blackorange
blackorange
Joined: Sep 17, 2010
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September 20th, 2010 at 3:02:33 AM permalink
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
Ayecarumba
Ayecarumba
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September 21st, 2010 at 11:09:57 AM permalink
You actually have 14 values. The mean is the sum divided by the count, so you would divide by 14.

If the 1 and 11 are the same card, you will have to assign one value or the other, you cannot use both. Sum the 13 values and divide by 13 to get the mean.

If you would like to represent the fact that an ace can be a 1 or 11 in the mean of 13 cards, assign it a value of 6.
Simplicity is the ultimate sophistication - Leonardo da Vinci
goatcabin
goatcabin
Joined: Feb 13, 2010
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September 21st, 2010 at 12:54:23 PM permalink
Quote: Ayecarumba

You actually have 14 values. The mean is the sum divided by the count, so you would divide by 14.

If the 1 and 11 are the same card, you will have to assign one value or the other, you cannot use both. Sum the 13 values and divide by 13 to get the mean.

If you would like to represent the fact that an ace can be a 1 or 11 in the mean of 13 cards, assign it a value of 6.



If ace high, then 95 / 13 = 7.31
else 85 / 13 = 6.54
endif

Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
blackorange
blackorange
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September 23rd, 2010 at 1:21:54 AM permalink
Thanks to both of you.

So if the ace is 6, and we have 2 3 4 5 6 7 8 9 10 10 10 10 =

91/13=

7.

I am just trying to estimate wether a card drawn will be over or under a certain value.

And a 10 is rated at 4/13.

Just trying to figure where a card may end up from say, a 5.

My thinking is that one can calculate what are the odds of throwing vs the odds of standing.

Eg if the dealer has 9/13 of winning, and we have say 5, well it would be worth drawing a card.
vice versa it would be pointless to try to improve odds.

I want to be able to calculate this with each hand.

And it doesnt matter what each other player has.

I figure being first player to receive card has best/closest to natural deck of cards.

Any takers on this train of throught?

I also believe more decks the merrier, and if they are continuous, then more consistent play. ie no need for card counting or memorising basic strategy. If one can calculate the odds, its obvious wether one should take advantage by doubling or splitting..
weaselman
weaselman
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September 23rd, 2010 at 4:45:52 AM permalink
I don't get what's your beef with the strategy cards. They give you exactly the results you are asking about.
Do you just think that you will be able to do the math better, then those countless pros who computed and re-computed the basic strategy times and again? Or is the thinking more in the direction that they are out to get you, and intentionally put errors into the cards to make you lose?
"When two people always agree one of them is unnecessary"
blackorange
blackorange
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September 28th, 2010 at 11:49:38 AM permalink
Yes, I have a suspicion about the basic strategy cards and the whole deal there. I appreciate there are 'best odds', but I have a different angle in mind, and I also believe that there has been so much manipulation by the casinos world wide regarding data about games, that I take it with a grain of salt. I believe that there is a best chance strategy, but I don't take it from that point. You would agree surely that if there was a simpler way, that they wouldn't want us to know, and given their power and might, they would do anything to manipulate. Wouldn't you pay a few bright professors to come up with something so very complicated that we would all bow down in respect, if you were a casino?

Sure, there are millions of independent minds working on the subject too, but I find it 'odd' that the history is a little shrouded. I have read that a french gambler approached two mathematicians about it for the answer regarding wether it would work. There is no other info or proof of what they actually calculated.


And regarding the endless pros and there computations, if we played with a shoe of 10,000 desks, and used a continous shufffler, we could pretty much rule out card counting, and the effect that removing a card from the deck might do.

ie 1 card out of the deck from a million still means that ace is 1/13 cards. Yes, it's less for a 6 or 8 deck, but you get my drift. Isn't all the math based around that concept? I believe that one really only needs to deal with whole numbers. At least thats what I'm trying to prove or disprove to myself. Right or wrong.


Regarding the above answer from a previous poster:

If ace is regarded as a 6 in terms of means averages, would 21 be considered standard average result for 3 cards thrown?

ie. 2 3 4 5 6 7 8 9 10 10 10 10 + 6= 91/13=7.
Kelmo
Kelmo
Joined: Aug 15, 2010
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October 6th, 2010 at 4:21:54 PM permalink
IMO

It depends on what hands you are considering. If it is only the first two cards in a hand, then the Ace is always counted as 11, so the value is 7.3077

If all possible blackjack hands are considered (multi-card hands), I would think that you would still divide the total by 13, but the value of the Ace would be an intermediate value between 1 and 11. That value would depend on the average value of the ace determined by the frequencies that aces are counted as 1 or 11. Interesting question, because they might not be 50/50. In other words, the ace could be counted as 11 way more often than 1.

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