ssho88
ssho88
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April 28th, 2014 at 9:06:34 PM permalink
The edge provided by an :-

1) In Blackajck, ACE(with a random card) : +50%, T/K/Q/J(with a random card) : +14%
2) In 3CP, ACE(with two random cards) : +64%, King(with two random cards) : +23%
3) In Caribbean Stud Poker, ACE(with four random cards) : ????%, King(with four random cards) : ????%


Anyone can help me ?
odiousgambit
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April 29th, 2014 at 3:20:43 AM permalink
You know what it is worth except in Carribean stud poker?
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
ssho88
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April 29th, 2014 at 3:37:03 AM permalink
You are asking a question instead of answering my question.

Do you think that it's funny ?
odiousgambit
odiousgambit
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April 29th, 2014 at 4:00:00 AM permalink
Quote: ssho88

You are asking a question instead of answering my question.

Do you think that it's funny ?



I think your response is now *quite* hilarious. That was an honest question.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
ssho88
ssho88
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April 29th, 2014 at 4:04:06 AM permalink
As in my first post, the ACE worth +50% and 64% for BLACKJACK and 3CP respectively.

How much an ACE worth in Caribbean Stud Poker ?
chaunceyb3
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April 29th, 2014 at 5:24:13 AM permalink
Quote: ssho88

As in my first post, the ACE worth +50% and 64% for BLACKJACK and 3CP respectively.

How much an ACE worth in Caribbean Stud Poker ?



I mean, to be exact, you would have to comb through all the combinations of Player winning between 1 and 201 units, pushes, and Player losing -1 or -3 units.

I'll take an educated guess and say it's around +20 to +30%.

Assuming you play all hands of one pair or better, and fold all Ace-King highs and worse, you would be folding about 50% of the time (compared to the basic strategy 48% fold percentage), but all your two pair and one pair hands get disproportionally stronger (about 40% of one pair hands are now Aces), and much more likely to be 5 or 3 unit winners.

So where in the typical game, the Player will lose 3 units about 13.7% of the time, and win 3 units about 11.7% of the time, if we are weighing one pair hands heavily toward Aces, this number may be closer to 8% (lose 3 units) and 17% (win 3 units).
ssho88
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April 29th, 2014 at 5:42:29 AM permalink
I am agree with you that player must comb through all the combinations in order to get the exact results.

I guess you should have the program to comb through all the combinations since you can give all the above numbers.

Do you have such program ?
teliot
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April 29th, 2014 at 12:02:10 PM permalink
Quote: ssho88

I am agree with you that player must comb through all the combinations in order to get the exact results.

I guess you should have the program to comb through all the combinations since you can give all the above numbers.

Do you have such program ?

I am running a simulation right now for you. I am not sure I fully understand the question. So, I am assuming that the player has top-card knowledge and can see if his *first card* is an Ace. You want that edge. Also, what is the edge if his *first card* is a King? This is not the same as knowing that there is at least one Ace(King) in the hand.
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teliot
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April 29th, 2014 at 12:15:51 PM permalink
Quote: ssho88

The edge provided by an :-

3) In Caribbean Stud Poker, ACE(with four random cards) : ????%, King(with four random cards) : ????%

The following results are based on simulating 100M hands of CS with the indicated first card.

* Given that the player's *first* card is an Ace, the player has about a 12.49% edge over the house.

* Given that the player's *first* card is a King, the player has about an 8.77% edge over the house.
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13s
13s
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May 1st, 2014 at 9:33:32 AM permalink
Maybe the simplest answer to this question is "one or eleven"?!
teliot
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May 6th, 2014 at 9:27:52 PM permalink
Results from a complete cycle:

Edge with A???? is 12.727992%
Edge with K???? is 9.032054%

These results assume perfect strategy. My previous post based on a simulation of 100M hands used Mike's simplified CS strategy. Each of the results above required a cycle consisting of 1,916,656,780,500 individual hands. Each took about 5 hours of computer time.
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tringlomane
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May 6th, 2014 at 10:20:47 PM permalink
A bit higher than I guessed. Thanks teliot. I hope the OP eventually thanks you for this though... :-\
teliot
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May 7th, 2014 at 11:16:06 AM permalink
Quote: tringlomane

A bit higher than I guessed. Thanks teliot. I hope the OP eventually thanks you for this though... :-\

I heard from the OP indirectly -- all I can say is that he should not have posted his question if he didn't want a public answer. Once I read the question, it didn't take long to figure out that the guy is a strong advantage player who had a specific play in mind.
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