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## Questions about Player Advantage and 'Next Card is An Ace' Math

Apologies as I have asked one of the 2 following questions previously on this Forum or another Forum, but I can't find what I am looking for.

1) What is my player advantage if I know that whenever there is a '6' card, an Ace will be the very next card 15.1% of the time? Isn't the answer something like (51% X .151) minus something?

2) What is my player advantage if I know that whenever there is a TEN, an Ace will be the very next card 15.1% of the time? I am wondering if this scenario is different from 1) above simply because there are 4 more times that I may be correct in predicting an Ace only due to chance because there are 4 times more 10s than there are 5s, or am i engaging in erroneous thinking?

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Originally Posted by Overkill
Apologies as I have asked one of the 2 following questions previously on this Forum or another Forum, but I can't find what I am looking for.

1) What is my player advantage if I know that whenever there is a '6' card, an Ace will be the very next card 15.1% of the time? Isn't the answer something like (51% X .151) minus something?

2) What is my player advantage if I know that whenever there is a TEN, an Ace will be the very next card 15.1% of the time? I am wondering if this scenario is different from 1) above simply because there are 4 more times that I may be correct in predicting an Ace only due to chance because there are 4 times more 10s than there are 5s, or am i engaging in erroneous thinking?

Much like economics: "It depends."

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Of course, sorry!

Double deck, h17, 3:2. Hopefully that is enough info. to get me a ballpark answer to my 2 questions above.

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Can anyone share some insight regarding my 2 questions above? It would be much appreciated.

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Originally Posted by Overkill
Can anyone share some insight regarding my 2 questions above? It would be much appreciated.
I can do the simulations as below :-

1) Remove an ACE from double deck, shuffle those remaining 103 cards.
2) Insert randomly that ACE between 1st and 6th card.
3) Play head up with dealer using basic strategy and record the win/loss
4) Shuffle after each round.
5) Repeat above for million rounds to calculate the EV

Is that what you want ? Please note that I did the similar sim for my own game.

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Originally Posted by James989
Is that what you want ? Please note that I did the similar sim for my own game.
No. He said whenever there is the RANK of 6 as a card appearing, an ace follows it as the next card. Or, whenever there is a 10-value card card appearing. But, it isn't at all clear to me what he means by "an ace will be the very next card." Where does the 6 appear? Am I playing alone at the table? Was the 6 the last card out of the pack, such that I now know I'm going to get the ace? The question needs to be clarified.

Don

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Originally Posted by DSchles
No. He said whenever there is the RANK of 6 as a card appearing, an ace follows it as the next card. Or, whenever there is a 10-value card card appearing. But, it isn't at all clear to me what he means by "an ace will be the very next card." Where does the 6 appear? Am I playing alone at the table? Was the 6 the last card out of the pack, such that I now know I'm going to get the ace? The question needs to be clarified.

Don
Thanks for clarification.

Then I agree with your question : Where does the 6 appear?

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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.

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I guess you're playing a S17 game ... (?)

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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.

This would surely have to be simulated. I can't imagine how to go about solving it analytically. It's one thing to be playing alone and to have the last card of the prior round be a 6, thereby knowing that the Ace could be yours on the next round. But, what you're describing goes WAY beyond that.

Don

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