Ace side w/HiLo?

[Blitzkrieg]

Interesting BK. 3 of your 4 examples represented 30% blackjacks which is in line with the full single deck composition. Your composition in example 2 to seems a little light even at 40%.

30% is what I also noticed for 3 out of the 4 examples for short-term results. What do you mean that the composition in example 2 seems to be light at 40%?

[Tarzan]

You're flawed right out of the starting gate, beginning with "I used a single deck of cards", Blitzkrieg.
A) To have those deck compositions with 52 cards remaining to be played you have to start with more than one deck. The 2 examples 8-0-4-4r @1 and 0-10-2-4r @1 indicate these deck compositions with one deck remaining to be played.

B) If you are stating that based on observations of 40 hands (even if the parameters of the experiment were accurate) or any similar short term observation of empirical data is enough to claim the standard formula for determining EV for the insurance bet is flawed then I suggest you rewrite Theory of Blackjack and prove out why Peter Griffin was wrong. It would probably get about the same notoriety as the works of Ion Saliu though.

[Three]

I did an experiment at the house today where I used a single deck of cards with the 2 examples and the discard information you provided above. I took the recommended cards out of the deck and put them in a fake discard tray. Since there were 4 aces remaining in the deck I kept the dealer up card as the constant controlled variable, an ACE.

In the first experiment I simulated and dealt out 2 player hands with the dealer hand showing the Ace (the constant). Only 10 total hands per example, No washing of the deck was done, each separate deal consisted of about 7 riffle shuffles with several cuts, a few more riffles and then a final cut on the deck to get randomization of the deck, player cards and the dealer hole card were plugged back into the deck at random spots after checking for the dealer Blackjack before the next deal.

What in the world are you trying to do here? Insurance needs no experiment to prove when it is advantageous to take it. All that matters is the number of face cards to every other card or as Moses likes to look at the percentage of aces to other cards. If there are at most twice as many other cards as T's then insurance is not a losing proposition in the long run. Counting any other card with the faces will give an inaccurate assessment of the ratio. The index used in such a case will assume a "normal nmber of aces"s seen. Normal can only be exact on every 13th card dealt. When you have many decks unseen the affect of this is minimal but with less than a deck left it is not. The non 10 value cards can be anything even jokers.

Out of 40 total hands even though it is short-term the method of:
One deck remains, exactly 52 cards. Two more {T} are in the discard than {2-5}, ten more {6-9} are in the discard than {2-5}, four (A) in the remainder of the deck.
Hi-Lo player / Elitist Asshole / House Edge on insurance bet
TC0, no insurance / IC+6, take insurance / -3.84%

52 cards left: 4 aces
48 non aces left
T + {2-5} + {6-9} = 48
T = 48 - {2-5 left} - {6-9 left}
{2-5 left} = T + 2
{6-9 left} = {2-5 left} -10 = T + 2 - 10 = T - 8
T = 48 - (T + 2) - (T - 8)= 42- 2*T
3T = 42
T = 14
non T = 48 - 14 = 34 half that is 17
Last I checked 17 was more than 14 so you do not insure.
14/52 = 26.923% the dealer will have BJ (33% is the break even point).
52/14 =3.714 (needs to be 3 or less to insure
10 rounds/(52/14) = 2.692 (the number of BJ expected in each of your 10 round tests if the deck is set up right)

It doesn't matter how many cards are on the table. If they are seen they are considered "in the discards" if not they are considered yet to be dealt.

At 2 player hands produced 1 more BJ. The results I got for heads up play against a dealer Ace for both sets of rules was even, even though 40 hands may not be enough hands to show a definitive result.

Assuming you set up your deck with the proper number of cards for each card group which it sounds like you did not, this is just random variance from this average and has noting to do with the number of players if all cards seen are considered in the discards and all cards unseen are considered undealt. 4 aces undealt (means 4 aces besides the dealer's upcard). T =14, {2-5} 12, {7-9} 22, total of 48 non-aces.

Finally, you aren't going to get statistically significant results wit such a infinitesimally small number of trials. We all know that so what is the point. Either use a simulator to get close to exact results hat are statistically significant or do the math as I did to get exact results.

PS: Sorry for initial errors that have now been fixed.(I think I got them all)

[Blitzkrieg]

You're flawed right out of the starting gate, beginning with "I used a single deck of cards", Blitzkrieg.
A) To have those deck compositions with 52 cards remaining to be played you have to start with more than one deck. The 2 examples 8-0-4-4r @1 and 0-10-2-4r @1 indicate these deck compositions with one deck remaining to be played.

Ion is long gone although people keep bringing up remnants of his name for some reason, he is not used to taking heat unlike myself. I guess I'm slightly wrong because he did spend some time in an asylum under Nicolae Ceausescu. I think he was a programmer anyways, I probably should have asked him to create a certain program/game for me before he checked out of ZenGrifters site for good. It's the least he could have done to help me out along my way. Oh well. By the way when you get your book completed give me a heads up so I can view your work. I'm more of a short story writer myself.

So your saying that I cannot use a complete single deck of cards, okay. Would I have to use a complete single deck plus 12 cards? Enlighten me, how would I have to set the deck up to duplicate what you are saying with the minimal amount of cards necessary with a single deck left to be played?

[Blitzkrieg]

What in the world are you trying to do here? Insurance needs no experiment to prove when it is advantageous to take it. All that matters is the number of face cards to every other card or as Moses likes to look at the percentage of aces to other cards. If there are at most twice as many other cards as T's then insurance is not a losing proposition in the long run. Counting any other card with the faces will give an inaccurate assessment of the ratio. The index used in such a case will assume a "normal nmber of aces"s seen. Normal can only be exact on every 13th card dealt. When you have many decks unseen the affect of this is minimal but with less than a deck left it is not. The non 10 value cards can be anything even jokers.

52 cards left: 4 aces
48 non aces left
T + {2-5} + {6-9} = 48
T = 48 - {2-5 left} - {6-9 left}
{2-5 left} = T + 2
{6-9 left} = {2-5 left} -10 = T + 2 - 10 = T - 8
T = 48 - (T + 2) - (T - 8)= 42- 2*T
3T = 42
T = 14
non T = 48 - 14 = 34 half that is 17
Last I checked 17 was more than 14 so you do not insure.
14/52 = 26.923% the dealer will have BJ (33% is the break even point).
52/14 =3.714 (needs to be 3 or less to insure
10 rounds/(52/14) = 2.692 (the number of BJ expected in each of your 10 round tests if the deck is set up right)

It doesn't matter how many cards are on the table. If they are seen they are considered "in the discards" if not they are considered yet to be dealt.


Assuming you set up your deck with the proper number of cards for each card group which it sounds like you did not, this is just random variance from this average and has noting to do with the number of players if all cards seen are considered in the discards and all cards unseen are considered undealt. 4 aces undealt (means 4 aces besides the dealer's upcard). T =14, {2-5} 12, {7-9} 22, total of 48 non-aces.

Finally, you aren't going to get statistically significant results wit such a infinitesimally small number of trials. We all know that so what is the point. Either use a simulator to get close to exact results hat are statistically significant or do the math as I did to get exact results.

PS: Sorry for initial errors that have now been fixed.(I think I got them all)

I was trying to set the deck up in accordance with the info Tarzan provided to see what would happen in a short-term experiment. Now that I know I set up the deck composition incorrectly I am back to do it again. I want to see the edge that he is referring to. Do you have an idea on how I would have to set up the deck composition to run another short-term experiment? Would a complete single deck plus 12 additional cards be sufficient to carry out the exercise?

[Three]

I want to see the edge that he is referring to. Do you have an idea on how I would have to set up the deck composition to run another short-term experiment? Would a complete single deck plus 12 additional cards be sufficient to carry out the exercise?

All you need is an ace upcard and 14 T's and 38 other cards that are not T's. T's are the only card that will give the dealer a BJ with an ace up. You could use Uno cards if you wanted for the 38 non T's. Take 2 T's out of the deck, get another deck for the ace upcard and and to replace the 2 T's with non-T's. You can use the jokers if you like so as not to mix the two decks.

[Tarzan]

Blitzkrieg, you are obviously an intelligent, thinking and inquisitive individual. With that being the case, your time and efforts are better spent on more worthwhile projects. You want to argue with the bona fide math of it? I sure don't! Hahaha An examination of the short term data would be relatively meaningless and the understanding of an insurance count and how it works would be more fruitful for you. It's a waste of your time and there are more important things to study but if you wanted to recreate 8-0-4-4r @1 you would take two decks. From one complete deck take out eight {2-5} and four {T}. From the other deck replace these with four {2-5}, four {6-9} and four {T}. Use the (A) from the other deck as the dealer upcard to have four (A) remaining. This configuration has a house advantage on the insurance bet of 7.7%. There are sixteen {T} and thirty-six cards other than {T} with 52 total cards remaining. You don't need to know that exact percentage for each deck composition and only that you need 3 or 4 or more extra {T} per deck in the remainder to take insurance is all. A separate insurance count would give perfect insurance betting correlation compared to using your Hi-Lo TC as the basis of insurance betting which is far from perfect. I intentionally used these two deck compositions on the opposite ends of the fence due to surplus or deficit (7-9) to show the original poster that Hi-Lo even with factoring (A) accordingly is not all that pinpoint accurate without also factoring in (7-9) as far as the insurance bet goes. You can play around with the empirical short term results of it all but you're going to have to go a lot farther than 40 hands to show the difference between 3.85% advantage and 7.7% disadvantage... add some zeros to that 40! In other words, you'll be sitting at the kitchen table a long long time to try and discover known values that have already been long since discovered, a waste of your efforts that could be spent on more relevant information pertaining to insurance betting.

[Blitzkrieg]

Thanks for the response Tarzan and Tthree. Your probably right in that I should spend more time on a more worthwhile project. I understand that 40 hands is not enough to show the difference between 3.85% advantage and 7.7% disadvantage. Yesterday it didn't matter to me since I didn't have anything going on. When I read your original post I wanted to try to replicate what you had posted and decided to run a small experiment at the house so I busted out the cards.