Floating Advantage and Wonging

[Grisly Dreams]

I was just taking issue with the "it IS a single deck" part. If it's a 52 card pack that has the same count as a deck, the advantage will not always be identical (which one would assume from the "IS") to that in a true single deck. What if all the aces are gone? Then the blackjack advantage to the player is shot, no matter how rich the "deck" may be in tens.

 

[k_c]

certainly no expert, just i wonder for one thing if maybe not a lot of other things. how is it one if they have an absence of knowing anything other than the number of cards and and the count and that there is a cut card present, again, how is it that one would assume those additional 9 cards are equally distributed across whats left?:whip:
i mean is there some physical reason or some mathematical reason that makes a mixture of things in stuff such as cards with numbers and pictures printed on them to be mixed in some particular way?
in other words if stuff is mixed in a random way or even a quasi random way then are we supposed to assume equal distribution of those things that were mixed?

Let's try to start with the most simple case we can.

Let's say we're using Hi-Lo and know 1 card remains:
If running count = 0, the card is a 7, 8, or 9. If we have no other information then the probability card is 7 = probability card is 8 = probability card is 9 = 1/3 = .33333....

If running count = -1, the card is a 2, 3, 4, 5, or 6. If we have no other information then the probability card is 2 = probability card is 3 = probability card is 4 = probability card is 5 = probability card is 6 = 1/5 = .2.

If running count = +1, the card is a 10 or ace. If we have no other information then the probability card is ace = .2. Probability card is 10 = 4*.2 = .8.

Now let's say 2 cards remain and we have no info other than the count. There are more possibilities:
If running count = -2, 2 low cards remain
If running count = -1, 1 low card and 1 neutral card remain
If running count = 0, either 2 neutral cards or 1 low card & 1 high card remain
If running count = +1, 1 high card and 1 neutral card remain
If running count = +2, 2 high cards remain

Notice that for a running count of 0 there are 2 possible subsets of cards. But are each of the subsets just as likely to occur?
No! For the time being, let's assume we started with an infinite number of decks to make the calculations less messy.
Probability of 2 neutral cards = 3/13 * 3/13 = 9/169
Probability of 1 high & 1 low card = 5/13 * 5/13 * 2 = 50/169.
For a finite number of decks the numbers will be different, but not significantly so that the infinite shoe numbers won't serve as a reasonable illustration.

So the chance that a 2 card Hi-Lo count of 0 consists of 1 high and 1 low card are more than 5 times as likely as 2 neutral cards. Therefore the probability of drawing a neutral card from a 2 card running Hi-Lo count of 0 is significantly less than 1/13, the probability of drawing a 2,3,4,5,6,ace are more than 1/13, and the probability of drawing a 10 is more than 4/13.

As more and more cards are added, more and more possible subsets exist relative to a given count. But at some point this turns around and as more cards are added we finally get to a full shoe, which is the only possibility given no cards have beeen removed. For a full shoe the probability of drawing a non-ten = 1/13 and the probability of drawing a ten = 4/13.

Curiously enough it turns out that for any possible running count and exactly a half shoe remaining the probability of drawing a neutral card is exactly equal to 1/13. The probability of drawing a neutral card from a full shoe is also equal to exactly 1/13. If we have no other info besides running count and cards remaining those are the only 2 cases where probability of drawing a neutral card is exactly 1/13.

I don't know if this answers your question. You can assume an even distribution to some extent but in cases where there is more than one possibility each possibility must be weighted as well to get to the final probability of drawing a given rank based solely upon a given count and cards remaining.

 

[sagefr0g]

Let's try to start with the most simple case we can.

Let's say we're using Hi-Lo and know 1 card remains:
If running count = 0, the card is a 7, 8, or 9. If we have no other information then the probability card is 7 = probability card is 8 = probability card is 9 = 1/3 = .33333....

If running count = -1, the card is a 2, 3, 4, 5, or 6. If we have no other information then the probability card is 2 = probability card is 3 = probability card is 4 = probability card is 5 = probability card is 6 = 1/5 = .2.

If running count = +1, the card is a 10 or ace. If we have no other information then the probability card is ace = .2. Probability card is 10 = 4*.2 = .8.

Now let's say 2 cards remain and we have no info other than the count. There are more possibilities:
If running count = -2, 2 low cards remain
If running count = -1, 1 low card and 1 neutral card remain
If running count = 0, either 2 neutral cards or 1 low card & 1 high card remain
If running count = +1, 1 high card and 1 neutral card remain
If running count = +2, 2 high cards remain

Notice that for a running count of 0 there are 2 possible subsets of cards. But are each of the subsets just as likely to occur?
No! For the time being, let's assume we started with an infinite number of decks to make the calculations less messy.
Probability of 2 neutral cards = 3/13 * 3/13 = 9/169
Probability of 1 high & 1 low card = 5/13 * 5/13 * 2 = 50/169.
For a finite number of decks the numbers will be different, but not significantly so that the infinite shoe numbers won't serve as a reasonable illustration.

So the chance that a 2 card Hi-Lo count of 0 consists of 1 high and 1 low card are more than 5 times as likely as 2 neutral cards. Therefore the probability of drawing a neutral card from a 2 card running Hi-Lo count of 0 is significantly less than 1/13, the probability of drawing a 2,3,4,5,6,ace are more than 1/13, and the probability of drawing a 10 is more than 4/13.

As more and more cards are added, more and more possible subsets exist relative to a given count. But at some point this turns around and as more cards are added we finally get to a full shoe, which is the only possibility given no cards have beeen removed. For a full shoe the probability of drawing a non-ten = 1/13 and the probability of drawing a ten = 4/13.

Curiously enough it turns out that for any possible running count and exactly a half shoe remaining the probability of drawing a neutral card is exactly equal to 1/13. The probability of drawing a neutral card from a full shoe is also equal to exactly 1/13. If we have no other info besides running count and cards remaining those are the only 2 cases where probability of drawing a neutral card is exactly 1/13.

I don't know if this answers your question. You can assume an even distribution to some extent but in cases where there is more than one possibility each possibility must be weighted as well to get to the final probability of drawing a given rank based solely upon a given count and cards remaining.

thanks k_c, what your saying there i think gives me some insight into the issue. i gotta go over what you wrote some more, though as i'm slow witted, lol.

like ok, before i asked you to consider this issue, well i tried to think about it some and i was going to post the following, but i got stuck and confused, lol
the stuff quoted is what i was thinking about:

still this is confusing to me.

ok, regarding those nine additional high cards say, i would guess that what is being said is that over the long run, that the probability that those cards will be evenly distributed over the cards left to be dealt is higher than say for those cards to be asymmetrically distributed in all sorts of ways given those cards distributed in the same number of cards left to be dealt?

so ok if you only had nine cards left to be dealt then definitely the high cards are going to be evenly distributed, lol.

so ok if you had 18 cards left to be dealt and nine of those cards were high cards and nine of the cards were low cards, is it expected that the distribution will be either
lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo ,hi
or
hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi,lo,hi ,lo ?
in other words those two distributions have a higher probability of presenting over other possible distributions?

and then if you had nine more high cards to be mixed in with those eighteen cards would the distribution of highest probability then be:
lo,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi ,hi,lo,hi,hi,lo,hi,hi,lo,hi,hi
or
hi,lo,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi,hi,lo ,hi,hi,lo,hi,hi,lo,hi,hi,lo,hi
as opposed to other possible distributions of those extra nine high cards?
such as say:
lo,lo,hi,hi,lo,hi,lo,hi,hi,hi,lo,hi,lo,hi,hi,lo,hi ,lo,hi,lo,hi,hi,hi,hi,hi,hi,hi or what ever other combination's one could come up with.

i guess what i'm asking is, is it mathematically valid that certain of the possible combination's of distributions would be able to happen more often than others?

so this is where i was wondering if that permutation, or maybe combination maths stuff would have the answer?

 

[k_c]

thanks k_c, what your saying there i think gives me some insight into the issue. i gotta go over what you wrote some more, though as i'm slow witted, lol.

like ok, before i asked you to consider this issue, well i tried to think about it some and i was going to post the following, but i got stuck and confused, lol
the stuff quoted is what i was thinking about:


so this is where i was wondering if that permutation, or maybe combination maths stuff would have the answer?

What if you had 18 cards with nothing printed on them? If truly randomly ordered then any of the cards could appear in any position. Why would it be any different if High or Low was printed on them?

 

[sagefr0g]

What if you had 18 cards with nothing printed on them? If truly randomly ordered then any of the cards could appear in any position. Why would it be any different if High or Low was printed on them?

pretty much that was essentially my outlook of this issue.
then i started wondering about the permutation and or combination stuff, not really understanding that stuff i was wondering if that stuff had any bearing on the issue or if that sort of stuff could shed light on the issue.:whip:

i think my gut instinct on the issue is that since you don't know the order of the cards it would be best to assume they could be ordered in any manner of ways possible instead of one set of order being predominate.