1 2 3 4 5 6 7 8 9 10 10 10 10 11 =
Do we divide by 13 cards = 7.38
or 14 values = 6.85.
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...
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.
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%
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)..
10% 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.
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.
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.
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.