Ace Prediction Math

[aceside]

Why are you insulting a fox?

Did I say something wrong? Freightman sounds not pleased with what I said. Bosox once said I should read more and post less. Maybe he is correct.

[DSchles]

Bosox once said I should read more and post less. Maybe he is correct.

Ya think? And I said God gave you one mouth and two ears so that you could listen twice as much as you talk. Guess you missed that. I also said no one ever learns anything by talking. You learn by listening. Guess you missed that too.

Don

[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.

[Overkill]

Thanks so much to all that replied thus far.

Yes, I will need to read that 42.08% article.

[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

[21forme]

Better to remain silent and be thought a fool than to speak and to remove all doubt.
Abraham Lincoln

[Cohiba]

the ace is a powerful card for the player and the 51% expected return is real. In addition to the 3/2 payoff on 20/21 hands that dont push a dealers 21, you have to determine the results for every possible hand starting with an ace in the players hand against every possible dealers up card. Play each hand out and add the results. Hard to do with a paper and pencil but simple for a computer and a program like CVData. Many studies and results have been published supporting the 51% edge - no longer a topic for debate in polite circles. A ten in hand is worth about 13% as Don noted - again not a topic for discussion. I had a ten exposed by the dealer once accidentally at the end of the round and he placed under the shoe as my first card to be dealt - no brainer table max bet despite the slight negative count.

I think its good to ask questions trying to understand the logic and reasoning behind that math. That said i would suggest that a response from someone like Don S, who can be argued is a recognized and credible source on math behind 21, should be considered accurate and would focus my attention on seeking to understand. Sometimes the follow up responses have a tone of disputing and challenging the answer as opposed to seeking to understand.

With that i will follow 21's advice and be quiet. Bring on the aces!

Cohiba

[aceside]

With that i will follow 21's advice and be quiet. Bring on the aces!

I just double checked these numbers. Actually , the ace card is a double-edged sword, because it gives an expected return of 51% for the player but also 37% for the dealer. So, I am confident that we can bet table max when the player's ace is accidentally exposed. However, how much should we bet when the player is only 50% certain that he will get an ace? This comes back to the original question on this topic.

[DSchles]

I just double checked these numbers. Actually , the ace card is a double-edged sword, because it gives an expected return of 51% for the player but also 37% for the dealer. So, I am confident that we can bet table max when the player's ace is accidentally exposed. However, how much should we bet when the player is only 50% certain that he will get an ace? This comes back to the original question on this topic.

You figure that having "only" a ~25% edge is cause for being more conservative with your bet??

Don

[Cohiba]

Loosely translated Don is saying "max bet, max bet!'

WRt to your question about sending the ace to the dealer; not getting into specifics but by all means take proper precautions to avoid this calamity. Practice safe sequencing.

Cohiba