Ace Prediction Math

[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

[aceside]

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

Don

I have thought about this again. The player’s 25.5% advantage is based on the assumption that the dealer’s probability of getting an ace is 1/13, but now the situation is the dealer’s probability of getting an ace is 50%. What is the expected return for this particular situation? This has been bothering me for two days. Thank you in advance.

[DSchles]

I have thought about this again. The player’s 25.5% advantage is based on the assumption that the dealer’s probability of getting an ace is 1/13, but now the situation is the dealer’s probability of getting an ace is 50%. What is the expected return for this particular situation? This has been bothering me for two days. Thank you in advance.

That's a completely different question. Your original version was that you could predict, with 50% accuracy, that YOU would receive the ace. You didn't say that if you didn't get it, the DEALER would! But that math is simple, no? 50% of the time, you have a 51% edge and 50% of the time, the dealer has a 37% edge. So, your overall edge is simply .50*.51 + .50*(-.37) = +.07 = 7%.

Don

[aceside]

That's a completely different question. Your original version was that you could predict, with 50% accuracy, that YOU would receive the ace. You didn't say that if you didn't get it, the DEALER would! But that math is simple, no? 50% of the time, you have a 51% edge and 50% of the time, the dealer has a 37% edge. So, your overall edge is simply .50*.51 + .50*(-.37) = +.07 = 7%.

Don

Very nice of you. Now I can take a rest. Thank you again.