September 17th, 2010 at 3:25:25 AM
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What is the mean average value of 13 cards?

1 2 3 4 5 6 7 8 9 10 10 10 10 11 =

96

--

Do we divide by 13 cards = 7.38

or 14 values = 6.85.

?

thanks

blackorange

1 2 3 4 5 6 7 8 9 10 10 10 10 11 =

96

--

Do we divide by 13 cards = 7.38

or 14 values = 6.85.

?

thanks

blackorange

September 17th, 2010 at 4:06:34 AM
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You cannot answer that question unless the value of an ace is defined. I am also not sure how this information will help you in any way. For example, if you are considering hitting a 16, the mean is not relevant. What is relevant is that you will get 17 1/13 of the time, 18 1/13 of the time.... bust 8/13 of the time. But if you forced me to answer your question, you could say the mean has a 'range' of 85/13 to 95/13, depending on the frequency that an ace is 1 versus 11.

September 18th, 2010 at 6:19:40 AM
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I would only consider hitting on a 16, relevant to what the dealer has got. If Im gonna lose 8/13 by throwing, so be it, if the dealer has an 8/13 of winning. If the dealer was looking to lose 9/13, I wouldnt need to throw as that would only be worsening my own odds.

And so to answer: The value of the ace is defined isnt it, by definition it has a designation, albeit, two designations. If it can be defined two ways, does it count as an extra unit, this 14 values? But, no, since there are only 13 cards, its therefore necessarily got to be divided by 13. So do we consider the average between the two? 90/13, which is 6.92.?

I am trying to figure the average of what a 50/50 average may be, so one could estimate wether theres a 50/50 chance of being under or over a certain figure. wether it be 6 or 7 or 8, or a fraction of that.

Then I can estimate where the dealer may lie 50% of the time, or better, and knowing the option of 4/13 for a 10, one can try to ascertain wether the dealer will land in the stand zone.

My train of thinking is that there are only 13 cards, no matter how many decks. Even if there an infinity of decks, as you know only 13 cards.

I say a guy play once, and he won like crazy.

He only ever bet a 10th of his purse, thus eliminatig wild betting, and being able to track his winnings, and he always bet the same way, no matter how high the stakes. He also didnt seem perturbed when he lost, because he knew when to double, and his 21's ended up putting him way in front. It got me thinking a lot about his strategy. Also he didnt look at anyone else cards, because in my estimation, with 306 cards in the shoe, and only 10 removed, its hardly worth the headache of calculating the .002 change in odds. I'm sure he wasn't card counting, and it would be quite useless as a mechanical shuffler was in play after every hand.

So I'd rather stick with whole numbers.

And 13 being odd, it seems that one can eitehr be on the high side or low, ie 7/13 comp. to 6/13.

I figure if the dealerr only has 5 places to land that he can stand on, surely there must be potentially 8/13 games to be won. ?

The other thing to consider is that there are a number of extra games that can be won and lost at the ends of the bell curve. ONe may forget that on 12, theres 4/13 games to throw away, and many players would unnecessarily lose these thinkingthey are 'low', with out evaluating the dealers potential. Similarly, to be on 18, one would think that to throw would be insanity, but, if the dealer had an 11/13 of wining, well, it would be a loss of a game not to throw, as there are 3/13 games to be picked up. And so it goes against common sense, but these extra games make up the winnings edge, couples with doubling and blackjack payout.

Hope you got this far. I havent seen anything written anywhere re this stuff, every talks about card counting...

And so to answer: The value of the ace is defined isnt it, by definition it has a designation, albeit, two designations. If it can be defined two ways, does it count as an extra unit, this 14 values? But, no, since there are only 13 cards, its therefore necessarily got to be divided by 13. So do we consider the average between the two? 90/13, which is 6.92.?

I am trying to figure the average of what a 50/50 average may be, so one could estimate wether theres a 50/50 chance of being under or over a certain figure. wether it be 6 or 7 or 8, or a fraction of that.

Then I can estimate where the dealer may lie 50% of the time, or better, and knowing the option of 4/13 for a 10, one can try to ascertain wether the dealer will land in the stand zone.

My train of thinking is that there are only 13 cards, no matter how many decks. Even if there an infinity of decks, as you know only 13 cards.

I say a guy play once, and he won like crazy.

He only ever bet a 10th of his purse, thus eliminatig wild betting, and being able to track his winnings, and he always bet the same way, no matter how high the stakes. He also didnt seem perturbed when he lost, because he knew when to double, and his 21's ended up putting him way in front. It got me thinking a lot about his strategy. Also he didnt look at anyone else cards, because in my estimation, with 306 cards in the shoe, and only 10 removed, its hardly worth the headache of calculating the .002 change in odds. I'm sure he wasn't card counting, and it would be quite useless as a mechanical shuffler was in play after every hand.

So I'd rather stick with whole numbers.

And 13 being odd, it seems that one can eitehr be on the high side or low, ie 7/13 comp. to 6/13.

I figure if the dealerr only has 5 places to land that he can stand on, surely there must be potentially 8/13 games to be won. ?

The other thing to consider is that there are a number of extra games that can be won and lost at the ends of the bell curve. ONe may forget that on 12, theres 4/13 games to throw away, and many players would unnecessarily lose these thinkingthey are 'low', with out evaluating the dealers potential. Similarly, to be on 18, one would think that to throw would be insanity, but, if the dealer had an 11/13 of wining, well, it would be a loss of a game not to throw, as there are 3/13 games to be picked up. And so it goes against common sense, but these extra games make up the winnings edge, couples with doubling and blackjack payout.

Hope you got this far. I havent seen anything written anywhere re this stuff, every talks about card counting...

September 18th, 2010 at 7:55:56 AM
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Wow! This was intense ...

Just look here: http://wizardofodds.com/blackjack/strategy/calculator.html

Or buy a strategy card in a casino gift store, it will just tell you what to do.

I'll just comment on one thing - betting 1/10th of "your purse" *is* pretty wild.

Just look here: http://wizardofodds.com/blackjack/strategy/calculator.html

Or buy a strategy card in a casino gift store, it will just tell you what to do.

I'll just comment on one thing - betting 1/10th of "your purse" *is* pretty wild.

"When two people always agree one of them is unnecessary"

September 18th, 2010 at 8:50:11 AM
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I figured that while you could lose 10 games in a row, as your purse goes down, so does your bet, so in some ways its still running at a constant. Its too easy to put it all upon a hand and lose it all. No one knows of they are gonna win or lose the next hand, so why go 'hunching'? riduculous. So eliminate that by betting a constant value. 10% is not actaully so wild. i found 20% sometimes is safer, but any less and it takes too long to make progress...

please forgive me for the poor typing in the previous post.

Thanks for the tip on betting cards, but I dont believe they are 100%

blackorange

please forgive me for the poor typing in the previous post.

Thanks for the tip on betting cards, but I dont believe they are 100%

blackorange

September 18th, 2010 at 9:08:33 AM
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Does the calculator shows house margin of winning per 100 games, or coin margin, based on a constant bet? if it only showed win margin, does it necessarily follow that the house will also make more money, even there is potential for splits and bj payouts?

I wasn't really trying to calculate the house margin, as opposed to calculating the potential for the dealer to land safe, or be forced to throw.

How, under wizards simple strategy, does 'cost for incorrect play at 0.53%'? Cost for really incorrect play can be a hell of a lot more costly if one seriously deviates from his suggested strategy.

If he reveals the process by which he formulated the table, there may be a way of memorising the process, so one can calculate thus with each hand, thus not 'playing incorrectly one hand per 12 hours of play' (seriously)..

I wasn't really trying to calculate the house margin, as opposed to calculating the potential for the dealer to land safe, or be forced to throw.

How, under wizards simple strategy, does 'cost for incorrect play at 0.53%'? Cost for really incorrect play can be a hell of a lot more costly if one seriously deviates from his suggested strategy.

If he reveals the process by which he formulated the table, there may be a way of memorising the process, so one can calculate thus with each hand, thus not 'playing incorrectly one hand per 12 hours of play' (seriously)..

September 18th, 2010 at 10:09:31 AM
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Quote:blackorange10% is not actaully so wild. i found 20% sometimes is safer, but any less and it takes too long to make progress...

It has been shown that overbetting your bankroll like this will cause you to lose all (read: most) of your money pretty quickly, even if you have an advantage! The Kelly bet criterion usually involves you making a bet around 1-2% of your bankroll, with an advantage. You should make bare minimum bets without an advantage, or enough that you have fun.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

September 18th, 2010 at 10:52:05 AM
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Quote:blackorangeI would only consider hitting on a 16, relevant to what the dealer has got. If Im gonna lose 8/13 by throwing, so be it, if the dealer has an 8/13 of winning. If the dealer was looking to lose 9/13, I wouldnt need to throw as that would only be worsening my own odds.

You're chasing your own tail. For one thing, the dealer will win more than 3/4 of the time if you stand on a stiff vs. a 7 thru Ace. That alone would suggest why you hit that 16, but you don't WIN 5/13 of the time; very often you make a 17 or 18 and lose anyway.

What hitting a hard 16 vs. a 7 or higher actually does is improve your chances of winning from lousy to slightly less lousy. That's all. Hard 16 is a bad hand.

The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw

September 20th, 2010 at 2:38:34 AM
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Sorry, what I meant earlier was not 20%, but 1/20th. If the overall purse goes down, so too does the wager. Why would one want to vary the gamble? Seeing that odds for any particluar round of play do not change in the overall scheme of things, irrespective of how much money is at stake.

And yes, 5/13 is not a guaranteed win, thats for sure, but it's also a certainty that we will defintely bust on 8/13 hands.

If the dealer has 10, and we've got 16, well the dealer has got an 8/13 of winning straight up, and if we throw we've got an 8/13 of losing straight up.

So do we throw the game by handing it over to the dealer, or try to inprove our own odds?

I would say it's safer to not lose more than we already will in that scenario, and let the dealer get the 5/13 cards that will force him to throw, and then throw the game.

Seeing that the dealer has a 5/13 of getting 10 or 11, we would really likely need to get a 4 or 5 to still be in the game, therefore we are reducing our chances to around 2/13.

So standing, and letting the dealer get 5/13 of bad cards, then does give a better than 50/50 of busting out for us.

that to me is better probability.

Any takers?

And yes, 5/13 is not a guaranteed win, thats for sure, but it's also a certainty that we will defintely bust on 8/13 hands.

If the dealer has 10, and we've got 16, well the dealer has got an 8/13 of winning straight up, and if we throw we've got an 8/13 of losing straight up.

So do we throw the game by handing it over to the dealer, or try to inprove our own odds?

I would say it's safer to not lose more than we already will in that scenario, and let the dealer get the 5/13 cards that will force him to throw, and then throw the game.

Seeing that the dealer has a 5/13 of getting 10 or 11, we would really likely need to get a 4 or 5 to still be in the game, therefore we are reducing our chances to around 2/13.

So standing, and letting the dealer get 5/13 of bad cards, then does give a better than 50/50 of busting out for us.

that to me is better probability.

Any takers?

September 20th, 2010 at 4:52:21 AM
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After reading your subsequent posts I now think I understand what you are asking. You are asking what is the best way to play to maximize your chance of winning. There is a concept, called basic strategy, that answers your question. There are slight variations depending on the rules of the table. For example you will play slightly differently if the dealer hits a soft 17 versus standing on a soft 17. The companion web site, wizardofodds.com, has a section on blackjack, and there are many small laminated cards you can buy with basic strategy. If you are even more interested and have the time and inclination, you can learn to count cards and thus increase your bets when there is an advantage for you, or decrease when the opposite is true. As far as your math goes, it is too simplistic. You will have to trust me (and those that have developed it) that the full math analysis of all possible combinations is what has lead to what is now called 'basic strategy'. Good luck.