Ace Prediction Math

[Overkill]

On average, one expects to see an Ace about once every 13 cards.

1) Assuming 6-deck Strip rules, if one knows with 100% certainty on a specific 'target' hand that (at least) one of her 2 dealt cards will be an Ace, the probability of winning the hand is about 51%, correct?.

But let's now say that rather than knowing with 100% certainty, she 'only' knows with 38% certainty (vs. the probability with which a non-Ace sequencer can expect to encounter, on average, an Ace: 1 out of every 13 cards, or about 7.69%) when, on average, at least one of her 2 dealt cards in 2 (on average) specific 'target' rounds (out of about 35 rounds of heads up play within one 6-deck shoe, Strip rules) will contain an Ace.

2) So, IF WONGING IN FOR ONLY THOSE 2 SPECIFIC TARGET ROUNDS PER SHOE, will the player advantage FOR JUST THOSE 2 ROUNDS per shoe be .51 X .38 = 19.38%?

3) If so, how does one now calculate proper bet size with a 19.38% advantage on 2 rounds per shoe? Should one bet table max? What if one's bankroll is only $200, what would the proper bet be? Do we use Kelly to calculate this bet size as a function of our bankroll, especially because we only have, on average, 2 opportunities to bet big per shoe? As such, variance is very large, correct?

4) So, for 2 specific 'target' rounds per shoe, if she thinks she can predict with, on average, 38% certainty when at least 1 of her 2 dealt cards will be an Ace, is her OVERALL (for ALL rounds; i.e., playing every hand - no Wonging) player advantage something like 2 rounds of betting big at about a 19.38% player advantage per 6-deck shoe offset by about 33 rounds of flat betting at about a -.55% player advantage per 6-deck shoe?

Thank you in advance!

[Cohiba]

On average, one expects to see an Ace about once every 13 cards.

1) Assuming 6-deck Strip rules, if one knows with 100% certainty on a specific 'target' hand that (at least) one of her 2 dealt cards will be an Ace, the probability of winning the hand is about 51%, correct?.

But let's now say that rather than knowing with 100% certainty, she 'only' knows with 38% certainty (vs. the probability with which a non-Ace sequencer can expect to encounter, on average, an Ace: 1 out of every 13 cards, or about 7.69%) when, on average, at least one of her 2 dealt cards in 2 (on average) specific 'target' rounds (out of about 35 rounds of heads up play within one 6-deck shoe, Strip rules) will contain an Ace.

2) So, IF WONGING IN FOR ONLY THOSE 2 SPECIFIC TARGET ROUNDS PER SHOE, will the player advantage FOR JUST THOSE 2 ROUNDS per shoe be .51 X .38 = 19.38%?

3) If so, how does one now calculate proper bet size with a 19.38% advantage on 2 rounds per shoe? Should one bet table max? What if one's bankroll is only $200, what would the proper bet be? Do we use Kelly to calculate this bet size as a function of our bankroll, especially because we only have, on average, 2 opportunities to bet big per shoe? As such, variance is very large, correct?

4) So, for 2 specific 'target' rounds per shoe, if she thinks she can predict with, on average, 38% certainty when at least 1 of her 2 dealt cards will be an Ace, is her OVERALL (for ALL rounds; i.e., playing every hand - no Wonging) player advantage something like 2 rounds of betting big at about a 19.38% player advantage per 6-deck shoe offset by about 33 rounds of flat betting at about a -.55% player advantage per 6-deck shoe?

Thank you in advance!

Overkill - nice game when you can find it. WRt to your questions:

1. not quite - 51% reflects the total expected return on the bet not the number of times you win a hand. Remember a minority of hands that start with an ace pair nicely with a ten to return 150% of the initial bet and so forth. So you will win 51 cents for every dollar you wager if the first card is an ace.

2. Close dont forget to subtract for the 62% of hands where the dealer has an approximate edge of 0.5%. Over 100 hands 38% will have a 51% edge and 62 will have a minus 0.5 edge. If you run the numbers the negative hands dont have a huge impact for the sake of discussion have to be considered.

3. Cant answer. Can calculate things like ev and ror for a given size bank but its up to the you as to what is deemed acceptable. if it were me i would table max bet unless - undue heat and want to make sure in advance that you can split and double above table limits. Regardless of what you find acceptable my advice is always leave enough in your pocket to split (maybe even re-split) and/or double. Saw a guy once get A,A with a high bet (he just went all in) and rejected my offer to partner on the hand and then got to observe his joy of hitting a 10,10 to bust his original A,A.

4. Close remember the 19% is spread over the two hands - what not clear is is 38% edge of spiking an ace of each of the two hands or 38% of an ace on one of the two hands? makes a difference in total edge. The rest you have right - highlights why in a play like this a shallow cut shoe game is preferable.


Hope this helps.

Cohiba

[aceside]

This explanation is exceptionally clear. Are you two talking about ace sequencing? How do you spot the situations when aces are likely to come out in the next two hands? This skill is important.

[Cohiba]

This explanation is exceptionally clear. Are you two talking about ace sequencing? How do you spot the situations when aces are likely to come out in the next two hands? This skill is important.

I wasnt talking about anything other than trying to respond to Kill's comments/questions. My only question is where is the play? and what is the technique? and can i join the party?

This is on of the games where shallow pen rules!

Cohiba

[aceside]

I wasnt talking about anything other than trying to respond to Kill's comments/questions. My only question is where is the play? and what is the technique? and can i join the party?

This is on of the games where shallow pen rules!

Cohiba

Thank you for helping. I guess this ace prediction applies only to hand shuffled shoes. If anybody can predict aces, he must be very skilled.

[aceside]

Overkill - nice game when you can find it. WRt to your questions:

1. not quite - 51% reflects the total expected return on the bet not the number of times you win a hand. Remember a minority of hands that start with an ace pair nicely with a ten to return 150% of the initial bet and so forth. So you will win 51 cents for every dollar you wager if the first card is an ace.

Cohiba

I have thought about this again and become more confused with the huge advantage of ace prediction. The expected return of the player’s blackjack should be exactly 150%-4.7%*4.7%=149.7%, the expected return for a hand containing an ace is 51% (which includes blackjacks), but the expected return for a hand containing a ten is 60%. If we exclude the benefit of catching blackjacks, there is no benefit of ace prediction. Is this correct? We can just do ten prediction because a hand containing a ten gives a higher expected return. Is this logic correct?

[DSchles]

I have thought about this again and become more confused with the huge advantage of ace prediction. The expected return of the player’s blackjack should be exactly 150%-4.7%*4.7%=149.7%, the expected return for a hand containing an ace is 51% (which includes blackjacks), but the expected return for a hand containing a ten is 60%. If we exclude the benefit of catching blackjacks, there is no benefit of ace prediction. Is this correct? We can just do ten prediction because a hand containing a ten gives a higher expected return. Is this logic correct?

Virtually everything you've written above is wrong. I wish you'd leave the math to the mathematicians. Hopefully, no one takes the numbers you provide seriously. Your methodology for calculating the return for a natural is doubly flawed. First, you don't subtract, you multiply the edge (150%) by the probability that the hand is untied. Second, you don't multiply the 4.7% twice; you already have the natural, so you multiply the 150% by (approximately) 100% - 4.7% = 95.3% just once. The return is about 143%.

Whatever gave you the ridiculous notion that a hand containing a ten has a 60% edge? It's about 13%-14%, depending on rules!

Don

[aceside]

I was about to run my program for the numbers but my computer got stuck this morning. Thank you for clarifying.

[aceside]

Virtually everything you've written above is wrong. I wish you'd leave the math to the mathematicians. Hopefully, no one takes the numbers you provide seriously. Your methodology for calculating the return for a natural is doubly flawed. First, you don't subtract, you multiply the edge (150%) by the probability that the hand is untied. Second, you don't multiply the 4.7% twice; you already have the natural, so you multiply the 150% by (approximately) 100% - 4.7% = 95.3% just once. The return is about 143%.

Whatever gave you the ridiculous notion that a hand containing a ten has a 60% edge? It's about 13%-14%, depending on rules!

Don

Let me dig this a little more. The expected return for a player's natural is 150%*(1-4.7%)=143%, and therefore the expected return for a player hand containing an ace (from these natural blackjacks only) is 143%*(4/13)=44%. Therefore, of the 51% expected return of the player's ace-containing hand, 44% is from catching a natural. This just means ace prediction is really all about catching blackjacks, because a ten-containing hand can have an expected return of 14%, larger than 51%-44%=7% . I hope this is correct.

[DSchles]

Let me dig this a little more. The expected return for a player's ace is 150%*(1-4.7%)=143%,

No! The expected return for a BLACKJACK, before the dealer checks his hand for one, is 143%. It isn't the return for a player ace, which we've already established is 51%; it's the return for a player natural. It's not 150%, because if the dealer also has a natural, the hand pushes. Clear?

therefore the expected return for this blackjack is 143%*(4/13)=44%.

Backwards again. You're trying to determine how much of the 51% edge from the player ace in the hand is due to receiving a subsequent blackjack. And, yes, the answer is most of it!

Don

[DSchles]

Remember one other thing. When calculating Kelly bet for the hand with an ace, the variance is considerably larger than for a normal hand, so the bet size is smaller than you might expect. Still, it is so large that, for practical purposes, you just bet as much of your bankroll as you feel comfortable doing.

Don

[aceside]

Remember one other thing. When calculating Kelly bet for the hand with an ace, the variance is considerably larger than for a normal hand, so the bet size is smaller than you might expect. Still, it is so large that, for practical purposes, you just bet as much of your bankroll as you feel comfortable doing.

Don

The variance of getting a blackjack must be very small but that of getting a non-blackjack 21 should be larger, so overall, I guess the variance is not that very large.

[Cohiba]

The variance of getting a blackjack must be very small but that of getting a non-blackjack 21 should be larger, so overall, I guess the variance is not that very large.

The size of the bets on the hands where you are trolling for an ace will dwarf the size of your waiting bets and that drives variance as your results will be highly dependent on the few hands/shoe that matter to your overall results. Remember variance is a product of your original bet size.

Cohiba