + (12/13)x = -0.5%. Solving for x, we get -4.79%. So far, so good.

But now, the +51% enjoys an inflated 15.1% probability. So the new calculation for the overall edge becomes: .151(51%) + .849(-4.79%) = edge. Player edge is 3.63%.

Of course, this is the edge only when the 6 (or a ten, for that matter) appears. Naturally, it isn't your overall edge for the game, because, well, sixes appear only 1/13 of the time. So, that 3.63% edge would then be divided by 13 to give your flat-bet advantage of 0.28%. But, it goes without saying that, when the 6 (or ten) would appear, you'd bet a great deal more than your minimum bet. And, it's here that the frequency of the tens (4/13) would come in. That 0.28% would be multiplied by four (1.12%), for your flat-bet edge if you knew aces followed tens, instead of sixes, with 15.1% accuracy.

Think I've covered it all.

Don

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Good explanation, Don.

Moral of the story - improve your ace prediction skills.

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I am curious to know where the 15.1% prediction comes from, and if he bets more than 1 spot to catch the ace does he know the prediction of the Ace falling on the 2nd spot, 3rd spot, etc? The advantage of an Ace is 51-52% on the spot receiving the Ace as first card, when the predictability is 100%...if the predictability is only 15.1% then the advantage is just 15.1% of that.
If the question is framed in the realm of Ace-Location then probably it is best to examine it from that perspective. Usually in 6+deck games the Ace locator uses 2 key cards which he identifies by denomination and suit. It is feasible to use just 1 key card, knowing the denomination and suit, for 1-4 deck games...but at a much reduced advantage. To use just 1 key card without identifying it by suit reduces the advantage even more. Why would one do that?...well, perhaps for cover or for tables with insufficient maximums to take large bets, so might as well place many more smaller bets with less advantage. Whatever the reason, the single key 6 not identified by suit gives only 1/4 of predictable advantage for 1 deck, 1/8 for 2 decks and a miniscule 1/16 advantage for 4 decks....and the single key ten of course is 1/4 of all of those values.

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Thanks so much, Don! That's what I was looking for, and more!

But regarding the 10 cards, dont we need to adjust for the varying frequencies of 10 vs. Aces (whereas we dont have to adjust for the scenario of 6s and Aces). What I mean is, if I believe Aces tend to follow tens, because there arent enough Aces to 'keep up' with the greater number of 10s (vs. Aces), isn't the anticipated 10-Ace 'connection' not as likely to take place due only to an unequal amount of Aces relative to 10s, thereby warranting some sort of 'correction' (perhaps multiplication?)?

Also, Freightman, if we are 100% certain (not just 15.1% certain) our first card will be an Ace, isnt our Advantage 51%, not just 7.8%?

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